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bg-background","children":["$","$L3b",null,{"backHref":"/a-level/maths/wjec/questions","basePath":"/a-level/maths/wjec/integration/questions","formulaUrl":"https://www.wjec.co.uk/media/4jfdytcz/ss-w27-18-e.pdf","initialIndex":50,"initialQuestionWindow":[{"id":"1520e624-4946-44a1-ae9e-3fa2571618f3","revisionId":"d564f751-7b09-4bb6-bf1e-e3c129876e66","questionNumber":null,"questionType":"LA","specification":"","difficulty":null,"difficultyBand":"hard","tier":null,"totalMarks":4,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"1951ef06-ee7d-4b29-af8a-03bf7e169edb","label":null,"content":"Find the value of $k$ such that\n$\\displaystyle\\int_k^9 \\dfrac{10}{\\sqrt{x}}\\,dx = 20$","markscheme":"- Writes the integrand as $10x^{-\\frac{1}{2}}$ and integrates to give $20x^{\\frac{1}{2}}$ M1\n- Both the coefficient and the power correct A1\n- Substitutes both limits: $60 - 20\\sqrt{k} = 20$ M1\n- $\\sqrt{k} = 2$, so $k = 4$ A1\n\nThe unknown is the lower limit here, so the subtraction must be the right way round; $20\\sqrt k - 60 = 20$ gives $k = 16$ and scores M1A1M0A0.","marks":4,"order":0,"markingRubric":null,"tableImages":[],"explanation":null,"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"1810efea-c282-4b7b-8588-f452e1094917","revisionId":"d768725d-b402-49d5-8f16-256cd2c2e391","questionNumber":null,"questionType":"LA","specification":"The cross-section of a custom-designed drainage channel is modeled by the parametric equations\n\\[x = 2u^2 + 2u, \\quad y = \\dfrac{10}{u(7 - u)}, \\quad a \\le u \\le b\\]\nwhere $a$ and $b$ are constants. The surface of the water in the channel is represented by the horizontal line $y = 1$, which connects the two endpoints of the curve.","difficulty":7,"difficultyBand":"hard","tier":null,"totalMarks":14,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"6ff203b9-c986-4fd9-9ac4-ccc438a40a7a","label":"a","content":"Determine the values of the constants $a$ and $b$, where $b > a$.","markscheme":"- Sets $y = 1$ to find intersections: $\\dfrac{10}{7u - u^2} = 1 \\Rightarrow u^2 - 7u + 10 = 0$ M1\n- Factorises to $(u - 2)(u - 5) = 0$ M1\n- States $a = 2$ and $b = 5$ A1","marks":3,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"algebraic_factoring","label":"Algebraic Factoring","steps":[{"type":"text","order":0,"title":"Set up the equation","content":"The question states that the surface of the water in the channel is represented by the horizontal line $y = 1$, which connects the two endpoints of the curve corresponding to $u = a$ and $u = b$.\n\nTo find these endpoints, we set the parametric equation for $y$ equal to $1$:\n\n$$\\frac{10}{u(7 - u)} = 1$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$10 = \\htmlData{anchor=u-expand}{u(7 - u)}$"},{"latex":"$$10 = \\htmlData{anchor=expanded}{7u - u^2}$"},{"latex":"$$u^2 - 7u + 10 = 0$"}],"highlights":[{"color":"sky","style":"box","anchor":"u-expand"},{"color":"sky","style":"box","anchor":"expanded"}],"connections":[{"to":"expanded","from":"u-expand","color":"sky","label":"expand","style":"solid"}]},"order":1,"title":"Rearrange to a quadratic","description":"We multiply both sides of the equation by $u(7 - u)$ to clear the fraction, and then rearrange it into the standard form of a quadratic equation, $Au^2 + Bu + C = 0$."},{"type":"text","order":2,"title":"Factorise the quadratic equation","content":"Now we solve the quadratic equation $u^2 - 7u + 10 = 0$ by factoring.\n\nWe need to find two numbers that multiply to $+10$ and add up to $-7$. These numbers are $-2$ and $-5$:\n\n$$(u - 2)(u - 5) = 0$$\n\nThis gives two possible values for $u$:\n\n$$u = 2 \\quad \\text{or} \\quad u = 5$$"},{"type":"text","order":3,"title":"Identify constants a and b","content":"Since $a$ and $b$ are the boundary values of $u$ with $b > a$, we assign the smaller value to $a$ and the larger value to $b$:\n\n$$a = 2, \\quad b = 5$$"}]},{"id":"quadratic_formula","label":"Quadratic Formula","steps":[{"type":"text","order":0,"title":"Set $y = 1$","content":"The surface of the water is represented by the horizontal line $y = 1$. Since this line connects the two endpoints of the curve corresponding to $u = a$ and $u = b$, we find these endpoints by setting the parametric equation for $y$ equal to $1$:\n\n$$\\frac{10}{u(7 - u)} = 1$$","highlights":[{"text":"horizontal line y = 1","location":"specification"}]},{"type":"text","order":1,"title":"Form a quadratic equation","content":"Next, we rearrange the equation to form a standard quadratic equation. Multiply both sides by the denominator $u(7 - u)$:\n\n$$10 = u(7 - u)$$\n\nExpand the right-hand side:\n\n$$10 = 7u - u^2$$\n\nRearrange all terms to one side to get a quadratic equation in the form $Au^2 + Bu + C = 0$:\n\n$$u^2 - 7u + 10 = 0$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\htmlData{anchor=quad-eq}{\\color{@sky}{1}}u^2 \\htmlData{anchor=coeff-b}{\\color{@amber}{-7}}u + \\htmlData{anchor=coeff-c}{\\color{@purple}{10}} = 0$$"},{"latex":"$$$u = \\frac{-(\\color{@amber}{-7}) \\pm \\sqrt{(\\color{@amber}{-7})^2 - 4(\\color{@sky}{1})(\\color{@purple}{10})}}{2(\\color{@sky}{1})}$$"},{"latex":"$$$u = \\frac{7 \\pm \\sqrt{49 - 40}}{2} = \\frac{7 \\pm \\sqrt{9}}{2}$$"},{"latex":"$$$u = \\frac{7 \\pm 3}{2}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"quad-eq"},{"color":"amber","style":"text-color","anchor":"coeff-b"},{"color":"purple","style":"text-color","anchor":"coeff-c"}],"connections":[{"to":"coeff-b","from":"quad-eq","color":"sky","label":"a = 1","style":"dashed"}]},"order":2,"title":"Apply the quadratic formula","description":"We use the quadratic formula $$u = \\frac{-B \\pm \\sqrt{B^2 - 4AC}}{2A}$$ with coefficients $A = 1$, $B = -7$, and $C = 10$ to solve for $u$."},{"type":"text","order":3,"title":"Find $a$ and $b$","content":"Solving for $u$ gives two possible values:\n\n$$u = \\frac{7 + 3}{2} = 5 \\quad \\text{or} \\quad u = \\frac{7 - 3}{2} = 2$$\n\nSince we are given that $b > a$ for the interval $a \\le u \\le b$, we assign the smaller value to $a$ and the larger value to $b$:\n\n$$a = 2, \\quad b = 5$$","highlights":[{"text":"b > a","location":"specification"}]}]},{"id":"graphical_calculator","label":"Using a Graphing Calculator","steps":[{"type":"text","order":0,"title":"Set up the equation","content":"To find where the curve meets the water surface, we set the parametric equation for $y$ equal to $1$, which represents the water level: $$\\frac{10}{u(7-u)} = 1$$ To find this intersection using a graphing calculator, we can define the parameter $u$ as the variable $x$ and look for the points of intersection between: $$Y_1 = \\frac{10}{x(7-x)}$$ and $$Y_2 = 1$$","highlights":[{"text":"y = 1","location":"specification"}]},{"type":"text","order":1,"title":"Find intersections on GDC","content":"We can now plot these two equations on our graphing calculator and find their intersection points.","calculator":[{"gdc":{"expression":"Y_1 = 10 / (X * (7 - X)), Y_2 = 1","instructions":"Enter the Graph mode. Input Y1 = 10 / (X * (7 - X)) and Y2 = 1. Adjust the window settings so both intersections are visible (e.g., X from 0 to 8, Y from -1 to 5). Use the Graph Solver (G-Solv) and select 'Intersection' (INTSECT) to find the x-coordinates of the intersections: x = 2 and x = 5."},"ti84":{"expression":"Y_1 = 10 / (X * (7 - X)), Y_2 = 1","instructions":"Press [Y=] and enter Y1 = 10 / (X * (7 - X)) and Y2 = 1. Press [WINDOW] and set Xmin = 0, Xmax = 8, Ymin = -1, Ymax = 5. Press [GRAPH]. To find the first intersection, press [2nd] [CALC] and select '5: intersect'. Move the cursor near the first intersection point and press [ENTER] three times to find x = 2. Repeat this process near the second intersection point to find x = 5."},"desmos":{"expression":"y = 10 / (x * (7 - x)), y = 1","instructions":"Type y = 10 / (x * (7 - x)) in the first expression bar, and y = 1 in the second expression bar. Click on the gray intersection points on the graph to display their coordinates, which are (2, 1) and (5, 1)."},"description":"Graph the curve and the horizontal line to find their intersection points."}]},{"type":"text","order":2,"title":"Determine constants a and b","content":"The calculator gives the two intersection $x$-coordinates (which represent our parameter $u$) as: $$u = 2 \\quad \\text{and} \\quad u = 5$$ Since the domain of the curve is defined for $a \\le u \\le b$ with the condition $b > a$, we assign the smaller value to $a$ and the larger value to $b$: $$a = 2$$ $$b = 5$$","highlights":[{"text":"b > a","location":"specification"}]}]}],"generatedAt":"2026-06-09T03:43:12.384Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"d57d1eb0-ad2a-4542-8193-254dc48ffd4a","label":"b","content":"The region $R$ represents the area of the cross-section of the water, bounded by the curve and the line $y = 1$. Show that the area of $R$ is given by\n\\[M - k \\int_{a}^{b} \\frac{2u+1}{u(7-u)} \\, du\\]\nwhere $M$ and $k$ are constants to be determined.","markscheme":"- Calculates $x$ at the boundaries: $x(5) = 2(25) + 2(5) = 60$ and $x(2) = 2(4) + 2(2) = 12$ B1\n- Identifies the area under the line $y=1$ as $1 \\times (60 - 12) = 48$ B1\n- Differentiates $x$ with respect to $u$: $\\dfrac{dx}{du} = 4u + 2$ M1\n- Sets up $\\text{Area} = 48 - \\int_{2}^{5} \\dfrac{10}{u(7-u)} \\cdot (4u + 2) \\, du$ and factors out $20$ to show $M = 48, k = 20$ A1","marks":4,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"geometric_splitting","label":"Geometric Splitting Approach","steps":[{"type":"text","order":0,"title":"Find the limits for $u$","content":"First, let's find the values of $a$ and $b$ where the curve intersects the line $y=1$. We do this by setting $y = 1$:\n\n$$\\frac{10}{u(7-u)} = 1$$\n\nSolving for $u$:\n\n$$\\begin{aligned} 10 &= u(7-u) \\\\ 10 &= 7u - u^2 \\\\ u^2 - 7u + 10 &= 0 \\\\ (u-2)(u-5) &= 0 \\end{aligned}$$\n\nThis gives $u = 2$ and $u = 5$. Since $a \\le u \\le b$, we have $a = 2$ and $b = 5$.","highlights":[{"text":"The surface of the water in the channel is represented by the horizontal line y = 1, which connects the two endpoints of the curve.","location":"specification"}]},{"type":"text","order":1,"title":"Calculate boundary $x$ values","content":"Next, let's find the corresponding horizontal coordinates $x$ at these boundaries by substituting $u = 2$ and $u = 5$ into the parametric equation for $x$:\n\nFor $u = 2$:\n$$x(2) = 2(2)^2 + 2(2) = 8 + 4 = 12$$\n\nFor $u = 5$:\n$$x(5) = 2(5)^2 + 2(5) = 50 + 10 = 60$$"},{"type":"text","order":2,"title":"Calculate the rectangular area","content":"We can think of the region $R$ as the area of a large rectangle minus the area under the curve.\n\nThe rectangle is bounded from $x = 12$ to $x = 60$ horizontally, and has a height of $y = 1$ vertically.\n\nLet's calculate this rectangular area:\n$$\\text{Area}_{\\text{rectangle}} = \\text{width} \\times \\text{height} = (60 - 12) \\times 1 = 48$$"},{"type":"text","order":3,"title":"Find $dx/du$","content":"To find the area under the curve using parametric integration, we use the formula:\n$$\\text{Area}_{\\text{under curve}} = \\int_{a}^{b} y \\frac{dx}{du} \\, du$$\n\nFirst, let's find $\\frac{dx}{du}$ by differentiating $x = 2u^2 + 2u$ with respect to $u$:\n$$\\frac{dx}{du} = 4u + 2$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Area}_{\\text{under curve}} = \\int_{2}^{5} \\htmlData{anchor=y-part}{\\color{@sky}{\\frac{10}{u(7-u)}}} \\cdot \\htmlData{anchor=dx-part}{\\color{@amber}{(4u + 2)}} \\, du$"},{"latex":"$$= \\int_{2}^{5} \\frac{10(4u+2)}{u(7-u)} \\, du$"},{"latex":"$$= \\int_{2}^{5} \\frac{20(2u+1)}{u(7-u)} \\, du$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"y-part"},{"color":"amber","style":"text-color","anchor":"dx-part"}],"connections":[{"to":"dx-part","from":"y-part","color":"sky","label":"y * dx/du","style":"solid"}]},"order":4,"title":"Set up parametric integration","description":"We substitute our expressions for $y$ and $\\frac{dx}{du}$ into the parametric area integral, then factor out a common factor of $2$ from $(4u+2)$ to align with the required form."},{"type":"text","order":5,"title":"Determine $M$ and $k$","content":"Subtracting the area under the curve from the area of the rectangle gives the area of $R$:\n\n$$\\text{Area of } R = \\text{Area}_{\\text{rectangle}} - \\text{Area}_{\\text{under curve}}$$\n\n$$\\text{Area of } R = 48 - \\int_{2}^{5} \\frac{20(2u+1)}{u(7-u)} \\, du$$\n\nFactor out the constant $20$ from the integral to match the required format:\n\n$$\\text{Area of } R = 48 - 20 \\int_{a}^{b} \\frac{2u+1}{u(7-u)} \\, du$$\n\nBy comparing this with the required template, we find:\n* $M = 48$\n* $k = 20$\n* $a = 2$ and $b = 5$"}]},{"id":"direct_parametric_integration","label":"Direct Parametric Integration","steps":[{"type":"text","order":0,"title":"Set up the area integral","content":"The region $R$ is bounded above by the horizontal line $y = 1$ and below by the parametric curve.\n\nThe area of a region bounded between $y = 1$ and a curve $y(x)$ from $x_1$ to $x_2$ is given by the integral:\n$$\\text{Area} = \\int_{x_1}^{x_2} (1 - y) \\, dx$$"},{"type":"text","order":1,"title":"Find the integration limits","content":"First, let's find the values of $u$ at the boundaries where the water surface meets the channel curve, which is where $y = 1$:\n\n$$\\frac{10}{u(7-u)} = 1 \\implies u(7-u) = 10 \\implies u^2 - 7u + 10 = 0$$\n\nFactoring this quadratic gives:\n$$(u-2)(u-5) = 0 \\implies u = 2 \\text{ or } u = 5$$\n\nThus, our limits are $a = 2$ and $b = 5$.\n\nNow, we find the corresponding $x$ limits using $x = 2u^2 + 2u$:\n- At $u = 2$: $x_1 = 2(2)^2 + 2(2) = 12$\n- At $u = 5$: $x_2 = 2(5)^2 + 2(5) = 60$"},{"type":"text","order":2,"title":"Find the differential dx/du","content":"To change the variable of integration from $x$ to $u$, we need to differentiate $x = 2u^2 + 2u$ with respect to $u$:\n\n$$\\frac{dx}{du} = 4u + 2 \\implies dx = (4u + 2)\\,du$$"},{"type":"text","order":3,"title":"Substitute and expand","content":"Now substitute $y = \\dfrac{10}{u(7-u)}$ and $dx = (4u + 2)\\,du$ into our area integral:\n\n$$\\text{Area} = \\int_{2}^{5} \\left(1 - \\frac{10}{u(7-u)}\\right) (4u + 2)\\,du$$\n\nWe expand this single integral to separate the integrable polynomial part from the fractional term:\n\n$$\\text{Area} = \\int_{2}^{5} (4u + 2)\\,du - \\int_{2}^{5} \\frac{10(4u+2)}{u(7-u)}\\,du$$"},{"type":"text","order":4,"title":"Calculate the constant term M","content":"Let's evaluate the first term, which represents the area under the line $y = 1$:\n\n$$\\int_{2}^{5} (4u + 2)\\,du = \\left[ 2u^2 + 2u \\right]_{2}^{5}$$\n\nNotice that this is simply the difference in our $x$ limits, $[x]_{12}^{60}$:\n\n$$\\left( 2(5)^2 + 2(5) \\right) - \\left( 2(2)^2 + 2(2) \\right) = 60 - 12 = 48$$\n\nSo, the constant $M = 48$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\text{Area} = 48 - \\int_{2}^{5} \\frac{10(\\htmlData{anchor=factor-this}{4u+2})}{u(7-u)}\\,du$$"},{"latex":"$$$\\text{Area} = 48 - \\int_{2}^{5} \\frac{10 \\cdot \\htmlData{anchor=factored}{\\color{@orange}{2(2u+1)}}}{u(7-u)}\\,du$$"},{"latex":"$$$\\text{Area} = 48 - \\htmlData{anchor=k-val}{\\color{@pink}{20}} \\int_{2}^{5} \\frac{\\htmlData{anchor=target-fraction}{2u+1}}{u(7-u)}\\,du$$"}],"highlights":[{"color":"orange","style":"underline","anchor":"factor-this"},{"color":"orange","style":"text-color","anchor":"factored"},{"color":"pink","style":"box","anchor":"k-val"}],"connections":[{"to":"factored","from":"factor-this","color":"orange","label":"factor out 2","style":"solid"},{"to":"k-val","from":"factored","color":"pink","label":"multiply 10 by 2","style":"solid"}]},"order":5,"title":"Rearrange to match the target form","description":"We factor $2$ out of $(4u+2)$ inside the numerator of the second integral to match the required form of $\\frac{2u+1}{u(7-u)}$."},{"type":"text","order":6,"title":"State the final constants","content":"Comparing our expression to the target form:\n\n$$\\text{Area} = M - k \\int_{a}^{b} \\frac{2u+1}{u(7-u)} \\, du$$\n\nWe determine that:\n- $M = 48$\n- $k = 20$\n- $a = 2$\n- $b = 5$"}]}],"generatedAt":"2026-06-09T03:44:58.327Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"031fb07d-8392-4b31-b95d-4611e966e569","label":"ci","content":"Express $\\dfrac{2u+1}{u(7-u)}$ in partial fractions.","markscheme":"- Sets up partial fractions: $\\dfrac{2u+1}{u(7-u)} = \\dfrac{P}{u} + \\dfrac{Q}{7-u} \\Rightarrow 2u+1 = P(7-u) + Qu$ M1\n- Substitutes $u = 0$ to find $P = \\dfrac{1}{7}$ A1\n- Substitutes $u = 7$ to find $Q = \\dfrac{15}{7}$ A1","marks":3,"order":2,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"substitution","label":"Substitution Method","steps":[{"type":"text","order":0,"title":"Set up the partial fractions","content":"Since the denominator $u(7-u)$ consists of distinct linear factors, we can express the fraction in the form:\n\n$$\\frac{2u+1}{u(7-u)} = \\frac{P}{u} + \\frac{Q}{7-u}$$\n\nTo eliminate the fractions, multiply both sides by the common denominator $u(7-u)$. This gives us the cleared-fraction identity:\n\n$$2u+1 = P(7-u) + Qu$$\n\nSetting this up correctly earns the first method mark (**M1**)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$2(\\htmlData{anchor=u-zero}{\\color{@sky}{0}}) + 1 = P(7 - \\htmlData{anchor=u-zero-sub}{\\color{@sky}{0}}) + Q(\\htmlData{anchor=u-zero-q}{\\color{@sky}{0}})$$"},{"latex":"$$$1 = 7\\htmlData{anchor=p-val}{\\color{@amber}{P}}$$"},{"latex":"$$$\\htmlData{anchor=p-result}{\\color{@amber}{P}} = \\frac{1}{7}$$"}],"highlights":[{"color":"sky","style":"box","anchor":"u-zero"},{"color":"amber","style":"text-color","anchor":"p-result"}],"connections":[{"to":"u-zero-sub","from":"u-zero","color":"sky","label":"substitute","style":"solid"},{"to":"p-result","from":"p-val","color":"amber","label":"solve","style":"solid"}]},"order":1,"title":"Substitute $u = 0$","description":"To solve for $P$, we choose a value of $u$ that eliminates the term containing $Q$. Setting $u = 0$ does this directly. This calculation earns the second mark (**A1**)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$2(\\htmlData{anchor=u-seven}{\\color{@purple}{7}}) + 1 = P(7 - \\htmlData{anchor=u-seven-sub}{\\color{@purple}{7}}) + Q(\\htmlData{anchor=u-seven-q}{\\color{@purple}{7}})$$"},{"latex":"$$$15 = 7\\htmlData{anchor=q-val}{\\color{@teal}{Q}}$$"},{"latex":"$$$\\htmlData{anchor=q-result}{\\color{@teal}{Q}} = \\frac{15}{7}$$"}],"highlights":[{"color":"purple","style":"box","anchor":"u-seven"},{"color":"teal","style":"text-color","anchor":"q-result"}],"connections":[{"to":"u-seven-sub","from":"u-seven","color":"purple","label":"substitute","style":"solid"},{"to":"q-result","from":"q-val","color":"teal","label":"solve","style":"solid"}]},"order":2,"title":"Substitute $u = 7$","description":"To solve for $Q$, we choose a value of $u$ that eliminates the term containing $P$. Setting $u = 7$ makes $(7-u) = 0$. This calculation earns the third mark (**A1**)."},{"type":"text","order":3,"title":"State final expression","content":"Now we substitute our values of $P = \\frac{1}{7}$ and $Q = \\frac{15}{7}$ back into the original partial fraction setup:\n\n$$\\frac{2u+1}{u(7-u)} = \\frac{1}{7u} + \\frac{15}{7(7-u)}$$"}]},{"id":"equating_coefficients","label":"Equating Coefficients","steps":[{"type":"text","order":0,"title":"Set up the identity","content":"To express the algebraic fraction in partial fractions, we begin by setting up an identity with two unknown constants, $A$ and $B$:\n\n$$\\frac{2u+1}{u(7-u)} \\equiv \\frac{A}{u} + \\frac{B}{7-u}$$\n\nMultiply both sides by the common denominator $u(7-u)$ to clear the fractions. This gives us the linear identity:\n\n$$2u + 1 \\equiv A(7 - u) + Bu$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$2u + 1 \\equiv \\htmlData{anchor=expand}{A(7 - u) + Bu}$"},{"latex":"$$2u + 1 \\equiv \\htmlData{anchor=collected}{(B - A)u + 7A}$"}],"highlights":[{"color":"sky","style":"box","anchor":"expand"},{"color":"sky","style":"box","anchor":"collected"}],"connections":[{"to":"collected","from":"expand","color":"sky","label":"expand & collect","style":"solid"}]},"order":1,"title":"Expand and collect terms","description":"Expand the right-hand side and collect terms with $u$ together so we can compare both sides of the identity."},{"type":"text","order":2,"title":"Equate the coefficients","content":"Now we equate the coefficients of corresponding terms on both sides of the identity:\n\n- **Constant terms:**\n $$7A = 1 \\implies A = \\frac{1}{7}$$\n\n- **$u$ terms:**\n $$B - A = 2$$"},{"type":"text","order":3,"title":"Solve for B","content":"Substitute our value of $A = \\frac{1}{7}$ into the second equation to find $B$:\n\n$$\\begin{aligned} B - \\frac{1}{7} &= 2 \\\\ B &= 2 + \\frac{1}{7} \\\\ B &= \\frac{15}{7} \\end{aligned}$$"},{"type":"text","order":4,"title":"Write final expression","content":"Substitute the values of $A = \\frac{1}{7}$ and $B = \\frac{15}{7}$ back into our partial fractions setup:\n\n$$\\frac{2u+1}{u(7-u)} = \\frac{1}{7u} + \\frac{15}{7(7-u)}$$"}]}],"generatedAt":"2026-06-09T03:46:18.886Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"55830931-b474-4459-a267-e04ece8619e8","label":"cii","content":"Use algebraic integration to find the exact area of $R$, giving your answer in the form $A - B \\ln(2.5)$, where $A$ and $B$ are constants.","markscheme":"- Integrates the partial fractions: $\\int \\left( \\dfrac{1/7}{u} + \\dfrac{15/7}{7-u} \\right) du = \\dfrac{1}{7} \\ln|u| - \\dfrac{15}{7} \\ln|7-u|$ M1\n- Evaluates at limits $u=5$ and $u=2$: $\\left( \\frac{1}{7} \\ln 5 - \\frac{15}{7} \\ln 2 \\right) - \\left( \\frac{1}{7} \\ln 2 - \\frac{15}{7} \\ln 5 \\right)$ M1\n- Simplifies to $\\dfrac{16}{7} \\ln 5 - \\dfrac{16}{7} \\ln 2 = \\dfrac{16}{7} \\ln 2.5$ M1\n- Final area calculation: $48 - 20 \\left( \\dfrac{16}{7} \\ln 2.5 \\right) = 48 - \\dfrac{320}{7} \\ln 2.5$ A1","marks":4,"order":3,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"area_subtraction","label":"Subtracting Area under Curve from Rectangle","steps":[{"type":"text","order":0,"title":"Formulate the Area Strategy","content":"To find the exact area of the drainage channel $R$ (which lies between the water surface $y = 1$ and the curved bottom), we can subtract the area under the parametric curve from the area of the bounding rectangle.\n\nThe bounding rectangle is formed by the horizontal water surface line $y = 1$ and the endpoints of the curve:\n- At $u = 2$, $x = 2(2)^2 + 2(2) = 12$\n- At $u = 5$, $x = 2(5)^2 + 2(5) = 60$\n\nThe width of this rectangle is $60 - 12 = 48$, and its height is $1$. Therefore:\n\n$$\\text{Area of Rectangle} = 1 \\times 48 = 48$$\n\nTo find the area of $R$, we use:\n$$\\text{Area of } R = 48 - \\int_{12}^{60} y \\, dx$$"},{"type":"text","order":1,"title":"Set Up Parametric Integration","content":"We convert the integral $\\int y \\, dx$ into parametric form with respect to $u$ using the limits $u = 2$ to $u = 5$:\n\n$$\\int_{12}^{60} y \\, dx = \\int_{2}^{5} y \\frac{dx}{du} \\, du$$\n\nGiven $x = 2u^2 + 2u$:\n$$\\frac{dx}{du} = 4u + 2 = 2(2u + 1)$$\n\nSubstituting $y = \\dfrac{10}{u(7 - u)}$ and $\\frac{dx}{du}$ into the integral:\n\n$$\\int_{2}^{5} \\frac{10}{u(7-u)} \\cdot (4u + 2) \\, du = \\int_{2}^{5} \\frac{20(2u + 1)}{u(7 - u)} \\, du = 20 \\int_{2}^{5} \\frac{2u + 1}{u(7 - u)} \\, du$$"},{"type":"text","order":2,"title":"Decompose using Partial Fractions","content":"To integrate $\\dfrac{2u+1}{u(7-u)}$, we split it into partial fractions:\n\n$$\\frac{2u + 1}{u(7 - u)} \\equiv \\frac{P}{u} + \\frac{Q}{7 - u}$$\n\nMultiplying both sides by the denominator $u(7 - u)$ gives:\n$$2u + 1 \\equiv P(7 - u) + Qu$$\n\n- Let $u = 0$:\n$$1 = 7P \\implies P = \\frac{1}{7}$$\n\n- Let $u = 7$:\n$$15 = 7Q \\implies Q = \\frac{15}{7}$$\n\nSo, the integrand becomes:\n$$\\frac{2u + 1}{u(7 - u)} = \\frac{1/7}{u} + \\frac{15/7}{7-u}$$"},{"type":"text","order":3,"title":"Integrate the Expressions","content":"Now we integrate each term with respect to $u$:\n\n$$\\int \\left( \\frac{1/7}{u} + \\frac{15/7}{7-u} \\right) du = \\frac{1}{7} \\ln|u| - \\frac{15}{7} \\ln|7-u|$$\n\nNote that the negative sign in the second term arises from the chain rule applied to the linear denominator $(7-u)$, which earns a method mark."},{"type":"text","order":4,"title":"Evaluate the Limits","content":"We evaluate this antiderivative between the limits $u=2$ and $u=5$:\n\n$$\\left[ \\frac{1}{7} \\ln|u| - \\frac{15}{7} \\ln|7-u| \\right]_{2}^{5} = \\left( \\frac{1}{7} \\ln 5 - \\frac{15}{7} \\ln 2 \\right) - \\left( \\frac{1}{7} \\ln 2 - \\frac{15}{7} \\ln 5 \\right)$$\n\nGrouping the terms together:\n$$\\left( \\frac{1}{7} \\ln 5 + \\frac{15}{7} \\ln 5 \\right) - \\left( \\frac{15}{7} \\ln 2 + \\frac{1}{7} \\ln 2 \\right) = \\frac{16}{7} \\ln 5 - \\frac{16}{7} \\ln 2$$\n\nUsing the laws of logarithms:\n$$\\frac{16}{7} (\\ln 5 - \\ln 2) = \\frac{16}{7} \\ln\\left(\\frac{5}{2}\\right) = \\frac{16}{7} \\ln 2.5$$\n\nThis simplification step earns a method mark."},{"type":"text","order":5,"title":"Subtract from Rectangle Area","content":"Recall that the total area under the curve is scaled by the factor $k = 20$ from our parametric integral:\n\n$$\\text{Area under Curve} = 20 \\times \\left( \\frac{16}{7} \\ln 2.5 \\right) = \\frac{320}{7} \\ln 2.5$$\n\nNow, subtract this from our rectangle area ($48$) to find the exact area of $R$:\n\n$$\\text{Area of } R = 48 - \\frac{320}{7} \\ln 2.5$$\n\nThis is in the required form $A - B\\ln(2.5)$, where $A = 48$ and $B = \\frac{320}{7}$. This final result earns the accuracy mark."}]},{"id":"direct_integration","label":"Direct Integration of the Shaded Region","steps":[{"type":"text","order":0,"title":"Set up the integral","content":"The shaded region $R$ is bounded above by the line $y = 1$ (the water surface) and below by the parametric curve. The limits of integration for $u$ correspond to where the curve meets $y=1$, which gives $u = 2$ and $u = 5$.\n\nThe area of the shaded region $R$ is given by the integral of the top boundary minus the bottom boundary:\n$$\\text{Area} = \\int_{x(2)}^{x(5)} (1 - y) \\, dx$$\n\nTo integrate with respect to $u$, we use parametric integration:\n$$dx = \\frac{dx}{du} \\, du$$\n\nGiven $x = 2u^2 + 2u$, we differentiate to find:\n$$\\frac{dx}{du} = 4u + 2$$\n\nSubstituting this and the expression for $y$ into the integral, we get:\n$$\\text{Area} = \\int_{2}^{5} \\left(1 - \\frac{10}{u(7 - u)}\\right)(4u + 2) \\, du$$"},{"type":"text","order":1,"title":"Split into two integrals","content":"To evaluate this, we expand the integrand and split it into two simpler integrals:\n$$\\begin{aligned} \\text{Area} &= \\int_{2}^{5} (4u + 2) \\, du - \\int_{2}^{5} \\frac{10(4u + 2)}{u(7 - u)} \\, du \\\\ &= I_1 - I_2 \\end{aligned}$$\n\nLet's evaluate the first integral, $I_1$, directly:\n$$\\begin{aligned} I_1 &= \\int_{2}^{5} (4u + 2) \\, du \\\\ &= \\left[ 2u^2 + 2u \\right]_{2}^{5} \\\\ &= \\left( 2(5)^2 + 2(5) \\right) - \\left( 2(2)^2 + 2(2) \\right) \\\\ &= 60 - 12 \\\\ &= 48 \\end{aligned}$$"},{"type":"text","order":2,"title":"Simplify the second integrand","content":"Now let's simplify the integrand of $I_2$:\n$$I_2 = \\int_{2}^{5} \\frac{10(4u + 2)}{u(7 - u)} \\, du = \\int_{2}^{5} \\frac{40u + 20}{u(7 - u)} \\, du$$\n\nWe can decompose the integrand using partial fractions:\n$$\\frac{40u + 20}{u(7 - u)} = \\frac{P}{u} + \\frac{Q}{7 - u}$$\n\nMultiplying both sides by the denominator $u(7 - u)$ gives:\n$$40u + 20 = P(7 - u) + Qu$$\n\n- Setting $u = 0$:\n $$20 = 7P \\implies P = \\frac{20}{7}$$\n- Setting $u = 7$:\n $$40(7) + 20 = 7Q \\implies 300 = 7Q \\implies Q = \\frac{300}{7}$$\n\nSo the expression becomes:\n$$\\frac{40u + 20}{u(7 - u)} = \\frac{20}{7u} + \\frac{300}{7(7 - u)}$$\n\nWe can factor out a constant of $20$ to align with the markscheme:\n$$\\frac{40u + 20}{u(7 - u)} = 20 \\left( \\frac{1/7}{u} + \\frac{15/7}{7-u} \\right)$$"},{"type":"text","order":3,"title":"Integrate the partial fractions","content":"Now we integrate each term with respect to $u$. Remember to apply the chain rule for the linear denominator $(7-u)$, which introduces a negative sign:\n$$\\int \\left( \\frac{1/7}{u} + \\frac{15/7}{7-u} \\right) du = \\frac{1}{7} \\ln|u| - \\frac{15}{7} \\ln|7-u|$$"},{"type":"text","order":4,"title":"Evaluate the second integral","content":"Next, we evaluate this antiderivative at the limits $u=5$ and $u=2$:\n$$\\begin{aligned} I_2 &= 20 \\left[ \\frac{1}{7} \\ln|u| - \\frac{15}{7} \\ln|7-u| \\right]_{2}^{5} \\\\ &= 20 \\left[ \\left( \\frac{1}{7} \\ln 5 - \\frac{15}{7} \\ln 2 \\right) - \\left( \\frac{1}{7} \\ln 2 - \\frac{15}{7} \\ln 5 \\right) \\right] \\\\ &= 20 \\left[ \\left( \\frac{1}{7} \\ln 5 + \\frac{15}{7} \\ln 5 \\right) - \\left( \\frac{1}{7} \\ln 2 + \\frac{15}{7} \\ln 2 \\right) \\right] \\\\ &= 20 \\left[ \\frac{16}{7} \\ln 5 - \\frac{16}{7} \\ln 2 \\right] \\\\ &= 20 \\left( \\frac{16}{7} \\ln\\left(\\frac{5}{2}\\right) \\right) \\\\ &= \\frac{320}{7} \\ln(2.5) \\end{aligned}$$"},{"type":"text","order":5,"title":"Subtract to find final area","content":"Finally, we combine our results from $I_1$ and $I_2$ to find the exact area:\n$$\\text{Area} = I_1 - I_2 = 48 - \\frac{320}{7} \\ln(2.5)$$\n\nThis is in the required form $A - B \\ln(2.5)$, where:\n$$A = 48, \\quad B = \\frac{320}{7}$$"}]}],"generatedAt":"2026-06-09T03:48:18.592Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"6ff203b9-c986-4fd9-9ac4-ccc438a40a7a:algebraic_factoring","label":"(a) Algebraic Factoring","steps":[{"type":"text","order":0,"title":"Set up the equation","content":"The question states that the surface of the water in the channel is represented by the horizontal line $y = 1$, which connects the two endpoints of the curve corresponding to $u = a$ and $u = b$.\n\nTo find these endpoints, we set the parametric equation for $y$ equal to $1$:\n\n$$\\frac{10}{u(7 - u)} = 1$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$10 = \\htmlData{anchor=u-expand}{u(7 - u)}$"},{"latex":"$$10 = \\htmlData{anchor=expanded}{7u - u^2}$"},{"latex":"$$u^2 - 7u + 10 = 0$"}],"highlights":[{"color":"sky","style":"box","anchor":"u-expand"},{"color":"sky","style":"box","anchor":"expanded"}],"connections":[{"to":"expanded","from":"u-expand","color":"sky","label":"expand","style":"solid"}]},"order":1,"title":"Rearrange to a quadratic","description":"We multiply both sides of the equation by $u(7 - u)$ to clear the fraction, and then rearrange it into the standard form of a quadratic equation, $Au^2 + Bu + C = 0$."},{"type":"text","order":2,"title":"Factorise the quadratic equation","content":"Now we solve the quadratic equation $u^2 - 7u + 10 = 0$ by factoring.\n\nWe need to find two numbers that multiply to $+10$ and add up to $-7$. These numbers are $-2$ and $-5$:\n\n$$(u - 2)(u - 5) = 0$$\n\nThis gives two possible values for $u$:\n\n$$u = 2 \\quad \\text{or} \\quad u = 5$$"},{"type":"text","order":3,"title":"Identify constants a and b","content":"Since $a$ and $b$ are the boundary values of $u$ with $b > a$, we assign the smaller value to $a$ and the larger value to $b$:\n\n$$a = 2, \\quad b = 5$$"}]},{"id":"6ff203b9-c986-4fd9-9ac4-ccc438a40a7a:quadratic_formula","label":"(a) Quadratic Formula","steps":[{"type":"text","order":0,"title":"Set $y = 1$","content":"The surface of the water is represented by the horizontal line $y = 1$. Since this line connects the two endpoints of the curve corresponding to $u = a$ and $u = b$, we find these endpoints by setting the parametric equation for $y$ equal to $1$:\n\n$$\\frac{10}{u(7 - u)} = 1$$","highlights":[{"text":"horizontal line y = 1","location":"specification"}]},{"type":"text","order":1,"title":"Form a quadratic equation","content":"Next, we rearrange the equation to form a standard quadratic equation. Multiply both sides by the denominator $u(7 - u)$:\n\n$$10 = u(7 - u)$$\n\nExpand the right-hand side:\n\n$$10 = 7u - u^2$$\n\nRearrange all terms to one side to get a quadratic equation in the form $Au^2 + Bu + C = 0$:\n\n$$u^2 - 7u + 10 = 0$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\htmlData{anchor=quad-eq}{\\color{@sky}{1}}u^2 \\htmlData{anchor=coeff-b}{\\color{@amber}{-7}}u + \\htmlData{anchor=coeff-c}{\\color{@purple}{10}} = 0$$"},{"latex":"$$$u = \\frac{-(\\color{@amber}{-7}) \\pm \\sqrt{(\\color{@amber}{-7})^2 - 4(\\color{@sky}{1})(\\color{@purple}{10})}}{2(\\color{@sky}{1})}$$"},{"latex":"$$$u = \\frac{7 \\pm \\sqrt{49 - 40}}{2} = \\frac{7 \\pm \\sqrt{9}}{2}$$"},{"latex":"$$$u = \\frac{7 \\pm 3}{2}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"quad-eq"},{"color":"amber","style":"text-color","anchor":"coeff-b"},{"color":"purple","style":"text-color","anchor":"coeff-c"}],"connections":[{"to":"coeff-b","from":"quad-eq","color":"sky","label":"a = 1","style":"dashed"}]},"order":2,"title":"Apply the quadratic formula","description":"We use the quadratic formula $$u = \\frac{-B \\pm \\sqrt{B^2 - 4AC}}{2A}$$ with coefficients $A = 1$, $B = -7$, and $C = 10$ to solve for $u$."},{"type":"text","order":3,"title":"Find $a$ and $b$","content":"Solving for $u$ gives two possible values:\n\n$$u = \\frac{7 + 3}{2} = 5 \\quad \\text{or} \\quad u = \\frac{7 - 3}{2} = 2$$\n\nSince we are given that $b > a$ for the interval $a \\le u \\le b$, we assign the smaller value to $a$ and the larger value to $b$:\n\n$$a = 2, \\quad b = 5$$","highlights":[{"text":"b > a","location":"specification"}]}]},{"id":"6ff203b9-c986-4fd9-9ac4-ccc438a40a7a:graphical_calculator","label":"(a) Using a Graphing Calculator","steps":[{"type":"text","order":0,"title":"Set up the equation","content":"To find where the curve meets the water surface, we set the parametric equation for $y$ equal to $1$, which represents the water level: $$\\frac{10}{u(7-u)} = 1$$ To find this intersection using a graphing calculator, we can define the parameter $u$ as the variable $x$ and look for the points of intersection between: $$Y_1 = \\frac{10}{x(7-x)}$$ and $$Y_2 = 1$$","highlights":[{"text":"y = 1","location":"specification"}]},{"type":"text","order":1,"title":"Find intersections on GDC","content":"We can now plot these two equations on our graphing calculator and find their intersection points.","calculator":[{"gdc":{"expression":"Y_1 = 10 / (X * (7 - X)), Y_2 = 1","instructions":"Enter the Graph mode. Input Y1 = 10 / (X * (7 - X)) and Y2 = 1. Adjust the window settings so both intersections are visible (e.g., X from 0 to 8, Y from -1 to 5). Use the Graph Solver (G-Solv) and select 'Intersection' (INTSECT) to find the x-coordinates of the intersections: x = 2 and x = 5."},"ti84":{"expression":"Y_1 = 10 / (X * (7 - X)), Y_2 = 1","instructions":"Press [Y=] and enter Y1 = 10 / (X * (7 - X)) and Y2 = 1. Press [WINDOW] and set Xmin = 0, Xmax = 8, Ymin = -1, Ymax = 5. Press [GRAPH]. To find the first intersection, press [2nd] [CALC] and select '5: intersect'. Move the cursor near the first intersection point and press [ENTER] three times to find x = 2. Repeat this process near the second intersection point to find x = 5."},"desmos":{"expression":"y = 10 / (x * (7 - x)), y = 1","instructions":"Type y = 10 / (x * (7 - x)) in the first expression bar, and y = 1 in the second expression bar. Click on the gray intersection points on the graph to display their coordinates, which are (2, 1) and (5, 1)."},"description":"Graph the curve and the horizontal line to find their intersection points."}]},{"type":"text","order":2,"title":"Determine constants a and b","content":"The calculator gives the two intersection $x$-coordinates (which represent our parameter $u$) as: $$u = 2 \\quad \\text{and} \\quad u = 5$$ Since the domain of the curve is defined for $a \\le u \\le b$ with the condition $b > a$, we assign the smaller value to $a$ and the larger value to $b$: $$a = 2$$ $$b = 5$$","highlights":[{"text":"b > a","location":"specification"}]}]},{"id":"d57d1eb0-ad2a-4542-8193-254dc48ffd4a:geometric_splitting","label":"(b) Geometric Splitting Approach","steps":[{"type":"text","order":0,"title":"Find the limits for $u$","content":"First, let's find the values of $a$ and $b$ where the curve intersects the line $y=1$. We do this by setting $y = 1$:\n\n$$\\frac{10}{u(7-u)} = 1$$\n\nSolving for $u$:\n\n$$\\begin{aligned} 10 &= u(7-u) \\\\ 10 &= 7u - u^2 \\\\ u^2 - 7u + 10 &= 0 \\\\ (u-2)(u-5) &= 0 \\end{aligned}$$\n\nThis gives $u = 2$ and $u = 5$. Since $a \\le u \\le b$, we have $a = 2$ and $b = 5$.","highlights":[{"text":"The surface of the water in the channel is represented by the horizontal line y = 1, which connects the two endpoints of the curve.","location":"specification"}]},{"type":"text","order":1,"title":"Calculate boundary $x$ values","content":"Next, let's find the corresponding horizontal coordinates $x$ at these boundaries by substituting $u = 2$ and $u = 5$ into the parametric equation for $x$:\n\nFor $u = 2$:\n$$x(2) = 2(2)^2 + 2(2) = 8 + 4 = 12$$\n\nFor $u = 5$:\n$$x(5) = 2(5)^2 + 2(5) = 50 + 10 = 60$$"},{"type":"text","order":2,"title":"Calculate the rectangular area","content":"We can think of the region $R$ as the area of a large rectangle minus the area under the curve.\n\nThe rectangle is bounded from $x = 12$ to $x = 60$ horizontally, and has a height of $y = 1$ vertically.\n\nLet's calculate this rectangular area:\n$$\\text{Area}_{\\text{rectangle}} = \\text{width} \\times \\text{height} = (60 - 12) \\times 1 = 48$$"},{"type":"text","order":3,"title":"Find $dx/du$","content":"To find the area under the curve using parametric integration, we use the formula:\n$$\\text{Area}_{\\text{under curve}} = \\int_{a}^{b} y \\frac{dx}{du} \\, du$$\n\nFirst, let's find $\\frac{dx}{du}$ by differentiating $x = 2u^2 + 2u$ with respect to $u$:\n$$\\frac{dx}{du} = 4u + 2$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Area}_{\\text{under curve}} = \\int_{2}^{5} \\htmlData{anchor=y-part}{\\color{@sky}{\\frac{10}{u(7-u)}}} \\cdot \\htmlData{anchor=dx-part}{\\color{@amber}{(4u + 2)}} \\, du$"},{"latex":"$$= \\int_{2}^{5} \\frac{10(4u+2)}{u(7-u)} \\, du$"},{"latex":"$$= \\int_{2}^{5} \\frac{20(2u+1)}{u(7-u)} \\, du$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"y-part"},{"color":"amber","style":"text-color","anchor":"dx-part"}],"connections":[{"to":"dx-part","from":"y-part","color":"sky","label":"y * dx/du","style":"solid"}]},"order":4,"title":"Set up parametric integration","description":"We substitute our expressions for $y$ and $\\frac{dx}{du}$ into the parametric area integral, then factor out a common factor of $2$ from $(4u+2)$ to align with the required form."},{"type":"text","order":5,"title":"Determine $M$ and $k$","content":"Subtracting the area under the curve from the area of the rectangle gives the area of $R$:\n\n$$\\text{Area of } R = \\text{Area}_{\\text{rectangle}} - \\text{Area}_{\\text{under curve}}$$\n\n$$\\text{Area of } R = 48 - \\int_{2}^{5} \\frac{20(2u+1)}{u(7-u)} \\, du$$\n\nFactor out the constant $20$ from the integral to match the required format:\n\n$$\\text{Area of } R = 48 - 20 \\int_{a}^{b} \\frac{2u+1}{u(7-u)} \\, du$$\n\nBy comparing this with the required template, we find:\n* $M = 48$\n* $k = 20$\n* $a = 2$ and $b = 5$"}]},{"id":"d57d1eb0-ad2a-4542-8193-254dc48ffd4a:direct_parametric_integration","label":"(b) Direct Parametric Integration","steps":[{"type":"text","order":0,"title":"Set up the area integral","content":"The region $R$ is bounded above by the horizontal line $y = 1$ and below by the parametric curve.\n\nThe area of a region bounded between $y = 1$ and a curve $y(x)$ from $x_1$ to $x_2$ is given by the integral:\n$$\\text{Area} = \\int_{x_1}^{x_2} (1 - y) \\, dx$$"},{"type":"text","order":1,"title":"Find the integration limits","content":"First, let's find the values of $u$ at the boundaries where the water surface meets the channel curve, which is where $y = 1$:\n\n$$\\frac{10}{u(7-u)} = 1 \\implies u(7-u) = 10 \\implies u^2 - 7u + 10 = 0$$\n\nFactoring this quadratic gives:\n$$(u-2)(u-5) = 0 \\implies u = 2 \\text{ or } u = 5$$\n\nThus, our limits are $a = 2$ and $b = 5$.\n\nNow, we find the corresponding $x$ limits using $x = 2u^2 + 2u$:\n- At $u = 2$: $x_1 = 2(2)^2 + 2(2) = 12$\n- At $u = 5$: $x_2 = 2(5)^2 + 2(5) = 60$"},{"type":"text","order":2,"title":"Find the differential dx/du","content":"To change the variable of integration from $x$ to $u$, we need to differentiate $x = 2u^2 + 2u$ with respect to $u$:\n\n$$\\frac{dx}{du} = 4u + 2 \\implies dx = (4u + 2)\\,du$$"},{"type":"text","order":3,"title":"Substitute and expand","content":"Now substitute $y = \\dfrac{10}{u(7-u)}$ and $dx = (4u + 2)\\,du$ into our area integral:\n\n$$\\text{Area} = \\int_{2}^{5} \\left(1 - \\frac{10}{u(7-u)}\\right) (4u + 2)\\,du$$\n\nWe expand this single integral to separate the integrable polynomial part from the fractional term:\n\n$$\\text{Area} = \\int_{2}^{5} (4u + 2)\\,du - \\int_{2}^{5} \\frac{10(4u+2)}{u(7-u)}\\,du$$"},{"type":"text","order":4,"title":"Calculate the constant term M","content":"Let's evaluate the first term, which represents the area under the line $y = 1$:\n\n$$\\int_{2}^{5} (4u + 2)\\,du = \\left[ 2u^2 + 2u \\right]_{2}^{5}$$\n\nNotice that this is simply the difference in our $x$ limits, $[x]_{12}^{60}$:\n\n$$\\left( 2(5)^2 + 2(5) \\right) - \\left( 2(2)^2 + 2(2) \\right) = 60 - 12 = 48$$\n\nSo, the constant $M = 48$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\text{Area} = 48 - \\int_{2}^{5} \\frac{10(\\htmlData{anchor=factor-this}{4u+2})}{u(7-u)}\\,du$$"},{"latex":"$$$\\text{Area} = 48 - \\int_{2}^{5} \\frac{10 \\cdot \\htmlData{anchor=factored}{\\color{@orange}{2(2u+1)}}}{u(7-u)}\\,du$$"},{"latex":"$$$\\text{Area} = 48 - \\htmlData{anchor=k-val}{\\color{@pink}{20}} \\int_{2}^{5} \\frac{\\htmlData{anchor=target-fraction}{2u+1}}{u(7-u)}\\,du$$"}],"highlights":[{"color":"orange","style":"underline","anchor":"factor-this"},{"color":"orange","style":"text-color","anchor":"factored"},{"color":"pink","style":"box","anchor":"k-val"}],"connections":[{"to":"factored","from":"factor-this","color":"orange","label":"factor out 2","style":"solid"},{"to":"k-val","from":"factored","color":"pink","label":"multiply 10 by 2","style":"solid"}]},"order":5,"title":"Rearrange to match the target form","description":"We factor $2$ out of $(4u+2)$ inside the numerator of the second integral to match the required form of $\\frac{2u+1}{u(7-u)}$."},{"type":"text","order":6,"title":"State the final constants","content":"Comparing our expression to the target form:\n\n$$\\text{Area} = M - k \\int_{a}^{b} \\frac{2u+1}{u(7-u)} \\, du$$\n\nWe determine that:\n- $M = 48$\n- $k = 20$\n- $a = 2$\n- $b = 5$"}]},{"id":"031fb07d-8392-4b31-b95d-4611e966e569:substitution","label":"(ci) Substitution Method","steps":[{"type":"text","order":0,"title":"Set up the partial fractions","content":"Since the denominator $u(7-u)$ consists of distinct linear factors, we can express the fraction in the form:\n\n$$\\frac{2u+1}{u(7-u)} = \\frac{P}{u} + \\frac{Q}{7-u}$$\n\nTo eliminate the fractions, multiply both sides by the common denominator $u(7-u)$. This gives us the cleared-fraction identity:\n\n$$2u+1 = P(7-u) + Qu$$\n\nSetting this up correctly earns the first method mark (**M1**)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$2(\\htmlData{anchor=u-zero}{\\color{@sky}{0}}) + 1 = P(7 - \\htmlData{anchor=u-zero-sub}{\\color{@sky}{0}}) + Q(\\htmlData{anchor=u-zero-q}{\\color{@sky}{0}})$$"},{"latex":"$$$1 = 7\\htmlData{anchor=p-val}{\\color{@amber}{P}}$$"},{"latex":"$$$\\htmlData{anchor=p-result}{\\color{@amber}{P}} = \\frac{1}{7}$$"}],"highlights":[{"color":"sky","style":"box","anchor":"u-zero"},{"color":"amber","style":"text-color","anchor":"p-result"}],"connections":[{"to":"u-zero-sub","from":"u-zero","color":"sky","label":"substitute","style":"solid"},{"to":"p-result","from":"p-val","color":"amber","label":"solve","style":"solid"}]},"order":1,"title":"Substitute $u = 0$","description":"To solve for $P$, we choose a value of $u$ that eliminates the term containing $Q$. Setting $u = 0$ does this directly. This calculation earns the second mark (**A1**)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$2(\\htmlData{anchor=u-seven}{\\color{@purple}{7}}) + 1 = P(7 - \\htmlData{anchor=u-seven-sub}{\\color{@purple}{7}}) + Q(\\htmlData{anchor=u-seven-q}{\\color{@purple}{7}})$$"},{"latex":"$$$15 = 7\\htmlData{anchor=q-val}{\\color{@teal}{Q}}$$"},{"latex":"$$$\\htmlData{anchor=q-result}{\\color{@teal}{Q}} = \\frac{15}{7}$$"}],"highlights":[{"color":"purple","style":"box","anchor":"u-seven"},{"color":"teal","style":"text-color","anchor":"q-result"}],"connections":[{"to":"u-seven-sub","from":"u-seven","color":"purple","label":"substitute","style":"solid"},{"to":"q-result","from":"q-val","color":"teal","label":"solve","style":"solid"}]},"order":2,"title":"Substitute $u = 7$","description":"To solve for $Q$, we choose a value of $u$ that eliminates the term containing $P$. Setting $u = 7$ makes $(7-u) = 0$. This calculation earns the third mark (**A1**)."},{"type":"text","order":3,"title":"State final expression","content":"Now we substitute our values of $P = \\frac{1}{7}$ and $Q = \\frac{15}{7}$ back into the original partial fraction setup:\n\n$$\\frac{2u+1}{u(7-u)} = \\frac{1}{7u} + \\frac{15}{7(7-u)}$$"}]},{"id":"031fb07d-8392-4b31-b95d-4611e966e569:equating_coefficients","label":"(ci) Equating Coefficients","steps":[{"type":"text","order":0,"title":"Set up the identity","content":"To express the algebraic fraction in partial fractions, we begin by setting up an identity with two unknown constants, $A$ and $B$:\n\n$$\\frac{2u+1}{u(7-u)} \\equiv \\frac{A}{u} + \\frac{B}{7-u}$$\n\nMultiply both sides by the common denominator $u(7-u)$ to clear the fractions. This gives us the linear identity:\n\n$$2u + 1 \\equiv A(7 - u) + Bu$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$2u + 1 \\equiv \\htmlData{anchor=expand}{A(7 - u) + Bu}$"},{"latex":"$$2u + 1 \\equiv \\htmlData{anchor=collected}{(B - A)u + 7A}$"}],"highlights":[{"color":"sky","style":"box","anchor":"expand"},{"color":"sky","style":"box","anchor":"collected"}],"connections":[{"to":"collected","from":"expand","color":"sky","label":"expand & collect","style":"solid"}]},"order":1,"title":"Expand and collect terms","description":"Expand the right-hand side and collect terms with $u$ together so we can compare both sides of the identity."},{"type":"text","order":2,"title":"Equate the coefficients","content":"Now we equate the coefficients of corresponding terms on both sides of the identity:\n\n- **Constant terms:**\n $$7A = 1 \\implies A = \\frac{1}{7}$$\n\n- **$u$ terms:**\n $$B - A = 2$$"},{"type":"text","order":3,"title":"Solve for B","content":"Substitute our value of $A = \\frac{1}{7}$ into the second equation to find $B$:\n\n$$\\begin{aligned} B - \\frac{1}{7} &= 2 \\\\ B &= 2 + \\frac{1}{7} \\\\ B &= \\frac{15}{7} \\end{aligned}$$"},{"type":"text","order":4,"title":"Write final expression","content":"Substitute the values of $A = \\frac{1}{7}$ and $B = \\frac{15}{7}$ back into our partial fractions setup:\n\n$$\\frac{2u+1}{u(7-u)} = \\frac{1}{7u} + \\frac{15}{7(7-u)}$$"}]},{"id":"55830931-b474-4459-a267-e04ece8619e8:area_subtraction","label":"(cii) Subtracting Area under Curve from Rectangle","steps":[{"type":"text","order":0,"title":"Formulate the Area Strategy","content":"To find the exact area of the drainage channel $R$ (which lies between the water surface $y = 1$ and the curved bottom), we can subtract the area under the parametric curve from the area of the bounding rectangle.\n\nThe bounding rectangle is formed by the horizontal water surface line $y = 1$ and the endpoints of the curve:\n- At $u = 2$, $x = 2(2)^2 + 2(2) = 12$\n- At $u = 5$, $x = 2(5)^2 + 2(5) = 60$\n\nThe width of this rectangle is $60 - 12 = 48$, and its height is $1$. Therefore:\n\n$$\\text{Area of Rectangle} = 1 \\times 48 = 48$$\n\nTo find the area of $R$, we use:\n$$\\text{Area of } R = 48 - \\int_{12}^{60} y \\, dx$$"},{"type":"text","order":1,"title":"Set Up Parametric Integration","content":"We convert the integral $\\int y \\, dx$ into parametric form with respect to $u$ using the limits $u = 2$ to $u = 5$:\n\n$$\\int_{12}^{60} y \\, dx = \\int_{2}^{5} y \\frac{dx}{du} \\, du$$\n\nGiven $x = 2u^2 + 2u$:\n$$\\frac{dx}{du} = 4u + 2 = 2(2u + 1)$$\n\nSubstituting $y = \\dfrac{10}{u(7 - u)}$ and $\\frac{dx}{du}$ into the integral:\n\n$$\\int_{2}^{5} \\frac{10}{u(7-u)} \\cdot (4u + 2) \\, du = \\int_{2}^{5} \\frac{20(2u + 1)}{u(7 - u)} \\, du = 20 \\int_{2}^{5} \\frac{2u + 1}{u(7 - u)} \\, du$$"},{"type":"text","order":2,"title":"Decompose using Partial Fractions","content":"To integrate $\\dfrac{2u+1}{u(7-u)}$, we split it into partial fractions:\n\n$$\\frac{2u + 1}{u(7 - u)} \\equiv \\frac{P}{u} + \\frac{Q}{7 - u}$$\n\nMultiplying both sides by the denominator $u(7 - u)$ gives:\n$$2u + 1 \\equiv P(7 - u) + Qu$$\n\n- Let $u = 0$:\n$$1 = 7P \\implies P = \\frac{1}{7}$$\n\n- Let $u = 7$:\n$$15 = 7Q \\implies Q = \\frac{15}{7}$$\n\nSo, the integrand becomes:\n$$\\frac{2u + 1}{u(7 - u)} = \\frac{1/7}{u} + \\frac{15/7}{7-u}$$"},{"type":"text","order":3,"title":"Integrate the Expressions","content":"Now we integrate each term with respect to $u$:\n\n$$\\int \\left( \\frac{1/7}{u} + \\frac{15/7}{7-u} \\right) du = \\frac{1}{7} \\ln|u| - \\frac{15}{7} \\ln|7-u|$$\n\nNote that the negative sign in the second term arises from the chain rule applied to the linear denominator $(7-u)$, which earns a method mark."},{"type":"text","order":4,"title":"Evaluate the Limits","content":"We evaluate this antiderivative between the limits $u=2$ and $u=5$:\n\n$$\\left[ \\frac{1}{7} \\ln|u| - \\frac{15}{7} \\ln|7-u| \\right]_{2}^{5} = \\left( \\frac{1}{7} \\ln 5 - \\frac{15}{7} \\ln 2 \\right) - \\left( \\frac{1}{7} \\ln 2 - \\frac{15}{7} \\ln 5 \\right)$$\n\nGrouping the terms together:\n$$\\left( \\frac{1}{7} \\ln 5 + \\frac{15}{7} \\ln 5 \\right) - \\left( \\frac{15}{7} \\ln 2 + \\frac{1}{7} \\ln 2 \\right) = \\frac{16}{7} \\ln 5 - \\frac{16}{7} \\ln 2$$\n\nUsing the laws of logarithms:\n$$\\frac{16}{7} (\\ln 5 - \\ln 2) = \\frac{16}{7} \\ln\\left(\\frac{5}{2}\\right) = \\frac{16}{7} \\ln 2.5$$\n\nThis simplification step earns a method mark."},{"type":"text","order":5,"title":"Subtract from Rectangle Area","content":"Recall that the total area under the curve is scaled by the factor $k = 20$ from our parametric integral:\n\n$$\\text{Area under Curve} = 20 \\times \\left( \\frac{16}{7} \\ln 2.5 \\right) = \\frac{320}{7} \\ln 2.5$$\n\nNow, subtract this from our rectangle area ($48$) to find the exact area of $R$:\n\n$$\\text{Area of } R = 48 - \\frac{320}{7} \\ln 2.5$$\n\nThis is in the required form $A - B\\ln(2.5)$, where $A = 48$ and $B = \\frac{320}{7}$. This final result earns the accuracy mark."}]},{"id":"55830931-b474-4459-a267-e04ece8619e8:direct_integration","label":"(cii) Direct Integration of the Shaded Region","steps":[{"type":"text","order":0,"title":"Set up the integral","content":"The shaded region $R$ is bounded above by the line $y = 1$ (the water surface) and below by the parametric curve. The limits of integration for $u$ correspond to where the curve meets $y=1$, which gives $u = 2$ and $u = 5$.\n\nThe area of the shaded region $R$ is given by the integral of the top boundary minus the bottom boundary:\n$$\\text{Area} = \\int_{x(2)}^{x(5)} (1 - y) \\, dx$$\n\nTo integrate with respect to $u$, we use parametric integration:\n$$dx = \\frac{dx}{du} \\, du$$\n\nGiven $x = 2u^2 + 2u$, we differentiate to find:\n$$\\frac{dx}{du} = 4u + 2$$\n\nSubstituting this and the expression for $y$ into the integral, we get:\n$$\\text{Area} = \\int_{2}^{5} \\left(1 - \\frac{10}{u(7 - u)}\\right)(4u + 2) \\, du$$"},{"type":"text","order":1,"title":"Split into two integrals","content":"To evaluate this, we expand the integrand and split it into two simpler integrals:\n$$\\begin{aligned} \\text{Area} &= \\int_{2}^{5} (4u + 2) \\, du - \\int_{2}^{5} \\frac{10(4u + 2)}{u(7 - u)} \\, du \\\\ &= I_1 - I_2 \\end{aligned}$$\n\nLet's evaluate the first integral, $I_1$, directly:\n$$\\begin{aligned} I_1 &= \\int_{2}^{5} (4u + 2) \\, du \\\\ &= \\left[ 2u^2 + 2u \\right]_{2}^{5} \\\\ &= \\left( 2(5)^2 + 2(5) \\right) - \\left( 2(2)^2 + 2(2) \\right) \\\\ &= 60 - 12 \\\\ &= 48 \\end{aligned}$$"},{"type":"text","order":2,"title":"Simplify the second integrand","content":"Now let's simplify the integrand of $I_2$:\n$$I_2 = \\int_{2}^{5} \\frac{10(4u + 2)}{u(7 - u)} \\, du = \\int_{2}^{5} \\frac{40u + 20}{u(7 - u)} \\, du$$\n\nWe can decompose the integrand using partial fractions:\n$$\\frac{40u + 20}{u(7 - u)} = \\frac{P}{u} + \\frac{Q}{7 - u}$$\n\nMultiplying both sides by the denominator $u(7 - u)$ gives:\n$$40u + 20 = P(7 - u) + Qu$$\n\n- Setting $u = 0$:\n $$20 = 7P \\implies P = \\frac{20}{7}$$\n- Setting $u = 7$:\n $$40(7) + 20 = 7Q \\implies 300 = 7Q \\implies Q = \\frac{300}{7}$$\n\nSo the expression becomes:\n$$\\frac{40u + 20}{u(7 - u)} = \\frac{20}{7u} + \\frac{300}{7(7 - u)}$$\n\nWe can factor out a constant of $20$ to align with the markscheme:\n$$\\frac{40u + 20}{u(7 - u)} = 20 \\left( \\frac{1/7}{u} + \\frac{15/7}{7-u} \\right)$$"},{"type":"text","order":3,"title":"Integrate the partial fractions","content":"Now we integrate each term with respect to $u$. Remember to apply the chain rule for the linear denominator $(7-u)$, which introduces a negative sign:\n$$\\int \\left( \\frac{1/7}{u} + \\frac{15/7}{7-u} \\right) du = \\frac{1}{7} \\ln|u| - \\frac{15}{7} \\ln|7-u|$$"},{"type":"text","order":4,"title":"Evaluate the second integral","content":"Next, we evaluate this antiderivative at the limits $u=5$ and $u=2$:\n$$\\begin{aligned} I_2 &= 20 \\left[ \\frac{1}{7} \\ln|u| - \\frac{15}{7} \\ln|7-u| \\right]_{2}^{5} \\\\ &= 20 \\left[ \\left( \\frac{1}{7} \\ln 5 - \\frac{15}{7} \\ln 2 \\right) - \\left( \\frac{1}{7} \\ln 2 - \\frac{15}{7} \\ln 5 \\right) \\right] \\\\ &= 20 \\left[ \\left( \\frac{1}{7} \\ln 5 + \\frac{15}{7} \\ln 5 \\right) - \\left( \\frac{1}{7} \\ln 2 + \\frac{15}{7} \\ln 2 \\right) \\right] \\\\ &= 20 \\left[ \\frac{16}{7} \\ln 5 - \\frac{16}{7} \\ln 2 \\right] \\\\ &= 20 \\left( \\frac{16}{7} \\ln\\left(\\frac{5}{2}\\right) \\right) \\\\ &= \\frac{320}{7} \\ln(2.5) \\end{aligned}$$"},{"type":"text","order":5,"title":"Subtract to find final area","content":"Finally, we combine our results from $I_1$ and $I_2$ to find the exact area:\n$$\\text{Area} = I_1 - I_2 = 48 - \\frac{320}{7} \\ln(2.5)$$\n\nThis is in the required form $A - B \\ln(2.5)$, where:\n$$A = 48, \\quad B = \\frac{320}{7}$$"}]}],"generatedAt":"2026-06-09T03:48:33.446Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null}],"subjectCode":"a_level_maths__wjec","summaries":[{"id":"019f9d33-ebbe-48df-b8a8-8d326a253ef1","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":false},{"id":"19b1d184-7b2e-4eab-8e8e-e0f619c14e25","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"91ca954b-1bdb-47fa-96fa-5f9ac25fb5f7","difficultyBand":"easy","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"9477a240-9ae6-4f66-bab4-556745c10e10","difficultyBand":"easy","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"af51146c-eb8e-481b-b344-e5fc3a1c1086","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"8826c5f2-c30f-49dd-b9a4-e3829062d186","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":false},{"id":"33eaf0c2-c8d6-4e1c-bb48-6c1ef08aac6d","difficultyBand":"easy","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"3831be23-9d8f-4c4f-b16f-263d62d79627","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"e7420bc2-5fd8-4afa-b008-b3f8d851a4bd","difficultyBand":"easy","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"770ea004-466a-49e0-b684-0442a91a6bfa","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":false},{"id":"79c30fa9-a5dd-42d9-abf9-eab78bc78177","difficultyBand":"easy","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"dc6d338f-b9f1-44d0-8427-04c5102e74d8","difficultyBand":"easy","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"51b90478-fd26-4384-8c46-fd55f1d16431","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"c954092c-c7bb-45f7-ac1a-d8d588662c85","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":false},{"id":"c7f0f4b9-941f-42fb-9929-432471f4a385","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"9b97554a-477a-4c87-8682-27bb8cd580fe","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"28c70014-d099-484b-8f5b-15c7ece2e133","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"1bae46b7-1eb5-4619-b72a-d5ef21a020d2","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":false},{"id":"15300de1-c9d6-4a6d-a178-8f9d8d9037ac","difficultyBand":"easy","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"5803523f-befb-4046-9229-885b54335540","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"1cf8d963-e234-4012-a2c2-83bfe85b27a2","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"6f30dcc9-3366-423b-9143-62632352c99e","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"a6bf2db0-a5bb-4e3f-bf62-e725cce44414","difficultyBand":"easy","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"909617f3-b963-4729-8ed3-97986871ef85","difficultyBand":"easy","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"f582775f-85c9-460f-9698-6998f64bc1e1","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":false},{"id":"e3a53aed-9072-40c0-9758-062c1493d764","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"24300108-9cd9-4576-bda9-8f7160cb0d03","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"64faf4d7-8ca9-4988-a80d-f641b25ef87a","difficultyBand":"medium","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"5827f271-bad7-474d-9d07-65d9a8fa3c67","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":false},{"id":"1009e034-bf8c-41fb-a3bb-f1bb1d489e4e","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"32fb424e-19aa-4633-b8a6-f129ef453446","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"0378f4f8-c73b-432c-a8ff-baec9957f572","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"b8373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1.8 Integration

Question 2

1.8 Integration Questions

1.8 Integration

Question 2

1.8 Integration Questions