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\int v \, du$: **First Integration by Parts:** - Let $u = \theta^2 \implies du = 2\theta \, d\theta$ - Let $dv = \cos 4\theta \, d\theta \implies v = \frac{1}{4}\sin 4\theta$ $$\int \theta^2 \cos 4\theta \, d\theta = \frac{1}{4}\theta^2 \sin 4\theta - \int \frac{1}{2}\theta \sin 4\theta \, d\theta$$ **Second Integration by Parts** (on the remaining integral $\int \theta \sin 4\theta \, d\theta$): - Let $u = \theta \implies du = d\theta$ - Let $dv = \sin 4\theta \, d\theta \implies v = -\frac{1}{4}\cos 4\theta$ $$\int \theta \sin 4\theta \, d\theta = -\frac{1}{4}\theta \cos 4\theta - \int \left(-\frac{1}{4}\cos 4\theta\right) d\theta = -\frac{1}{4}\theta \cos 4\theta + \frac{1}{16}\sin 4\theta$$ Substituting this back into our expression for $I$: $$I = \frac{1}{4}\theta^2 \sin 4\theta - \frac{1}{2}\left(-\frac{1}{4}\theta \cos 4\theta + \frac{1}{16}\sin 4\theta\right) = \frac{1}{4}\theta^2 \sin 4\theta + \frac{1}{8}\theta \cos 4\theta - \frac{1}{32}\sin 4\theta$$94:T456,Next, we find the second integral, $I = \int \theta^2 \cos 4\theta \, d\theta$, using integration by parts $\int u \, dv = uv - \int v \, du$: **First Integration by Parts:** - Let $u = \theta^2 \implies du = 2\theta \, d\theta$ - Let $dv = \cos 4\theta \, d\theta \implies v = \frac{1}{4}\sin 4\theta$ $$\int \theta^2 \cos 4\theta \, d\theta = \frac{1}{4}\theta^2 \sin 4\theta - \int \frac{1}{2}\theta \sin 4\theta \, d\theta$$ **Second Integration by Parts** (on the remaining integral $\int \theta \sin 4\theta \, d\theta$): - Let $u = \theta \implies du = d\theta$ - Let $dv = \sin 4\theta \, d\theta \implies v = -\frac{1}{4}\cos 4\theta$ $$\int \theta \sin 4\theta \, d\theta = -\frac{1}{4}\theta \cos 4\theta - \int \left(-\frac{1}{4}\cos 4\theta\right) d\theta = -\frac{1}{4}\theta \cos 4\theta + \frac{1}{16}\sin 4\theta$$ Substituting this back into our expression for $I$: $$I = \frac{1}{4}\theta^2 \sin 4\theta - \frac{1}{2}\left(-\frac{1}{4}\theta \cos 4\theta + \frac{1}{16}\sin 4\theta\right) = \frac{1}{4}\theta^2 \sin 4\theta + \frac{1}{8}\theta \cos 4\theta - \frac{1}{32}\sin 4\theta$$72:["$","$L8a",null,{"bundle":{"topicIds":["bda7b9da-346d-4ee2-a401-85bfd067aa82","6aec5e2a-0583-4e0c-a6ec-55316eea8b36","82a5e1ed-0a88-43e3-9bf6-e242a8d8b317","053492e1-c402-4158-af07-bb52b29684d7","97cb7c2e-f32f-45ff-ada1-651a91e4cf22","07681e27-ffd1-457b-97e1-6d9f7376045a","38009166-cd9e-451b-9fe8-9d5abe66fa70","d6a2b67f-1818-43ab-94fc-f534e818a847"],"paperIds":[],"stageIds":[],"stageTopicIds":[],"stageAssessmentIds":[],"topicNoteIds":[]},"children":[["$","$L8b",null,{"feature":"question-bank"}],"$L8c",["$","$L8e",null,{"subjectCode":"a_level_maths__ccea","subjectName":"Maths","topicSlug":"unit-a2-1-pure-mathematics","topicTitle":"3 Unit A2 1: Pure Mathematics (A-level only)","sectionTitle":null,"activity":"worksheet","resourceTitle":"3 Unit A2 1: Pure Mathematics (A-level only) Questions","resourceHref":"/a-level/maths/ccea/unit-a2-1-pure-mathematics/questions/q1","saveTrail":false}],["$","$L90",null,{"href":"/a-level/maths/ccea/unit-a2-1-pure-mathematics/questions/q1","label":"3 Unit A2 1: Pure Mathematics (A-level only)","subjectCode":"a_level_maths__ccea","surface":"questions"}],["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"LearningResource\",\"name\":\"Question 715 – 3 Unit A2 1: Pure Mathematics (A-level only) - Questions\",\"description\":\"Practise 3 Unit A2 1: Pure Mathematics (A-level only) A Level Maths exam-style questions with worked solutions on MathsGenie.\",\"learningResourceType\":\"Practice questions\",\"educationalLevel\":\"A Level\",\"about\":{\"@type\":\"Thing\",\"name\":\"3 Unit A2 1: Pure Mathematics (A-level only)\"},\"url\":\"https://mathsgenie.co.uk/a-level/maths/ccea/unit-a2-1-pure-mathematics/questions/q715\",\"inLanguage\":\"en-GB\",\"provider\":{\"@id\":\"https://mathsgenie.co.uk/#organization\"}}"}}],["$","div",null,{"className":"min-h-0 w-full flex-1 overflow-hidden bg-background","children":["$","$L92",null,{"backHref":"/a-level/maths/ccea/questions","basePath":"/a-level/maths/ccea/unit-a2-1-pure-mathematics/questions","formulaUrl":"https://ccea.org.uk/downloads/docs/Support/General/2019/Mathematical%20Formulae%20and%20Tables.pdf","initialIndex":714,"initialQuestionWindow":[{"id":"1ff67f62-316d-4587-9231-6eadf902fae5","revisionId":"3a532ebc-fee6-4f9d-a90d-12f3603a433e","questionNumber":null,"questionType":"LA","specification":"The path of a high-precision laser cutter across a titanium plate is defined by the parametric equations\n\\[x = t^2 + 8t, \\quad y = \\frac{12}{t(8-t)}, \\quad a \\le t \\le b\\]\nwhere \$t\$ is the time in seconds, and \$a\$ and \$b\$ are constants. The cutter's path intersects the reference line \$y = 1\$ at two points representing the start and end of a specific cooling phase.","difficultyBand":"hard","tier":null,"totalMarks":13,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"7032d491-a09a-485f-96ee-f8d9b625baf1","label":"a","content":"Find the value of \$a\$ and the value of \$b\$, where \$b > a\$.","metadata":null,"markscheme":"- Sets $y = 1$ to find intersections: $\\frac{12}{8t - t^2} = 1 \\Rightarrow t^2 - 8t + 12 = 0$ M1\n- Solves quadratic $(t-2)(t-6) = 0$ to find $a = 2$ and $b = 6$ A1","marks":2,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"algebraic","label":"Algebraic Approach","steps":[{"type":"text","order":0,"title":"Set up the equation","content":"To find where the path of the laser cutter intersects the reference line, we set the parametric equation for $y$ equal to $1$:\n\n$$\\frac{12}{t(8-t)} = 1$$\n\nThis will allow us to find the boundary times $a$ and $b$ for the cooling phase.","highlights":[{"text":"intersects the reference line y = 1","location":"specification"}]},{"type":"text","order":1,"title":"Rearrange to a quadratic","content":"First, multiply both sides of the equation by the denominator $t(8-t)$:\n\n$$12 = t(8-t)$$\n\nNow, expand the right-hand side:\n\n$$12 = 8t - t^2$$\n\nRearrange all terms to the left-hand side to form a standard quadratic equation of the form $At^2 + Bt + C = 0$:\n\n$$t^2 - 8t + 12 = 0$$\n\nSetting $y = 1$ and obtaining this quadratic equation earns the first mark M1."},{"type":"text","order":2,"title":"Solve by factoring","content":"To solve $t^2 - 8t + 12 = 0$, we look for two numbers that multiply to $12$ and add to $-8$. These numbers are $-2$ and $-6$.\n\nFactoring the quadratic equation gives:\n\n$$(t - 2)(t - 6) = 0$$\n\nSetting each factor to zero, we get the two solutions:\n\n$$t = 2 \\quad \\text{or} \\quad t = 6$$"},{"type":"text","order":3,"title":"Identify the parameters","content":"The question defines the domain as $a \\le t \\le b$ with $b > a$.\n\nMatching our solutions to these constants:\n- The smaller value is $a = 2$\n- The larger value is $b = 6$\n\nFinding these two correct values earns the final accuracy mark A1.","highlights":[{"text":"b > a","location":"specification"}]}]},{"id":"graphical","label":"Using Graphing Calculator","steps":[{"type":"text","order":0,"title":"Identify the intersection condition","content":"We are given that the laser cutter's path intersects the reference line $y = 1$. To find the times at which this occurs, we set the parametric equation for $y$ equal to $1$:\n\n$$\\frac{12}{t(8-t)} = 1$$","highlights":[{"text":"intersects the reference line y = 1","location":"specification"}]},{"type":"text","order":1,"title":"Graph and find intersections","content":"To solve this using a graphing utility, we graph the coordinate equation for $y$ (substituting $x$ for the parameter $t$) and the horizontal reference line $y = 1$:\n\n$$y_1 = \\frac{12}{x(8-x)}$$\n$$y_2 = 1$$\n\nBy finding the intersection points of these two curves, we can directly read the values of $t$ from the $x$-coordinates on the screen.","calculator":[{"gdc":{"expression":"y = 12 / (x * (8 - x)), y = 1","instructions":"1. Go to the Graph menu.\n2. Enter the first function as f1(x) = 12 / (x * (8 - x)) and the second function as f2(x) = 1.\n3. Choose a suitable viewing window (e.g., 0 to 10 for x).\n4. Press G-Solv (or Analyze Graph) and select Intersection.\n5. The calculator will display the intersection coordinates: (2, 1) and (6, 1)."},"ti84":{"expression":"Y1 = 12 / (X * (8 - X))\nY2 = 1","instructions":"1. Press [Y=].\n2. Enter Y1 = 12 / (X * (8 - X)) and Y2 = 1.\n3. Press [WINDOW] and set Xmin = 0, Xmax = 10, Ymin = -2, Ymax = 5.\n4. Press [GRAPH].\n5. Press [2nd] [CALC] and select 5: intersect.\n6. Move the cursor near the first intersection point and press [ENTER] three times to find the first root: X = 2.\n7. Repeat the steps, moving the cursor near the second intersection point to find the second root: X = 6."},"desmos":{"expression":"y = 12 / (x * (8 - x))\ny = 1","instructions":"1. Enter the equation y = 12 / (x * (8 - x)) in the first input box.\n2. Enter the equation y = 1 in the second input box.\n3. Click on the gray intersection points on the graph to display their coordinates.\n4. The intersection points are located at (2, 1) and (6, 1)."},"description":"Graph the functions and find their intersection points."}]},{"type":"text","order":2,"title":"Assign values to parameters","content":"The graphing utility displays two intersection points representing the start and end of the cooling phase at times:\n\n$$t = 2 \\quad \\text{and} \\quad t = 6$$\n\nSince the phase occurs on the interval $a \\le t \\le b$ where $b > a$, we assign:\n\n$$a = 2$$\n$$b = 6$$","highlights":[{"text":"b > a","location":"specification"}]}]}],"generatedAt":"2026-06-09T03:43:33.258Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"37d1d972-4862-4a92-b270-68387dab73e6","label":"b","content":"The region \$R\$ is bounded by the path and the line \$y = 1\$. Show that the area of region \$R\$ is given by\n\\[M - k \\int_{a}^{b} \\frac{t+4}{t(8-t)} \\, dt\\]\nwhere \$M\$ and \$k\$ are constants to be found.","metadata":null,"markscheme":"- Finds $\\frac{dx}{dt} = 2t + 8 = 2(t+4)$ B1\n- Area under the line $y = 1$ is $1 \\times (x(6) - x(2)) = (6^2 + 8(6)) - (2^2 + 8(2))$ M1\n- Evaluates $M = 84 - 20 = 64$ A1\n- Sets up area $R = \\int_a^b (1 - y) \\frac{dx}{dt} dt = 64 - \\int_2^6 \\frac{12}{t(8-t)} \\cdot 2(t+4) dt$ and identifies $k = 24$ A1","marks":4,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"geometric_split","label":"Geometric and Parametric Split","steps":[{"type":"text","order":0,"title":"Understand the split approach","content":"To find the area of the region $R$ bounded by the parametric curve and the horizontal line $y = 1$, we can split the area calculation into two components:\n\n1. The **geometric area** of the rectangle under the line $y = 1$ from $x(a)$ to $x(b)$.\n2. The **parametric area** under the curve $y(t)$ from $t = a$ to $t = b$.\n\nSubtracting the area under the curve from the area of the rectangle yields the area of region $R$:\n\n$$\\text{Area}(R) = \\text{Area}_{\\text{rectangle}} - \\text{Area}_{\\text{curve}}$$"},{"type":"text","order":1,"title":"Find the derivative dx/dt","content":"For parametric integration, we need the differential relationship $dx = \\frac{dx}{dt} \\, dt$. Let's differentiate $x = t^2 + 8t$ with respect to $t$:\n\n$$\\frac{dx}{dt} = 2t + 8 = 2(t + 4)$$\n\nThis earns the first mark B1."},{"type":"text","order":2,"title":"Find the bounds a and b","content":"The intersection points of the curve with the reference line $y = 1$ define our limits $a$ and $b$. We solve $y = 1$:\n\n$$1 = \\frac{12}{t(8-t)} \\implies t(8-t) = 12$$\n\n$$8t - t^2 = 12 \\implies t^2 - 8t + 12 = 0$$\n\n$$(t-2)(t-6) = 0$$\n\nThus, the limits of integration are $a = 2$ and $b = 6$."},{"type":"text","order":3,"title":"Calculate the geometric area M","content":"The area under the line $y = 1$ between $t = 2$ and $t = 6$ forms a rectangle of height $1$ and width $x(6) - x(2)$:\n\n$$x(6) = 6^2 + 8(6) = 36 + 48 = 84$$\n$$x(2) = 2^2 + 8(2) = 4 + 16 = 20$$\n\nThus, the geometric area is:\n\n$$M = 1 \\times (x(6) - x(2)) = 84 - 20 = 64$$\n\nThis completes the steps for the M1 and A1 marks."},{"type":"text","order":4,"title":"Set up parametric area subtraction","content":"Now we represent the area under the curve as a parametric integral:\n\n$$\\text{Area}_{\\text{curve}} = \\int_{a}^{b} y \\frac{dx}{dt} \\, dt$$\n\nSubstituting $y = \\frac{12}{t(8-t)}$ and $\\frac{dx}{dt} = 2(t + 4)$:\n\n$$\\text{Area}_{\\text{curve}} = \\int_{a}^{b} \\frac{12}{t(8-t)} \\cdot 2(t + 4) \\, dt = \\int_{a}^{b} \\frac{24(t+4)}{t(8-t)} \\, dt$$\n\nSubtracting this from the geometric area $M = 64$ gives:\n\n$$\\text{Area}(R) = 64 - 24 \\int_{a}^{b} \\frac{t+4}{t(8-t)} \\, dt$$\n\nComparing this to the given form $M - k \\int_{a}^{b} \\frac{t+4}{t(8-t)} \\, dt$, we find the constants:\n\n$$M = 64 \\quad \\text{and} \\quad k = 24$$\n\nThis earns the final mark A1."}]},{"id":"parametric_formulation","label":"Full Parametric Formulation","steps":[{"type":"text","order":0,"title":"Find the limits of integration","content":"To find the limits of integration $a$ and $b$, we need to find the values of $t$ where the curve intersects the line $y = 1$.\n\nSet $y = 1$ using the parametric equation for $y$:\n\n$$\\frac{12}{t(8-t)} = 1$$\n\nMultiply both sides by $t(8-t)$:\n\n$$12 = 8t - t^2 \\implies t^2 - 8t + 12 = 0$$\n\nFactoring this quadratic equation gives:\n\n$$(t-2)(t-6) = 0$$\n\nThus, the limits of integration are $a = 2$ and $b = 6$."},{"type":"text","order":1,"title":"Find the derivative dx/dt","content":"We need the derivative of $x$ with respect to $t$ to set up our parametric integral.\n\nGiven:\n\n$$x = t^2 + 8t$$\n\nDifferentiating with respect to $t$:\n\n$$\\frac{dx}{dt} = 2t + 8 = 2(t+4)$$\n\nThis earns the first mark B1 of the solution."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\text{Area } R = \\int_{a}^{b} \\htmlData{anchor=int-integrand}{\\color{@sky}{(1 - y)}} \\htmlData{anchor=int-diff}{\\color{@amber}{\\frac{dx}{dt}}} \\, dt$$"},{"latex":"$$$\\text{Area } R = \\int_{2}^{6} \\color{@sky}{\\left(1 - \\frac{12}{t(8-t)}\\right)} \\color{@amber}{2(t+4)} \\, dt$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"int-integrand"},{"color":"amber","style":"text-color","anchor":"int-diff"}],"connections":[{"to":"int-diff","from":"int-integrand","color":"sky","label":null,"style":"solid"}]},"order":2,"title":"Set up the parametric integral","description":"We set up the area of the region $R$ bounded by the path and $y=1$ as a single parametric integral:\n\n$$\\text{Area } R = \\int_{a}^{b} (1 - y) \\frac{dx}{dt} \\, dt$$"},{"type":"text","order":3,"title":"Expand and split the integral","content":"Now we expand the integrand to separate the algebraic part from the fraction:\n\n$$\\text{Area } R = \\int_{2}^{6} \\left[ 2(t+4) - \\frac{24(t+4)}{t(8-t)} \\right] dt$$\n\n$$\\text{Area } R = \\int_{2}^{6} 2(t+4) \\, dt - \\int_{2}^{6} \\frac{24(t+4)}{t(8-t)} \\, dt$$\n\nThis represents the formula in the required form, where the first term will yield the constant $M$."},{"type":"text","order":4,"title":"Evaluate the first term","content":"Now we integrate the first term to find the constant $M$ (which represents the area under the line $y = 1$):\n\n$$M = \\int_{2}^{6} 2(t+4) \\, dt$$\n\n$$M = \\left[ t^2 + 8t \\right]_{2}^{6}$$\n\nEvaluate at the upper limit $t = 6$:\n\n$$6^2 + 8(6) = 36 + 48 = 84$$\n\nEvaluate at the lower limit $t = 2$:\n\n$$2^2 + 8(2) = 4 + 16 = 20$$\n\nSubtracting these values gives:\n\n$$M = 84 - 20 = 64$$\n\nThis step earns the method mark M1 and accuracy mark A1 for evaluating $M$."},{"type":"text","order":5,"title":"Formulate final expression","content":"Substituting $M = 64$ back into our split integral expression gives:\n\n$$\\text{Area } R = 64 - \\int_{2}^{6} \\frac{24(t+4)}{t(8-t)} \\, dt$$\n\nWe can factor out the constant $k$ from the numerator of the integrand:\n\n$$\\text{Area } R = 64 - 24 \\int_{2}^{6} \\frac{t+4}{t(8-t)} \\, dt$$\n\nComparing this with the target expression $M - k \\int_{a}^{b} \\frac{t+4}{t(8-t)} \\, dt$, we successfully show the result where:\n\n* $M = 64$\n* $k = 24$\n\nThis completes the final mark A1."}]}],"generatedAt":"2026-06-09T03:45:27.059Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"4278a5e1-5f89-4b0e-851a-c1a234e198d9","label":"ci","content":"Express \$\\frac{t+4}{t(8-t)}\$ in partial fractions.","metadata":null,"markscheme":"- Sets up partial fractions $\\frac{t+4}{t(8-t)} = \\frac{A}{t} + \\frac{B}{8-t}$ and finds $t+4 = A(8-t) + Bt$ M1\n- Substitutes $t=0$ to find $A = \\frac{1}{2}$ A1\n- Substitutes $t=8$ to find $B = \\frac{3}{2}$ A1","marks":3,"order":2,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"substitution","label":"Substitution Method","steps":[{"type":"text","order":0,"title":"Set up partial fractions","content":"Since the denominator has two distinct linear factors, $t$ and $8-t$, we can express the fraction in the form:\n\n$$\\frac{t+4}{t(8-t)} \\equiv \\frac{A}{t} + \\frac{B}{8-t}$$\n\nwhere $A$ and $B$ are constants we need to determine."},{"type":"text","order":1,"title":"Clear the denominators","content":"To clear the denominators, we multiply both sides of the equation by the common denominator $t(8-t)$ to form an identity:\n\n$$t+4 \\equiv A(8-t) + Bt$$"},{"type":"text","order":2,"title":"Substitute t = 0","content":"We can find $A$ by substituting $t = 0$ into the identity, which eliminates the $B$ term:\n\n$$\\begin{aligned} 0 + 4 &= A(8 - 0) + B(0) \\\\ 4 &= 8A \\\\ A &= \\frac{1}{2} \\end{aligned}$$"},{"type":"text","order":3,"title":"Substitute t = 8","content":"We can find $B$ by substituting $t = 8$ into the identity, which eliminates the $A$ term:\n\n$$\\begin{aligned} 8 + 4 &= A(8 - 8) + B(8) \\\\ 12 &= 8B \\\\ B &= \\frac{3}{2} \\end{aligned}$$"},{"type":"text","order":4,"title":"State final expression","content":"Substitute our values $A = \\frac{1}{2}$ and $B = \\frac{3}{2}$ back into the original partial fractions template:\n\n$$\\frac{t+4}{t(8-t)} = \\frac{1}{2t} + \\frac{3}{2(8-t)}$$"}]},{"id":"equating_coefficients","label":"Equating Coefficients","steps":[{"type":"text","order":0,"title":"Set up partial fractions","content":"To express the algebraic fraction $\\frac{t+4}{t(8-t)}$ in partial fractions, we begin by setting up a model with linear denominators. Since the denominator has two distinct linear factors, $t$ and $8-t$, we write:\n\n$$\\frac{t+4}{t(8-t)} \\equiv \\frac{A}{t} + \\frac{B}{8-t}$$\n\nwhere $A$ and $B$ are constants to be found."},{"type":"text","order":1,"title":"Multiply to eliminate denominators","content":"To eliminate the denominators, we multiply both sides of the equation by the common denominator $t(8-t)$:\n\n$$t+4 \\equiv A(8-t) + Bt$$"},{"type":"text","order":2,"title":"Expand and group the terms","content":"Using the **Equating Coefficients** approach, we expand the right-hand side of the identity and group terms with the same powers of $t$:\n\n$$\\begin{aligned} t+4 &\\equiv 8A - At + Bt \\\\ t+4 &\\equiv (-A + B)t + 8A \\end{aligned}$$"},{"type":"text","order":3,"title":"Equate the coefficients","content":"We compare the linear terms (coefficients of $t$) and the constant terms on both sides of the identity:\n\n- **Constant terms:**\n $$8A = 4$$\n\n- **Linear ($t$) terms:**\n $$-A + B = 1$$"},{"type":"text","order":4,"title":"Solve the system of equations","content":"From our comparison, we obtain the following system of linear equations:\n\n1. **For the constant terms:**\n $$8A = 4 \\implies A = \\frac{1}{2}$$\n\n2. **For the linear terms ($t$ terms):**\n $$-A + B = 1$$\n\nSubstituting $A = \\frac{1}{2}$ into the second equation yields:\n\n$$-\\frac{1}{2} + B = 1 \\implies B = \\frac{3}{2}$$"},{"type":"text","order":5,"title":"State the final expression","content":"Now we substitute our values of $A = \\frac{1}{2}$ and $B = \\frac{3}{2}$ back into our initial template:\n\n$$\\frac{t+4}{t(8-t)} = \\frac{1}{2t} + \\frac{3}{2(8-t)}$$\n\nThis is the expression written in partial fractions."}]}],"generatedAt":"2026-06-09T03:46:51.699Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"7e699626-31cb-4693-ae04-14ed9ae2ef49","label":"cii","content":"Hence, use algebraic integration to find the exact area of \$R\$, giving your answer in the form \$P - Q \\ln 3\$, where \$P\$ and \$Q\$ are integers.","metadata":null,"markscheme":"- Integrates: $\\int_2^6 \\left( \\frac{1}{2t} + \\frac{3}{2(8-t)} \\right) dt = [\\frac{1}{2} \\ln|t| - \\frac{3}{2} \\ln|8-t|]_2^6$ M1\n- Substitutes limits: $(\\frac{1}{2} \\ln 6 - \\frac{3}{2} \\ln 2) - (\\frac{1}{2} \\ln 2 - \\frac{3}{2} \\ln 6)$ M1\n- Simplifies using log laws: $2 \\ln 6 - 2 \\ln 2 = 2 \\ln \\left(\\frac{6}{2}\\right) = 2 \\ln 3$ M1\n- Final area $R = 64 - 24(2 \\ln 3) = 64 - 48 \\ln 3$, so $P=64, Q=48$ A1","marks":4,"order":3,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"parametric_integration","label":"Parametric Integration","steps":[{"type":"text","order":0,"title":"Set up the area integral","content":"The area $R$ of the region between the curve and the reference line $y = 1$ is given by:\n\n$$\\text{Area} = \\int_{x_1}^{x_2} (y - 1) \\, dx$$\n\nFrom the intersection points, we have the limits of integration $t = 2$ to $t = 6$.\n\nFirst, we find the differential $dx$ in terms of $dt$ using $x = t^2 + 8t$:\n\n$$\\frac{dx}{dt} = 2t + 8 \\implies dx = (2t + 8) \\, dt = 2(t + 4) \\, dt$$\n\nNow, substitute $y$ and $dx$ into the integral:\n\n$$\\text{Area} = \\int_{2}^{6} \\left( \\frac{12}{t(8-t)} - 1 \\right) 2(t+4) \\, dt$$"},{"type":"text","order":1,"title":"Simplify the integrand","content":"Let's expand and simplify the integrand before integrating. We can split the integral into two parts:\n\n$$\\begin{aligned} \\text{Area} &= \\int_{2}^{6} \\frac{24(t+4)}{t(8-t)} \\, dt - \\int_{2}^{6} 2(t+4) \\, dt \\end{aligned}$$\n\nLet's calculate the second integral, which is straightforward:\n\n$$\\begin{aligned} \\int_{2}^{6} (2t + 8) \\, dt &= \\left[ t^2 + 8t \\right]_{2}^{6} \\\\ &= (6^2 + 8(6)) - (2^2 + 8(2)) \\\\ &= (36 + 48) - (4 + 16) \\\\ &= 84 - 20 = 64 \\end{aligned}$$\n\nSo the area expression becomes:\n\n$$\\text{Area} = \\int_{2}^{6} \\frac{24(t+4)}{t(8-t)} \\, dt - 64$$"},{"type":"text","order":2,"title":"Apply partial fractions","content":"To integrate the fraction, we first express it using partial fractions. Let:\n\n$$\\frac{24(t+4)}{t(8-t)} = \\frac{A}{t} + \\frac{B}{8-t}$$\n\nMultiply both sides by the denominator $t(8-t)$:\n\n$$24(t+4) = A(8-t) + Bt$$\n\nTo find $A$ and $B$, substitute key values of $t$:\n- Let $t = 0$:\n $$24(4) = A(8) \\implies 96 = 8A \\implies A = 12$$\n- Let $t = 8$:\n $$24(12) = B(8) \\implies 288 = 8B \\implies B = 36$$\n\nTherefore, the fraction becomes:\n\n$$\\frac{24(t+4)}{t(8-t)} = \\frac{12}{t} + \\frac{36}{8-t} = 24 \\left( \\frac{1}{2t} + \\frac{3}{2(8-t)} \\right)$$ \n\nThis matches the decomposition in our markscheme."},{"type":"text","order":3,"title":"Integrate the partial fractions","content":"Now we integrate each term with respect to $t$:\n\n$$\\begin{aligned} \\int_{2}^{6} \\left( \\frac{12}{t} + \\frac{36}{8-t} \\right) dt &= \\left[ 12 \\ln|t| - 36 \\ln|8-t| \\right]_2^6 \\\\ &= 24 \\left[ \\frac{1}{2} \\ln|t| - \\frac{3}{2} \\ln|8-t| \\right]_2^6 \\end{aligned}$$\n\nNow substitute the limits $t=6$ and $t=2$ into this antiderivative:\n\n$$\\begin{aligned} \\text{Value} &= \\left( 12\\ln 6 - 36\\ln 2 \\right) - \\left( 12\\ln 2 - 36\\ln 6 \\right) \\\\ &= 48\\ln 6 - 48\\ln 2 \\end{aligned}$$"},{"type":"text","order":4,"title":"Simplify using logarithm laws","content":"Using the laws of logarithms, we can simplify the expression further:\n\n$$\\begin{aligned} 48\\ln 6 - 48\\ln 2 &= 48 \\left( \\ln 6 - \\ln 2 \\right) \\\\ &= 48 \\ln \\left( \\frac{6}{2} \\right) \\\\ &= 48\\ln 3 \\end{aligned}$$"},{"type":"text","order":5,"title":"Formulate final exact area","content":"Recall from Step 1 that the area is given by the curve integral minus the line boundary integral:\n\n$$\\text{Area} = 48\\ln 3 - 64$$\n\nHowever, since the curve $y = \\frac{12}{t(8-t)}$ lies below the line $y=1$ on this interval, the area integral $\\int (y-1) \\, dx$ yields a negative value:\n\n$$48 \\ln 3 - 64 < 0$$\n\nSince area must be positive, we take the absolute value:\n\n$$\\text{Area} = 64 - 48\\ln 3$$\n\nThis is in the requested form $P - Q \\ln 3$ where:\n- $P = 64$\n- $Q = 48$"}]}],"generatedAt":"2026-06-09T03:47:43.548Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"7032d491-a09a-485f-96ee-f8d9b625baf1:algebraic","label":"(a) Algebraic Approach","steps":[{"type":"text","order":0,"title":"Set up the equation","content":"To find where the path of the laser cutter intersects the reference line, we set the parametric equation for $y$ equal to $1$:\n\n$$\\frac{12}{t(8-t)} = 1$$\n\nThis will allow us to find the boundary times $a$ and $b$ for the cooling phase.","highlights":[{"text":"intersects the reference line y = 1","location":"specification"}]},{"type":"text","order":1,"title":"Rearrange to a quadratic","content":"First, multiply both sides of the equation by the denominator $t(8-t)$:\n\n$$12 = t(8-t)$$\n\nNow, expand the right-hand side:\n\n$$12 = 8t - t^2$$\n\nRearrange all terms to the left-hand side to form a standard quadratic equation of the form $At^2 + Bt + C = 0$:\n\n$$t^2 - 8t + 12 = 0$$\n\nSetting $y = 1$ and obtaining this quadratic equation earns the first mark M1."},{"type":"text","order":2,"title":"Solve by factoring","content":"To solve $t^2 - 8t + 12 = 0$, we look for two numbers that multiply to $12$ and add to $-8$. These numbers are $-2$ and $-6$.\n\nFactoring the quadratic equation gives:\n\n$$(t - 2)(t - 6) = 0$$\n\nSetting each factor to zero, we get the two solutions:\n\n$$t = 2 \\quad \\text{or} \\quad t = 6$$"},{"type":"text","order":3,"title":"Identify the parameters","content":"The question defines the domain as $a \\le t \\le b$ with $b > a$.\n\nMatching our solutions to these constants:\n- The smaller value is $a = 2$\n- The larger value is $b = 6$\n\nFinding these two correct values earns the final accuracy mark A1.","highlights":[{"text":"b > a","location":"specification"}]}]},{"id":"7032d491-a09a-485f-96ee-f8d9b625baf1:graphical","label":"(a) Using Graphing Calculator","steps":[{"type":"text","order":0,"title":"Identify the intersection condition","content":"We are given that the laser cutter's path intersects the reference line $y = 1$. To find the times at which this occurs, we set the parametric equation for $y$ equal to $1$:\n\n$$\\frac{12}{t(8-t)} = 1$$","highlights":[{"text":"intersects the reference line y = 1","location":"specification"}]},{"type":"text","order":1,"title":"Graph and find intersections","content":"To solve this using a graphing utility, we graph the coordinate equation for $y$ (substituting $x$ for the parameter $t$) and the horizontal reference line $y = 1$:\n\n$$y_1 = \\frac{12}{x(8-x)}$$\n$$y_2 = 1$$\n\nBy finding the intersection points of these two curves, we can directly read the values of $t$ from the $x$-coordinates on the screen.","calculator":[{"gdc":{"expression":"y = 12 / (x * (8 - x)), y = 1","instructions":"1. Go to the Graph menu.\n2. Enter the first function as f1(x) = 12 / (x * (8 - x)) and the second function as f2(x) = 1.\n3. Choose a suitable viewing window (e.g., 0 to 10 for x).\n4. Press G-Solv (or Analyze Graph) and select Intersection.\n5. The calculator will display the intersection coordinates: (2, 1) and (6, 1)."},"ti84":{"expression":"Y1 = 12 / (X * (8 - X))\nY2 = 1","instructions":"1. Press [Y=].\n2. Enter Y1 = 12 / (X * (8 - X)) and Y2 = 1.\n3. Press [WINDOW] and set Xmin = 0, Xmax = 10, Ymin = -2, Ymax = 5.\n4. Press [GRAPH].\n5. Press [2nd] [CALC] and select 5: intersect.\n6. Move the cursor near the first intersection point and press [ENTER] three times to find the first root: X = 2.\n7. Repeat the steps, moving the cursor near the second intersection point to find the second root: X = 6."},"desmos":{"expression":"y = 12 / (x * (8 - x))\ny = 1","instructions":"1. Enter the equation y = 12 / (x * (8 - x)) in the first input box.\n2. Enter the equation y = 1 in the second input box.\n3. Click on the gray intersection points on the graph to display their coordinates.\n4. The intersection points are located at (2, 1) and (6, 1)."},"description":"Graph the functions and find their intersection points."}]},{"type":"text","order":2,"title":"Assign values to parameters","content":"The graphing utility displays two intersection points representing the start and end of the cooling phase at times:\n\n$$t = 2 \\quad \\text{and} \\quad t = 6$$\n\nSince the phase occurs on the interval $a \\le t \\le b$ where $b > a$, we assign:\n\n$$a = 2$$\n$$b = 6$$","highlights":[{"text":"b > a","location":"specification"}]}]},{"id":"37d1d972-4862-4a92-b270-68387dab73e6:geometric_split","label":"(b) Geometric and Parametric Split","steps":[{"type":"text","order":0,"title":"Understand the split approach","content":"To find the area of the region $R$ bounded by the parametric curve and the horizontal line $y = 1$, we can split the area calculation into two components:\n\n1. The **geometric area** of the rectangle under the line $y = 1$ from $x(a)$ to $x(b)$.\n2. The **parametric area** under the curve $y(t)$ from $t = a$ to $t = b$.\n\nSubtracting the area under the curve from the area of the rectangle yields the area of region $R$:\n\n$$\\text{Area}(R) = \\text{Area}_{\\text{rectangle}} - \\text{Area}_{\\text{curve}}$$"},{"type":"text","order":1,"title":"Find the derivative dx/dt","content":"For parametric integration, we need the differential relationship $dx = \\frac{dx}{dt} \\, dt$. Let's differentiate $x = t^2 + 8t$ with respect to $t$:\n\n$$\\frac{dx}{dt} = 2t + 8 = 2(t + 4)$$\n\nThis earns the first mark B1."},{"type":"text","order":2,"title":"Find the bounds a and b","content":"The intersection points of the curve with the reference line $y = 1$ define our limits $a$ and $b$. We solve $y = 1$:\n\n$$1 = \\frac{12}{t(8-t)} \\implies t(8-t) = 12$$\n\n$$8t - t^2 = 12 \\implies t^2 - 8t + 12 = 0$$\n\n$$(t-2)(t-6) = 0$$\n\nThus, the limits of integration are $a = 2$ and $b = 6$."},{"type":"text","order":3,"title":"Calculate the geometric area M","content":"The area under the line $y = 1$ between $t = 2$ and $t = 6$ forms a rectangle of height $1$ and width $x(6) - x(2)$:\n\n$$x(6) = 6^2 + 8(6) = 36 + 48 = 84$$\n$$x(2) = 2^2 + 8(2) = 4 + 16 = 20$$\n\nThus, the geometric area is:\n\n$$M = 1 \\times (x(6) - x(2)) = 84 - 20 = 64$$\n\nThis completes the steps for the M1 and A1 marks."},{"type":"text","order":4,"title":"Set up parametric area subtraction","content":"Now we represent the area under the curve as a parametric integral:\n\n$$\\text{Area}_{\\text{curve}} = \\int_{a}^{b} y \\frac{dx}{dt} \\, dt$$\n\nSubstituting $y = \\frac{12}{t(8-t)}$ and $\\frac{dx}{dt} = 2(t + 4)$:\n\n$$\\text{Area}_{\\text{curve}} = \\int_{a}^{b} \\frac{12}{t(8-t)} \\cdot 2(t + 4) \\, dt = \\int_{a}^{b} \\frac{24(t+4)}{t(8-t)} \\, dt$$\n\nSubtracting this from the geometric area $M = 64$ gives:\n\n$$\\text{Area}(R) = 64 - 24 \\int_{a}^{b} \\frac{t+4}{t(8-t)} \\, dt$$\n\nComparing this to the given form $M - k \\int_{a}^{b} \\frac{t+4}{t(8-t)} \\, dt$, we find the constants:\n\n$$M = 64 \\quad \\text{and} \\quad k = 24$$\n\nThis earns the final mark A1."}]},{"id":"37d1d972-4862-4a92-b270-68387dab73e6:parametric_formulation","label":"(b) Full Parametric Formulation","steps":[{"type":"text","order":0,"title":"Find the limits of integration","content":"To find the limits of integration $a$ and $b$, we need to find the values of $t$ where the curve intersects the line $y = 1$.\n\nSet $y = 1$ using the parametric equation for $y$:\n\n$$\\frac{12}{t(8-t)} = 1$$\n\nMultiply both sides by $t(8-t)$:\n\n$$12 = 8t - t^2 \\implies t^2 - 8t + 12 = 0$$\n\nFactoring this quadratic equation gives:\n\n$$(t-2)(t-6) = 0$$\n\nThus, the limits of integration are $a = 2$ and $b = 6$."},{"type":"text","order":1,"title":"Find the derivative dx/dt","content":"We need the derivative of $x$ with respect to $t$ to set up our parametric integral.\n\nGiven:\n\n$$x = t^2 + 8t$$\n\nDifferentiating with respect to $t$:\n\n$$\\frac{dx}{dt} = 2t + 8 = 2(t+4)$$\n\nThis earns the first mark B1 of the solution."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\text{Area } R = \\int_{a}^{b} \\htmlData{anchor=int-integrand}{\\color{@sky}{(1 - y)}} \\htmlData{anchor=int-diff}{\\color{@amber}{\\frac{dx}{dt}}} \\, dt$$"},{"latex":"$$$\\text{Area } R = \\int_{2}^{6} \\color{@sky}{\\left(1 - \\frac{12}{t(8-t)}\\right)} \\color{@amber}{2(t+4)} \\, dt$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"int-integrand"},{"color":"amber","style":"text-color","anchor":"int-diff"}],"connections":[{"to":"int-diff","from":"int-integrand","color":"sky","label":null,"style":"solid"}]},"order":2,"title":"Set up the parametric integral","description":"We set up the area of the region $R$ bounded by the path and $y=1$ as a single parametric integral:\n\n$$\\text{Area } R = \\int_{a}^{b} (1 - y) \\frac{dx}{dt} \\, dt$$"},{"type":"text","order":3,"title":"Expand and split the integral","content":"Now we expand the integrand to separate the algebraic part from the fraction:\n\n$$\\text{Area } R = \\int_{2}^{6} \\left[ 2(t+4) - \\frac{24(t+4)}{t(8-t)} \\right] dt$$\n\n$$\\text{Area } R = \\int_{2}^{6} 2(t+4) \\, dt - \\int_{2}^{6} \\frac{24(t+4)}{t(8-t)} \\, dt$$\n\nThis represents the formula in the required form, where the first term will yield the constant $M$."},{"type":"text","order":4,"title":"Evaluate the first term","content":"Now we integrate the first term to find the constant $M$ (which represents the area under the line $y = 1$):\n\n$$M = \\int_{2}^{6} 2(t+4) \\, dt$$\n\n$$M = \\left[ t^2 + 8t \\right]_{2}^{6}$$\n\nEvaluate at the upper limit $t = 6$:\n\n$$6^2 + 8(6) = 36 + 48 = 84$$\n\nEvaluate at the lower limit $t = 2$:\n\n$$2^2 + 8(2) = 4 + 16 = 20$$\n\nSubtracting these values gives:\n\n$$M = 84 - 20 = 64$$\n\nThis step earns the method mark M1 and accuracy mark A1 for evaluating $M$."},{"type":"text","order":5,"title":"Formulate final expression","content":"Substituting $M = 64$ back into our split integral expression gives:\n\n$$\\text{Area } R = 64 - \\int_{2}^{6} \\frac{24(t+4)}{t(8-t)} \\, dt$$\n\nWe can factor out the constant $k$ from the numerator of the integrand:\n\n$$\\text{Area } R = 64 - 24 \\int_{2}^{6} \\frac{t+4}{t(8-t)} \\, dt$$\n\nComparing this with the target expression $M - k \\int_{a}^{b} \\frac{t+4}{t(8-t)} \\, dt$, we successfully show the result where:\n\n* $M = 64$\n* $k = 24$\n\nThis completes the final mark A1."}]},{"id":"4278a5e1-5f89-4b0e-851a-c1a234e198d9:substitution","label":"(ci) Substitution Method","steps":[{"type":"text","order":0,"title":"Set up partial fractions","content":"Since the denominator has two distinct linear factors, $t$ and $8-t$, we can express the fraction in the form:\n\n$$\\frac{t+4}{t(8-t)} \\equiv \\frac{A}{t} + \\frac{B}{8-t}$$\n\nwhere $A$ and $B$ are constants we need to determine."},{"type":"text","order":1,"title":"Clear the denominators","content":"To clear the denominators, we multiply both sides of the equation by the common denominator $t(8-t)$ to form an identity:\n\n$$t+4 \\equiv A(8-t) + Bt$$"},{"type":"text","order":2,"title":"Substitute t = 0","content":"We can find $A$ by substituting $t = 0$ into the identity, which eliminates the $B$ term:\n\n$$\\begin{aligned} 0 + 4 &= A(8 - 0) + B(0) \\\\ 4 &= 8A \\\\ A &= \\frac{1}{2} \\end{aligned}$$"},{"type":"text","order":3,"title":"Substitute t = 8","content":"We can find $B$ by substituting $t = 8$ into the identity, which eliminates the $A$ term:\n\n$$\\begin{aligned} 8 + 4 &= A(8 - 8) + B(8) \\\\ 12 &= 8B \\\\ B &= \\frac{3}{2} \\end{aligned}$$"},{"type":"text","order":4,"title":"State final expression","content":"Substitute our values $A = \\frac{1}{2}$ and $B = \\frac{3}{2}$ back into the original partial fractions template:\n\n$$\\frac{t+4}{t(8-t)} = \\frac{1}{2t} + \\frac{3}{2(8-t)}$$"}]},{"id":"4278a5e1-5f89-4b0e-851a-c1a234e198d9:equating_coefficients","label":"(ci) Equating Coefficients","steps":[{"type":"text","order":0,"title":"Set up partial fractions","content":"To express the algebraic fraction $\\frac{t+4}{t(8-t)}$ in partial fractions, we begin by setting up a model with linear denominators. Since the denominator has two distinct linear factors, $t$ and $8-t$, we write:\n\n$$\\frac{t+4}{t(8-t)} \\equiv \\frac{A}{t} + \\frac{B}{8-t}$$\n\nwhere $A$ and $B$ are constants to be found."},{"type":"text","order":1,"title":"Multiply to eliminate denominators","content":"To eliminate the denominators, we multiply both sides of the equation by the common denominator $t(8-t)$:\n\n$$t+4 \\equiv A(8-t) + Bt$$"},{"type":"text","order":2,"title":"Expand and group the terms","content":"Using the **Equating Coefficients** approach, we expand the right-hand side of the identity and group terms with the same powers of $t$:\n\n$$\\begin{aligned} t+4 &\\equiv 8A - At + Bt \\\\ t+4 &\\equiv (-A + B)t + 8A \\end{aligned}$$"},{"type":"text","order":3,"title":"Equate the coefficients","content":"We compare the linear terms (coefficients of $t$) and the constant terms on both sides of the identity:\n\n- **Constant terms:**\n $$8A = 4$$\n\n- **Linear ($t$) terms:**\n $$-A + B = 1$$"},{"type":"text","order":4,"title":"Solve the system of equations","content":"From our comparison, we obtain the following system of linear equations:\n\n1. **For the constant terms:**\n $$8A = 4 \\implies A = \\frac{1}{2}$$\n\n2. **For the linear terms ($t$ terms):**\n $$-A + B = 1$$\n\nSubstituting $A = \\frac{1}{2}$ into the second equation yields:\n\n$$-\\frac{1}{2} + B = 1 \\implies B = \\frac{3}{2}$$"},{"type":"text","order":5,"title":"State the final expression","content":"Now we substitute our values of $A = \\frac{1}{2}$ and $B = \\frac{3}{2}$ back into our initial template:\n\n$$\\frac{t+4}{t(8-t)} = \\frac{1}{2t} + \\frac{3}{2(8-t)}$$\n\nThis is the expression written in partial fractions."}]},{"id":"7e699626-31cb-4693-ae04-14ed9ae2ef49:parametric_integration","label":"(cii) Parametric Integration","steps":[{"type":"text","order":0,"title":"Set up the area integral","content":"The area $R$ of the region between the curve and the reference line $y = 1$ is given by:\n\n$$\\text{Area} = \\int_{x_1}^{x_2} (y - 1) \\, dx$$\n\nFrom the intersection points, we have the limits of integration $t = 2$ to $t = 6$.\n\nFirst, we find the differential $dx$ in terms of $dt$ using $x = t^2 + 8t$:\n\n$$\\frac{dx}{dt} = 2t + 8 \\implies dx = (2t + 8) \\, dt = 2(t + 4) \\, dt$$\n\nNow, substitute $y$ and $dx$ into the integral:\n\n$$\\text{Area} = \\int_{2}^{6} \\left( \\frac{12}{t(8-t)} - 1 \\right) 2(t+4) \\, dt$$"},{"type":"text","order":1,"title":"Simplify the integrand","content":"Let's expand and simplify the integrand before integrating. We can split the integral into two parts:\n\n$$\\begin{aligned} \\text{Area} &= \\int_{2}^{6} \\frac{24(t+4)}{t(8-t)} \\, dt - \\int_{2}^{6} 2(t+4) \\, dt \\end{aligned}$$\n\nLet's calculate the second integral, which is straightforward:\n\n$$\\begin{aligned} \\int_{2}^{6} (2t + 8) \\, dt &= \\left[ t^2 + 8t \\right]_{2}^{6} \\\\ &= (6^2 + 8(6)) - (2^2 + 8(2)) \\\\ &= (36 + 48) - (4 + 16) \\\\ &= 84 - 20 = 64 \\end{aligned}$$\n\nSo the area expression becomes:\n\n$$\\text{Area} = \\int_{2}^{6} \\frac{24(t+4)}{t(8-t)} \\, dt - 64$$"},{"type":"text","order":2,"title":"Apply partial fractions","content":"To integrate the fraction, we first express it using partial fractions. Let:\n\n$$\\frac{24(t+4)}{t(8-t)} = \\frac{A}{t} + \\frac{B}{8-t}$$\n\nMultiply both sides by the denominator $t(8-t)$:\n\n$$24(t+4) = A(8-t) + Bt$$\n\nTo find $A$ and $B$, substitute key values of $t$:\n- Let $t = 0$:\n $$24(4) = A(8) \\implies 96 = 8A \\implies A = 12$$\n- Let $t = 8$:\n $$24(12) = B(8) \\implies 288 = 8B \\implies B = 36$$\n\nTherefore, the fraction becomes:\n\n$$\\frac{24(t+4)}{t(8-t)} = \\frac{12}{t} + \\frac{36}{8-t} = 24 \\left( \\frac{1}{2t} + \\frac{3}{2(8-t)} \\right)$$ \n\nThis matches the decomposition in our markscheme."},{"type":"text","order":3,"title":"Integrate the partial fractions","content":"Now we integrate each term with respect to $t$:\n\n$$\\begin{aligned} \\int_{2}^{6} \\left( \\frac{12}{t} + \\frac{36}{8-t} \\right) dt &= \\left[ 12 \\ln|t| - 36 \\ln|8-t| \\right]_2^6 \\\\ &= 24 \\left[ \\frac{1}{2} \\ln|t| - \\frac{3}{2} \\ln|8-t| \\right]_2^6 \\end{aligned}$$\n\nNow substitute the limits $t=6$ and $t=2$ into this antiderivative:\n\n$$\\begin{aligned} \\text{Value} &= \\left( 12\\ln 6 - 36\\ln 2 \\right) - \\left( 12\\ln 2 - 36\\ln 6 \\right) \\\\ &= 48\\ln 6 - 48\\ln 2 \\end{aligned}$$"},{"type":"text","order":4,"title":"Simplify using logarithm laws","content":"Using the laws of logarithms, we can simplify the expression further:\n\n$$\\begin{aligned} 48\\ln 6 - 48\\ln 2 &= 48 \\left( \\ln 6 - \\ln 2 \\right) \\\\ &= 48 \\ln \\left( \\frac{6}{2} \\right) \\\\ &= 48\\ln 3 \\end{aligned}$$"},{"type":"text","order":5,"title":"Formulate final exact area","content":"Recall from Step 1 that the area is given by the curve integral minus the line boundary integral:\n\n$$\\text{Area} = 48\\ln 3 - 64$$\n\nHowever, since the curve $y = \\frac{12}{t(8-t)}$ lies below the line $y=1$ on this interval, the area integral $\\int (y-1) \\, dx$ yields a negative value:\n\n$$48 \\ln 3 - 64 < 0$$\n\nSince area must be positive, we take the absolute value:\n\n$$\\text{Area} = 64 - 48\\ln 3$$\n\nThis is in the requested form $P - Q \\ln 3$ where:\n- $P = 64$\n- $Q = 48$"}]}],"generatedAt":"2026-06-09T03:47:55.974Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"202cdea0-5891-48a6-8af1-80314a36eb55","revisionId":"1c6bb2b1-c01d-4526-8228-d1ab1365e1da","questionNumber":null,"questionType":"LA","specification":"","difficultyBand":"hard","tier":null,"totalMarks":9,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"c790676c-c862-4781-b2d7-750bf062d56c","label":"a","content":"Find \$ \\int x^2 \\cos 4x \\, dx \$","metadata":null,"markscheme":"- Attempts integration by parts with $u = x^2$ and $v' = \\cos 4x$ to get $\\frac{1}{4}x^2 \\sin 4x - \\int \\frac{1}{2}x \\sin 4x \\, dx$ M1\n- Attempts integration by parts on $\\int \\frac{1}{2}x \\sin 4x \\, dx$ with $u = \\frac{1}{2}x$ and $v' = \\sin 4x$ M1\n- Obtains $\\int \\frac{1}{2}x \\sin 4x \\, dx = -\\frac{1}{8}x \\cos 4x + \\frac{1}{32} \\sin 4x$ A1\n- Combines results to find $\\int x^2 \\cos 4x \\, dx = \\frac{1}{4}x^2 \\sin 4x + \\frac{1}{8}x \\cos 4x - \\frac{1}{32} \\sin 4x + C$ A1","marks":4,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"standard_ibp","label":"Repeated Integration by Parts","steps":[{"type":"text","order":0,"title":"Identify the method","content":"To integrate the product of two functions, $x^2$ and $\\cos 4x$, we use the method of **Integration by Parts**: $$\\int u \\, dv = uv - \\int v \\, du$$ Because we have an $x^2$ term, we will need to apply this formula twice to completely eliminate the algebraic term."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Let } \\htmlData{anchor=u1}{\\color{@sky}{u = x^2}} \\implies \\htmlData{anchor=du1}{\\color{@sky}{du = 2x \\, dx}}$"},{"latex":"$$\\text{Let } \\htmlData{anchor=dv1}{\\color{@amber}{dv = \\cos 4x \\, dx}} \\implies \\htmlData{anchor=v1}{\\color{@amber}{v = \\frac{1}{4} \\sin 4x}}$"},{"latex":"$$\\int \\htmlData{anchor=u-part}{\\color{@sky}{x^2}} \\htmlData{anchor=dv-part}{\\color{@amber}{\\cos 4x \\, dx}} = \\htmlData{anchor=uv-part}{\\color{@sky}{x^2}\\left(\\color{@amber}{\\frac{1}{4}\\sin 4x}\\right)} - \\int \\htmlData{anchor=vdu-part}{\\left(\\color{@amber}{\\frac{1}{4}\\sin 4x}\\right)\\left(\\color{@sky}{2x \\, dx}\\right)}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u1"},{"color":"amber","style":"text-color","anchor":"dv1"}],"connections":[{"to":"u-part","from":"u1","color":"sky","label":"u","style":"solid"},{"to":"dv-part","from":"dv1","color":"amber","label":"dv","style":"solid"}]},"order":1,"title":"First integration by parts","description":"For our first application, we choose the polynomial part as $u$ so that its derivative is of a lower degree, and the trigonometric part as $dv$."},{"type":"text","order":2,"title":"Simplify the first result","content":"Simplifying our expression, we obtain: $$\\int x^2 \\cos 4x \\, dx = \\frac{1}{4}x^2 \\sin 4x - \\int \\frac{1}{2}x \\sin 4x \\, dx$$ *(This earns the first method mark **M1**)*."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Let } \\htmlData{anchor=u2}{\\color{@purple}{u = \\frac{1}{2}x}} \\implies \\htmlData{anchor=du2}{\\color{@purple}{du = \\frac{1}{2} \\, dx}}$"},{"latex":"$$\\text{Let } \\htmlData{anchor=dv2}{\\color{@teal}{dv = \\sin 4x \\, dx}} \\implies \\htmlData{anchor=v2}{\\color{@teal}{v = -\\frac{1}{4} \\cos 4x}}$"},{"latex":"$$\\int \\htmlData{anchor=u2-part}{\\color{@purple}{\\frac{1}{2}x}} \\htmlData{anchor=dv2-part}{\\color{@teal}{\\sin 4x \\, dx}} = \\htmlData{anchor=uv2-part}{\\color{@purple}{\\frac{1}{2}x}\\left(\\color{@teal}{-\\frac{1}{4}\\cos 4x}\\right)} - \\int \\htmlData{anchor=vdu2-part}{\\left(\\color{@teal}{-\\frac{1}{4}\\cos 4x}\\right)\\left(\\color{@purple}{\\frac{1}{2} \\, dx}\\right)}$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"u2"},{"color":"teal","style":"text-color","anchor":"dv2"}],"connections":[{"to":"u2-part","from":"u2","color":"purple","label":"u","style":"solid"},{"to":"dv2-part","from":"dv2","color":"teal","label":"dv","style":"solid"}]},"order":3,"title":"Second integration by parts","description":"We now need to integrate $\\int \\frac{1}{2}x \\sin 4x \\, dx$. We apply integration by parts once more on this remaining integral."},{"type":"text","order":4,"title":"Evaluate the second integral","content":"Simplify this second integration by parts step: $$\\int \\frac{1}{2}x \\sin 4x \\, dx = -\\frac{1}{8}x \\cos 4x + \\frac{1}{8} \\int \\cos 4x \\, dx$$ Evaluating the final simple integral: $$\\int \\frac{1}{2}x \\sin 4x \\, dx = -\\frac{1}{8}x \\cos 4x + \\frac{1}{32} \\sin 4x$$ *(This earns a method mark **M1** for attempting the second parts step, and an accuracy mark **A1** for obtaining this intermediate result)*."},{"type":"text","order":5,"title":"Combine the results","content":"Now substitute this result back into our original expression, making sure to distribute the negative sign carefully: $$\\int x^2 \\cos 4x \\, dx = \\frac{1}{4}x^2 \\sin 4x - \\left( -\\frac{1}{8}x \\cos 4x + \\frac{1}{32} \\sin 4x \\right) + C$$ $$\\int x^2 \\cos 4x \\, dx = \\frac{1}{4}x^2 \\sin 4x + \\frac{1}{8}x \\cos 4x - \\frac{1}{32} \\sin 4x + C$$ *(This earns the final accuracy mark **A1**)*."}]},{"id":"tabular_method","label":"Tabular Method","steps":[{"type":"text","order":0,"title":"Introduce the Tabular Method","content":"To evaluate the integral $$\\int x^2 \\cos 4x \\, dx$$ we can use the **Tabular Method** (also known as the DI method) for integration by parts. This method is incredibly efficient when we have a polynomial term like $x^2$ that eventually differentiates to zero, paired with a trigonometric term like $\\cos 4x$ that is easy to integrate repeatedly. We set up a table with three columns: 1. **Sign**: Alternating signs starting with $+$ 2. **D (Differentiate)**: The polynomial part $x^2$ 3. **I (Integrate)**: The trigonometric part $\\cos 4x$"},{"type":"text","order":1,"title":"Set up the table","content":"We repeatedly differentiate the term in the **D** column until we reach $0$, and we repeatedly integrate the term in the **I** column the same number of times. Let's perform the operations step-by-step: * **Row 0** (Initial setup): Sign: $+$, D: $x^2$, I: $\\cos 4x$ * **Row 1**: Sign: $-$, D: $\\frac{d}{dx}(x^2) = 2x$, I: $\\int \\cos 4x \\, dx = \\frac{1}{4}\\sin 4x$ * **Row 2**: Sign: $+$, D: $\\frac{d}{dx}(2x) = 2$, I: $\\int \\frac{1}{4}\\sin 4x \\, dx = -\\frac{1}{16}\\cos 4x$ * **Row 3**: Sign: $-$, D: $\\frac{d}{dx}(2) = 0$, I: $\\int -\\frac{1}{16}\\cos 4x \\, dx = -\\frac{1}{64}\\sin 4x$ Let's summarize this in a clean table layout: $$\\begin{array}{c|c|c} \\text{Sign} & \\text{D (Differentiate)} & \\text{I (Integrate)} \\\\ \\hline + & x^2 & \\cos 4x \\\\ - & 2x & \\frac{1}{4}\\sin 4x \\\\ + & 2 & -\\frac{1}{16}\\cos 4x \\\\ - & 0 & -\\frac{1}{64}\\sin 4x \\end{array}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$I = \\htmlData{anchor=term1}{\\color{@sky}{+ (x^2) \\left( \\frac{1}{4}\\sin 4x \\right)}} \\htmlData{anchor=term2}{\\color{@amber}{- (2x) \\left( -\\frac{1}{16}\\cos 4x \\right)}} \\htmlData{anchor=term3}{\\color{@purple}{+ (2) \\left( -\\frac{1}{64}\\sin 4x \\right)}} + C$"},{"latex":"$$I = \\htmlData{anchor=term1-simp}{\\color{@sky}{\\frac{1}{4}x^2 \\sin 4x}} \\htmlData{anchor=term2-simp}{\\color{@amber}{+ \\frac{1}{8}x \\cos 4x}} \\htmlData{anchor=term3-simp}{\\color{@purple}{- \\frac{1}{32}\\sin 4x}} + C$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"term1"},{"color":"amber","style":"text-color","anchor":"term2"},{"color":"purple","style":"text-color","anchor":"term3"}],"connections":[{"to":"term1-simp","from":"term1","color":"sky","label":"simplify first term","style":"solid"},{"to":"term2-simp","from":"term2","color":"amber","label":"simplify second term","style":"solid"},{"to":"term3-simp","from":"term3","color":"purple","label":"simplify third term","style":"solid"}]},"order":2,"title":"Multiply along the diagonals","description":"To construct the solution, we multiply each entry in the **D** column by the entry that is one row down in the **I** column, applying the sign from the **D** row."},{"type":"text","order":3,"title":"Write the final answer","content":"Combining the simplified terms and adding the constant of integration $C$, we get our final result: $$\\int x^2 \\cos 4x \\, dx = \\frac{1}{4}x^2 \\sin 4x + \\frac{1}{8}x \\cos 4x - \\frac{1}{32}\\sin 4x + C$$"}]}],"generatedAt":"2026-06-09T04:01:40.245Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"04d2e3f6-8080-4ee8-9f86-344b2bd0268e","label":"b","content":"A satellite's distance \$ r \$ from a planet is modeled by the differential equation\n\n\\[ \\frac{dr}{d\\theta} = \\frac{(\\theta \\sin 2\\theta)^2}{r^2} \\]\n\nwhere \$ \\theta \$ is the angular displacement in radians.\n\nHence solve this differential equation to find an expression for \$ r^3 \$ in terms of \$ \\theta \$.","metadata":null,"markscheme":"- Separates variables to obtain $\\int r^2 \\, dr = \\int \\theta^2 \\sin^2 2\\theta \\, d\\theta$ B1\n- Uses the identity $\\sin^2 2\\theta = \\frac{1 - \\cos 4\\theta}{2}$ to rewrite the integral as $\\frac{1}{2} \\int (\\theta^2 - \\theta^2 \\cos 4\\theta) \\, d\\theta$ M1\n- Integrates $\\frac{1}{2} \\theta^2$ to get $\\frac{1}{6} \\theta^3$ M1\n- Substitutes the result from part (a) for $\\int \\theta^2 \\cos 4\\theta \\, d\\theta$ and integrates LHS to $\\frac{r^3}{3}$ A1\n- Rearranges to $r^3 = \\frac{1}{2}\\theta^3 - \\frac{3}{8}\\theta^2 \\sin 4\\theta - \\frac{3}{16}\\theta \\cos 4\\theta + \\frac{3}{64} \\sin 4\\theta + c$ A1","marks":5,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"separation_of_variables","label":"Separation of Variables","steps":[{"type":"text","order":0,"title":"Separate the variables","content":"To solve the differential equation \n\n$$\\frac{dr}{d\\theta} = \\frac{(\\theta \\sin 2\\theta)^2}{r^2}$$\n\nusing the separation of variables method, we group all terms involving $r$ on the left side and all terms involving $\\theta$ on the right side:\n\n$$r^2 \\, dr = (\\theta \\sin 2\\theta)^2 \\, d\\theta$$\n\nNow, write both sides with integration signs:\n\n$$\\int r^2 \\, dr = \\int \\theta^2 \\sin^2 2\\theta \\, d\\theta$$"},{"type":"text","order":1,"title":"Apply trigonometric identity","content":"The integral on the right-hand side has a $\\sin^2 2\\theta$ term. We can simplify this using the double-angle trigonometric identity:\n\n$$\\sin^2 A = \\frac{1 - \\cos 2A}{2}$$\n\nSetting $A = 2\\theta$, we get:\n\n$$\\sin^2 2\\theta = \\frac{1 - \\cos 4\\theta}{2}$$\n\nSubstituting this into our integral yields:\n\n$$\\int \\theta^2 \\sin^2 2\\theta \\, d\\theta = \\int \\theta^2 \\left( \\frac{1 - \\cos 4\\theta}{2} \\right) d\\theta = \\frac{1}{2} \\int (\\theta^2 - \\theta^2 \\cos 4\\theta) \\, d\\theta$$"},{"type":"text","order":2,"title":"Integrate the first part","content":"Now we split the integral on the right-hand side into two separate integrals:\n\n$$\\frac{1}{2} \\int \\theta^2 \\, d\\theta - \\frac{1}{2} \\int \\theta^2 \\cos 4\\theta \\, d\\theta$$\n\nIntegrating the first term is straightforward using the power rule:\n\n$$\\frac{1}{2} \\int \\theta^2 \\, d\\theta = \\frac{1}{2} \\left( \\frac{\\theta^3}{3} \\right) = \\frac{1}{6}\\theta^3$$"},{"type":"text","order":3,"title":"Integrate by parts","content":"$93"},{"type":"text","order":4,"title":"Combine and solve","content":"Now we integrate the left-hand side, $\\int r^2 \\, dr = \\frac{r^3}{3}$, and equate it to the integrated right-hand side, including a constant of integration $C_1$:\n\n$$\\frac{r^3}{3} = \\frac{1}{6}\\theta^3 - \\frac{1}{2} \\left( \\frac{1}{4}\\theta^2 \\sin 4\\theta + \\frac{1}{8}\\theta \\cos 4\\theta - \\frac{1}{32}\\sin 4\\theta \\right) + C_1$$\n\n$$\\frac{r^3}{3} = \\frac{1}{6}\\theta^3 - \\frac{1}{8}\\theta^2 \\sin 4\\theta - \\frac{1}{16}\\theta \\cos 4\\theta + \\frac{1}{64}\\sin 4\\theta + C_1$$\n\nMultiply the entire equation by $3$ to find the expression for $r^3$:\n\n$$r^3 = \\frac{1}{2}\\theta^3 - \\frac{3}{8}\\theta^2 \\sin 4\\theta - \\frac{3}{16}\\theta \\cos 4\\theta + \\frac{3}{64}\\sin 4\\theta + C$$"}]}],"generatedAt":"2026-06-09T04:02:24.424Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"c790676c-c862-4781-b2d7-750bf062d56c:standard_ibp","label":"(a) Repeated Integration by Parts","steps":[{"type":"text","order":0,"title":"Identify the method","content":"To integrate the product of two functions, $x^2$ and $\\cos 4x$, we use the method of **Integration by Parts**: $$\\int u \\, dv = uv - \\int v \\, du$$ Because we have an $x^2$ term, we will need to apply this formula twice to completely eliminate the algebraic term."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Let } \\htmlData{anchor=u1}{\\color{@sky}{u = x^2}} \\implies \\htmlData{anchor=du1}{\\color{@sky}{du = 2x \\, dx}}$"},{"latex":"$$\\text{Let } \\htmlData{anchor=dv1}{\\color{@amber}{dv = \\cos 4x \\, dx}} \\implies \\htmlData{anchor=v1}{\\color{@amber}{v = \\frac{1}{4} \\sin 4x}}$"},{"latex":"$$\\int \\htmlData{anchor=u-part}{\\color{@sky}{x^2}} \\htmlData{anchor=dv-part}{\\color{@amber}{\\cos 4x \\, dx}} = \\htmlData{anchor=uv-part}{\\color{@sky}{x^2}\\left(\\color{@amber}{\\frac{1}{4}\\sin 4x}\\right)} - \\int \\htmlData{anchor=vdu-part}{\\left(\\color{@amber}{\\frac{1}{4}\\sin 4x}\\right)\\left(\\color{@sky}{2x \\, dx}\\right)}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u1"},{"color":"amber","style":"text-color","anchor":"dv1"}],"connections":[{"to":"u-part","from":"u1","color":"sky","label":"u","style":"solid"},{"to":"dv-part","from":"dv1","color":"amber","label":"dv","style":"solid"}]},"order":1,"title":"First integration by parts","description":"For our first application, we choose the polynomial part as $u$ so that its derivative is of a lower degree, and the trigonometric part as $dv$."},{"type":"text","order":2,"title":"Simplify the first result","content":"Simplifying our expression, we obtain: $$\\int x^2 \\cos 4x \\, dx = \\frac{1}{4}x^2 \\sin 4x - \\int \\frac{1}{2}x \\sin 4x \\, dx$$ *(This earns the first method mark **M1**)*."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Let } \\htmlData{anchor=u2}{\\color{@purple}{u = \\frac{1}{2}x}} \\implies \\htmlData{anchor=du2}{\\color{@purple}{du = \\frac{1}{2} \\, dx}}$"},{"latex":"$$\\text{Let } \\htmlData{anchor=dv2}{\\color{@teal}{dv = \\sin 4x \\, dx}} \\implies \\htmlData{anchor=v2}{\\color{@teal}{v = -\\frac{1}{4} \\cos 4x}}$"},{"latex":"$$\\int \\htmlData{anchor=u2-part}{\\color{@purple}{\\frac{1}{2}x}} \\htmlData{anchor=dv2-part}{\\color{@teal}{\\sin 4x \\, dx}} = \\htmlData{anchor=uv2-part}{\\color{@purple}{\\frac{1}{2}x}\\left(\\color{@teal}{-\\frac{1}{4}\\cos 4x}\\right)} - \\int \\htmlData{anchor=vdu2-part}{\\left(\\color{@teal}{-\\frac{1}{4}\\cos 4x}\\right)\\left(\\color{@purple}{\\frac{1}{2} \\, dx}\\right)}$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"u2"},{"color":"teal","style":"text-color","anchor":"dv2"}],"connections":[{"to":"u2-part","from":"u2","color":"purple","label":"u","style":"solid"},{"to":"dv2-part","from":"dv2","color":"teal","label":"dv","style":"solid"}]},"order":3,"title":"Second integration by parts","description":"We now need to integrate $\\int \\frac{1}{2}x \\sin 4x \\, dx$. We apply integration by parts once more on this remaining integral."},{"type":"text","order":4,"title":"Evaluate the second integral","content":"Simplify this second integration by parts step: $$\\int \\frac{1}{2}x \\sin 4x \\, dx = -\\frac{1}{8}x \\cos 4x + \\frac{1}{8} \\int \\cos 4x \\, dx$$ Evaluating the final simple integral: $$\\int \\frac{1}{2}x \\sin 4x \\, dx = -\\frac{1}{8}x \\cos 4x + \\frac{1}{32} \\sin 4x$$ *(This earns a method mark **M1** for attempting the second parts step, and an accuracy mark **A1** for obtaining this intermediate result)*."},{"type":"text","order":5,"title":"Combine the results","content":"Now substitute this result back into our original expression, making sure to distribute the negative sign carefully: $$\\int x^2 \\cos 4x \\, dx = \\frac{1}{4}x^2 \\sin 4x - \\left( -\\frac{1}{8}x \\cos 4x + \\frac{1}{32} \\sin 4x \\right) + C$$ $$\\int x^2 \\cos 4x \\, dx = \\frac{1}{4}x^2 \\sin 4x + \\frac{1}{8}x \\cos 4x - \\frac{1}{32} \\sin 4x + C$$ *(This earns the final accuracy mark **A1**)*."}]},{"id":"c790676c-c862-4781-b2d7-750bf062d56c:tabular_method","label":"(a) Tabular Method","steps":[{"type":"text","order":0,"title":"Introduce the Tabular Method","content":"To evaluate the integral $$\\int x^2 \\cos 4x \\, dx$$ we can use the **Tabular Method** (also known as the DI method) for integration by parts. This method is incredibly efficient when we have a polynomial term like $x^2$ that eventually differentiates to zero, paired with a trigonometric term like $\\cos 4x$ that is easy to integrate repeatedly. We set up a table with three columns: 1. **Sign**: Alternating signs starting with $+$ 2. **D (Differentiate)**: The polynomial part $x^2$ 3. **I (Integrate)**: The trigonometric part $\\cos 4x$"},{"type":"text","order":1,"title":"Set up the table","content":"We repeatedly differentiate the term in the **D** column until we reach $0$, and we repeatedly integrate the term in the **I** column the same number of times. Let's perform the operations step-by-step: * **Row 0** (Initial setup): Sign: $+$, D: $x^2$, I: $\\cos 4x$ * **Row 1**: Sign: $-$, D: $\\frac{d}{dx}(x^2) = 2x$, I: $\\int \\cos 4x \\, dx = \\frac{1}{4}\\sin 4x$ * **Row 2**: Sign: $+$, D: $\\frac{d}{dx}(2x) = 2$, I: $\\int \\frac{1}{4}\\sin 4x \\, dx = -\\frac{1}{16}\\cos 4x$ * **Row 3**: Sign: $-$, D: $\\frac{d}{dx}(2) = 0$, I: $\\int -\\frac{1}{16}\\cos 4x \\, dx = -\\frac{1}{64}\\sin 4x$ Let's summarize this in a clean table layout: $$\\begin{array}{c|c|c} \\text{Sign} & \\text{D (Differentiate)} & \\text{I (Integrate)} \\\\ \\hline + & x^2 & \\cos 4x \\\\ - & 2x & \\frac{1}{4}\\sin 4x \\\\ + & 2 & -\\frac{1}{16}\\cos 4x \\\\ - & 0 & -\\frac{1}{64}\\sin 4x \\end{array}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$I = \\htmlData{anchor=term1}{\\color{@sky}{+ (x^2) \\left( \\frac{1}{4}\\sin 4x \\right)}} \\htmlData{anchor=term2}{\\color{@amber}{- (2x) \\left( -\\frac{1}{16}\\cos 4x \\right)}} \\htmlData{anchor=term3}{\\color{@purple}{+ (2) \\left( -\\frac{1}{64}\\sin 4x \\right)}} + C$"},{"latex":"$$I = \\htmlData{anchor=term1-simp}{\\color{@sky}{\\frac{1}{4}x^2 \\sin 4x}} \\htmlData{anchor=term2-simp}{\\color{@amber}{+ \\frac{1}{8}x \\cos 4x}} \\htmlData{anchor=term3-simp}{\\color{@purple}{- \\frac{1}{32}\\sin 4x}} + C$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"term1"},{"color":"amber","style":"text-color","anchor":"term2"},{"color":"purple","style":"text-color","anchor":"term3"}],"connections":[{"to":"term1-simp","from":"term1","color":"sky","label":"simplify first term","style":"solid"},{"to":"term2-simp","from":"term2","color":"amber","label":"simplify second term","style":"solid"},{"to":"term3-simp","from":"term3","color":"purple","label":"simplify third term","style":"solid"}]},"order":2,"title":"Multiply along the diagonals","description":"To construct the solution, we multiply each entry in the **D** column by the entry that is one row down in the **I** column, applying the sign from the **D** row."},{"type":"text","order":3,"title":"Write the final answer","content":"Combining the simplified terms and adding the constant of integration $C$, we get our final result: $$\\int x^2 \\cos 4x \\, dx = \\frac{1}{4}x^2 \\sin 4x + \\frac{1}{8}x \\cos 4x - \\frac{1}{32}\\sin 4x + C$$"}]},{"id":"04d2e3f6-8080-4ee8-9f86-344b2bd0268e:separation_of_variables","label":"(b) Separation of Variables","steps":[{"type":"text","order":0,"title":"Separate the variables","content":"To solve the differential equation \n\n$$\\frac{dr}{d\\theta} = \\frac{(\\theta \\sin 2\\theta)^2}{r^2}$$\n\nusing the separation of variables method, we group all terms involving $r$ on the left side and all terms involving $\\theta$ on the right side:\n\n$$r^2 \\, dr = (\\theta \\sin 2\\theta)^2 \\, d\\theta$$\n\nNow, write both sides with integration signs:\n\n$$\\int r^2 \\, dr = \\int \\theta^2 \\sin^2 2\\theta \\, d\\theta$$"},{"type":"text","order":1,"title":"Apply trigonometric identity","content":"The integral on the right-hand side has a $\\sin^2 2\\theta$ term. We can simplify this using the double-angle trigonometric identity:\n\n$$\\sin^2 A = \\frac{1 - \\cos 2A}{2}$$\n\nSetting $A = 2\\theta$, we get:\n\n$$\\sin^2 2\\theta = \\frac{1 - \\cos 4\\theta}{2}$$\n\nSubstituting this into our integral yields:\n\n$$\\int \\theta^2 \\sin^2 2\\theta \\, d\\theta = \\int \\theta^2 \\left( \\frac{1 - \\cos 4\\theta}{2} \\right) d\\theta = \\frac{1}{2} \\int (\\theta^2 - \\theta^2 \\cos 4\\theta) \\, d\\theta$$"},{"type":"text","order":2,"title":"Integrate the first part","content":"Now we split the integral on the right-hand side into two separate integrals:\n\n$$\\frac{1}{2} \\int \\theta^2 \\, d\\theta - \\frac{1}{2} \\int \\theta^2 \\cos 4\\theta \\, d\\theta$$\n\nIntegrating the first term is straightforward using the power rule:\n\n$$\\frac{1}{2} \\int \\theta^2 \\, d\\theta = \\frac{1}{2} \\left( \\frac{\\theta^3}{3} \\right) = \\frac{1}{6}\\theta^3$$"},{"type":"text","order":3,"title":"Integrate by parts","content":"$94"},{"type":"text","order":4,"title":"Combine and solve","content":"Now we integrate the left-hand side, $\\int r^2 \\, dr = \\frac{r^3}{3}$, and equate it to the integrated right-hand side, including a constant of integration $C_1$:\n\n$$\\frac{r^3}{3} = \\frac{1}{6}\\theta^3 - \\frac{1}{2} \\left( \\frac{1}{4}\\theta^2 \\sin 4\\theta + \\frac{1}{8}\\theta \\cos 4\\theta - \\frac{1}{32}\\sin 4\\theta \\right) + C_1$$\n\n$$\\frac{r^3}{3} = \\frac{1}{6}\\theta^3 - \\frac{1}{8}\\theta^2 \\sin 4\\theta - \\frac{1}{16}\\theta \\cos 4\\theta + \\frac{1}{64}\\sin 4\\theta + C_1$$\n\nMultiply the entire equation by $3$ to find the expression for $r^3$:\n\n$$r^3 = \\frac{1}{2}\\theta^3 - \\frac{3}{8}\\theta^2 \\sin 4\\theta - \\frac{3}{16}\\theta \\cos 4\\theta + \\frac{3}{64}\\sin 4\\theta + C$$"}]}],"generatedAt":"2026-06-09T04:02:31.881Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"20d68fa4-6176-4f40-986e-b8ecf780fc78","revisionId":"f2316250-c6df-48c8-be40-4d38d1c5e1e5","questionNumber":null,"questionType":"LA","specification":"A curve has the equation\n\n$y = a^{x^2}$\n\nwhere $a$ is a constant greater than 1.","difficultyBand":"hard","tier":null,"totalMarks":8,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"efa772ea-fb7c-4a04-9521-c4fe29779314","label":"a","content":"Show that $\\dfrac{dy}{dx} = 2xa^{x^2}\\ln a$.","metadata":null,"markscheme":"- Writes $a^{x^2} = e^{x^2\\ln a}$ M1\n- Differentiates by the chain rule: $e^{x^2\\ln a} \\times 2x\\ln a$ M1\n- $= 2xa^{x^2}\\ln a$ as required A1\n\nAccept taking logarithms of both sides and differentiating implicitly.","marks":3,"order":0,"markingRubric":null,"tableImages":[],"explanation":null,"options":[]},{"id":"b6034ea2-ea81-44b3-8097-45ce1cc72853","label":"b","content":"The tangent to the curve at the point $(1, a)$ passes through the point $\\left(\\frac{1}{2}, 0\\right)$. Find the value of $a$, giving your answer in exact form.","metadata":null,"markscheme":"- The gradient of the line through $\\left(\\frac{1}{2}, 0\\right)$ and $(1, a)$ is $\\frac{a - 0}{1 - \\frac{1}{2}} = 2a$ M1\n- From (a), the gradient at $(1, a)$ is $2a\\ln a$, so $2a\\ln a = 2a$ M1\n- $a > 1$, so $a \\neq 0$ and $\\ln a = 1$, giving $a = e$ A1","marks":3,"order":1,"markingRubric":null,"tableImages":[],"explanation":null,"options":[]},{"id":"55210b9a-b920-4867-bd4b-8f27883c803a","label":"c","content":"By considering $\\dfrac{d^2y}{dx^2}$, show that the curve is convex for all values of $x$.","metadata":null,"markscheme":"- $\\frac{d^2y}{dx^2} = 2a^{x^2}\\ln a\\left(1 + 2x^2\\ln a\\right)$ M1\n- $a > 1$ gives $\\ln a > 0$, and $a^{x^2} > 0$ and $1 + 2x^2\\ln a > 0$ for every $x$, so $\\frac{d^2y}{dx^2} > 0$ and the curve is convex throughout A1\n\nAll three factors must be shown positive for the A1; asserting convexity from the shape of the graph scores 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3 Unit A2 1: Pure Mathematics (A-level only)

Question 61

3 Unit A2 1: Pure Mathematics (A-level only) Questions

3 Unit A2 1: Pure Mathematics (A-level only)

Question 61

3 Unit A2 1: Pure Mathematics (A-level only) Questions