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Integrate kx^n A Level Maths exam-style questions with worked solutions on MathsGenie.\",\"learningResourceType\":\"Practice questions\",\"educationalLevel\":\"A Level\",\"about\":{\"@type\":\"Thing\",\"name\":\"1.9.20 Integrate kx^n\"},\"url\":\"https://mathsgenie.co.uk/a-level/maths/ocr/mei/integrate-kx-n/questions/q55\",\"inLanguage\":\"en-GB\",\"provider\":{\"@id\":\"https://mathsgenie.co.uk/#organization\"}}"}}],["$","div",null,{"className":"min-h-0 w-full flex-1 overflow-hidden bg-background","children":["$","$L39",null,{"backHref":"/a-level/maths/ocr/mei/questions","basePath":"/a-level/maths/ocr/mei/integrate-kx-n/questions","formulaUrl":null,"initialIndex":54,"initialQuestionWindow":[{"id":"92b0a1af-00ff-4561-9a48-961c1db57937","revisionId":"fc77325d-117a-448e-b05b-64ca2171b289","questionNumber":null,"questionType":"LA","specification":"A chemical reaction produces a foam whose volume $V$ (in $\\text{cm}^3$) changes at a rate $V'(t)$ where $t$ is the time in seconds since the start of observation, for $t > 0$.\n\nGiven that\n- $V'(t) = \\dfrac{(t - 9)^2}{t\\sqrt{t}}$\n- the observation point $P(4, 10)$ lies on the curve $V = g(t)$","difficultyBand":"medium","tier":null,"totalMarks":9,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"13fb003b-7e32-4f57-8d67-3e1b6206fa4a","label":"a","content":"(i) Find the value of the gradient of the curve at $P$.\n\n(ii) Hence find the equation of the tangent to the curve at $P$, giving your answer in the form $at + bV + c = 0$, where $a$, $b$ and $c$ are integers to be found.","metadata":null,"markscheme":"- Substitutes $t = 4$ into $V'(t) = \\frac{(4 - 9)^2}{4\\sqrt{4}}$ M1\n- Obtains gradient $m = \\frac{25}{8}$ or $3.125$ A1\n- Attempts to find tangent using $V - 10 = \\frac{25}{8}(t - 4)$ or $V = mt + c$ M1\n- Rearranges to $25t - 8V - 20 = 0$ (or any integer multiple) A1","marks":4,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"point_slope","label":"Point-Slope Formula","steps":[{"type":"text","order":0,"title":"Evaluate the gradient","content":"The gradient of the curve $V = g(t)$ at any point is given by the derivative $V'(t)$. \n\nTo find the gradient $m$ of the curve at $P(4, 10)$, we substitute $t = 4$ into the expression for $V'(t)$:\n\n$$V'(4) = \\frac{(4 - 9)^2}{4\\sqrt{4}}$$\n\nCalculating the numerator:\n$$(4 - 9)^2 = (-5)^2 = 25$$\n\nCalculating the denominator:\n$$4\\sqrt{4} = 4 \\times 2 = 8$$\n\nThus, the gradient of the curve at $P$ is:\n\n$$m = \\frac{25}{8} \\quad \\text{or} \\quad 3.125$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\text{Given: } P(\\htmlData{anchor=t1-val}{\\color{@sky}{4}}, \\; \\htmlData{anchor=V1-val}{\\color{@pink}{10}}) \\quad \\text{and} \\quad m = \\htmlData{anchor=m-val}{\\color{@amber}{\\frac{25}{8}}}$$"},{"latex":"$$$V - \\htmlData{anchor=V1-sub}{\\color{@pink}{V_1}} = \\htmlData{anchor=m-sub}{\\color{@amber}{m}}(t - \\htmlData{anchor=t1-sub}{\\color{@sky}{t_1}})$$"},{"latex":"$$$V - \\color{@pink}{10} = \\color{@amber}{\\frac{25}{8}}(t - \\color{@sky}{4})$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"t1-val"},{"color":"pink","style":"text-color","anchor":"V1-val"},{"color":"amber","style":"text-color","anchor":"m-val"}],"connections":[{"to":"t1-sub","from":"t1-val","color":"sky","label":"t-coordinate","style":"solid"},{"to":"V1-sub","from":"V1-val","color":"pink","label":"V-coordinate","style":"solid"},{"to":"m-sub","from":"m-val","color":"amber","label":"gradient","style":"solid"}]},"order":1,"title":"Use point-slope formula","description":"We can write the equation of the tangent line using the point-slope form: $V - V_1 = m(t - t_1)$, where $(t_1, V_1) = (4, 10)$ and the gradient is $m = \\frac{25}{8}$."},{"type":"text","order":2,"title":"Rearrange to general form","content":"Now we rearrange the equation into the general linear form $at + bV + c = 0$ with integer coefficients.\n\nFirst, multiply both sides of the equation by $8$ to eliminate the fraction:\n\n$$8(V - 10) = 25(t - 4)$$\n\nExpand both sides:\n\n$$8V - 80 = 25t - 100$$\n\nMove all terms to one side to get the form $at + bV + c = 0$:\n\n$$25t - 8V - 100 + 80 = 0$$\n\n$$25t - 8V - 20 = 0$$\n\nThus, the equation of the tangent line is:\n\n$$25t - 8V - 20 = 0$$"}]},{"id":"slope_intercept","label":"Slope-Intercept Formula","steps":[{"type":"text","order":0,"title":"Evaluate the gradient at $P$","content":"The gradient of the curve at any point is given by its derivative, $V'(t)$. To find the gradient of the curve at the point $P(4, 10)$, we need to evaluate $V'(t)$ at $t = 4$.\n\nSubstitute $t = 4$ into the given rate equation:\n\n$$V'(4) = \\frac{(4 - 9)^2}{4\\sqrt{4}}$$\n\nCalculating the numerator:\n$$(4 - 9)^2 = (-5)^2 = 25$$\n\nCalculating the denominator:\n$$4\\sqrt{4} = 4 \\times 2 = 8$$\n\nTherefore, the gradient of the curve at $P$ is:\n\n$$m = \\frac{25}{8}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$V = \\htmlData{anchor=grad}{\\color{@sky}{m}}t + \\htmlData{anchor=intercept}{\\color{@amber}{c}}$$"},{"latex":"$$$10 = \\left(\\htmlData{anchor=grad-val}{\\color{@sky}{\\frac{25}{8}}}\\right)(4) + \\color{@amber}{c}$$"}],"highlights":[{"color":"sky","style":"box","anchor":"grad-val"},{"color":"amber","style":"text-color","anchor":"intercept"}],"connections":[{"to":"grad","from":"grad-val","color":"sky","label":"substitute gradient","style":"solid"}]},"order":1,"title":"Set up slope-intercept form","description":"We use the slope-intercept equation of a straight line, $V = mt + c$. By substituting our gradient $m = \\frac{25}{8}$ and the coordinates of the point $P(4, 10)$ (where $t = 4$ and $V = 10$), we can solve for the constant $c$."},{"type":"text","order":2,"title":"Solve for constant $c$","content":"Multiply the terms on the right side:\n\n$$10 = \\frac{25}{2} + c$$\n\nSubtract $\\frac{25}{2}$ (which is $12.5$) from both sides to find $c$:\n\n$$c = 10 - 12.5 = -2.5 = -\\frac{5}{2}$$\n\nNow, write down the full equation in slope-intercept form:\n\n$$V = \\frac{25}{8}t - \\frac{5}{2}$$"},{"type":"text","order":3,"title":"Convert to general form","content":"To write the equation in the requested form $at + bV + c = 0$ with integer coefficients, we multiply the entire equation by $8$ to clear the fractions:\n\n$$8V = 25t - 20$$\n\nRearranging all terms to one side gives:\n\n$$25t - 8V - 20 = 0$$\n\nThus, the integer constants are $a = 25$, $b = -8$, and $c = -20$."}]}],"generatedAt":"2026-06-09T07:41:48.843Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"1b0e7fe1-8390-445f-8bc6-d96e7deb6e74","label":"b","content":"Find $g(t)$, simplifying your answer.","metadata":null,"markscheme":"- Expands the derivative to $V'(t) = \\frac{t^2 - 18t + 81}{t^{1.5}} = t^{\\frac{1}{2}} - 18t^{-\\frac{1}{2}} + 81t^{-\\frac{3}{2}}$ M1\n- Integrates term-by-term, increasing the power of at least two terms by 1 M1\n- Obtains correct integrated terms: $g(t) = \\frac{2}{3}t^{\\frac{3}{2}} - 36t^{\\frac{1}{2}} - 162t^{-\\frac{1}{2}} + C$ A1\n- Substitutes $t = 4, V = 10$ into their expression: $10 = \\frac{2}{3}(4)^{\\frac{3}{2}} - 36(4)^{\\frac{1}{2}} - 162(4)^{-\\frac{1}{2}} + C$ M1\n- Finds $C = \\frac{473}{3}$ and states $g(t) = \\frac{2}{3}t^{\\frac{3}{2}} - 36t^{\\frac{1}{2}} - 162t^{-\\frac{1}{2}} + \\frac{473}{3}$ A1","marks":5,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"term_by_term","label":"Indefinite Integration by Expansion","steps":[{"type":"text","order":0,"title":"Expand the numerator","content":"To find $g(t)$, we integrate the rate of change $V'(t) = \\dfrac{(t - 9)^2}{t\\sqrt{t}}$. First, we expand the numerator: $(t - 9)^2 = t^2 - 18t + 81$."},{"type":"text","order":1,"title":"Express as individual powers","content":"We write the denominator as $t\\sqrt{t} = t^{\\frac{3}{2}}$. Dividing each term in the numerator by $t^{\\frac{3}{2}}$, we get: $V'(t) = \\dfrac{t^2 - 18t + 81}{t^{\\frac{3}{2}}} = t^{\\frac{1}{2}} - 18t^{-\\frac{1}{2}} + 81t^{-\\frac{3}{2}}$."},{"type":"text","order":2,"title":"Integrate term-by-term","content":"Integrate $V'(t)$ with respect to $t$ using the power rule:\n\n$$g(t) = \\int \\left( t^{\\frac{1}{2}} - 18t^{-\\frac{1}{2}} + 81t^{-\\frac{3}{2}} \\right) dt = \\frac{2}{3}t^{\\frac{3}{2}} - 36t^{\\frac{1}{2}} - 162t^{-\\frac{1}{2}} + C$$"},{"type":"text","order":3,"title":"Substitute the given point","content":"Using the observation point $P(4, 10)$, we substitute $t = 4$ and $g(4) = 10$ into the expression:\n\n$$10 = \\frac{2}{3}(4)^{\\frac{3}{2}} - 36(4)^{\\frac{1}{2}} - 162(4)^{-\\frac{1}{2}} + C$$"},{"type":"text","order":4,"title":"Solve for C","content":"Evaluating the terms gives $\\frac{2}{3}(8) - 36(2) - 162\\left(\\frac{1}{2}\\right) = \\frac{16}{3} - 72 - 81 = -\\frac{443}{3}$. Setting this equal to $10 - C$, we find $C = 10 + \\frac{443}{3} = \\frac{473}{3}$."},{"type":"text","order":5,"title":"State the final function","content":"Substituting $C$ back gives the final equation:\n\n$$g(t) = \\frac{2}{3}t^{\\frac{3}{2}} - 36t^{\\frac{1}{2}} - 162t^{-\\frac{1}{2}} + \\frac{473}{3}$$"}]},{"id":"definite_integral","label":"Definite Integration with Variable Limit","steps":[{"type":"text","order":0,"title":"Set up the integral","content":"By the Fundamental Theorem of Calculus, we write $g(t)$ directly using the initial condition at $P(4, 10)$:\n\n$$g(t) = 10 + \\int_{4}^{t} \\frac{(x - 9)^2}{x\\sqrt{x}} \\, dx$$"},{"type":"text","order":1,"title":"Simplify the integrand","content":"Expand and divide the integrand by $x^{\\frac{3}{2}}$:\n\n$$V'(x) = \\frac{x^2 - 18x + 81}{x^{\\frac{3}{2}}} = x^{\\frac{1}{2}} - 18x^{-\\frac{1}{2}} + 81x^{-\\frac{3}{2}}$$"},{"type":"text","order":2,"title":"Integrate term-by-term","content":"Apply the power rule to find the antiderivative:\n\n$$\\int (x^{\\frac{1}{2}} - 18x^{-\\frac{1}{2}} + 81x^{-\\frac{3}{2}}) \\, dx = \\left[ \\frac{2}{3}x^{\\frac{3}{2}} - 36x^{\\frac{1}{2}} - 162x^{-\\frac{1}{2}} \\right]_{4}^{t}$$"},{"type":"text","order":3,"title":"Evaluate at the boundaries","content":"Evaluate at the boundaries $t$ and $4$. The value at the lower limit is:\n\n$$\\frac{2}{3}(4)^{\\frac{3}{2}} - 36(4)^{\\frac{1}{2}} - 162(4)^{-\\frac{1}{2}} = \\frac{16}{3} - 72 - 81 = -\\frac{443}{3}$$"},{"type":"text","order":4,"title":"State the final expression","content":"Combine this with the initial constant $10$:\n\n$$g(t) = 10 + \\left( \\frac{2}{3}t^{\\frac{3}{2}} - 36t^{\\frac{1}{2}} - 162t^{-\\frac{1}{2}} \\right) - \\left( -\\frac{443}{3} \\right) = \\frac{2}{3}t^{\\frac{3}{2}} - 36t^{\\frac{1}{2}} - 162t^{-\\frac{1}{2}} + \\frac{473}{3}$$"}]},{"id":"substitution","label":"Integration by Substitution","steps":[{"type":"text","order":0,"title":"Set up substitution","content":"We evaluate $g(t) = \\int \\frac{(t - 9)^2}{t\\sqrt{t}} \\, dt$ by letting $u = \\sqrt{t}$, which gives $t = u^2$ and $dt = 2u \\, du$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$t = \\htmlData{anchor=t-val}{\\color{@sky}{u^2}}, \\quad \\sqrt{t} = \\htmlData{anchor=sqrt-val}{\\color{@pink}{u}}, \\quad dt = \\htmlData{anchor=dt-val}{\\color{@amber}{2u \\, du}}$"},{"latex":"$$\\int \\frac{(\\htmlData{anchor=t-sub}{\\color{@sky}{t}} - 9)^2}{\\htmlData{anchor=t-sub2}{\\color{@sky}{t}}\\htmlData{anchor=sqrt-sub}{\\color{@pink}{\\sqrt{t}}}} \\, \\htmlData{anchor=dt-sub}{\\color{@amber}{dt}} = \\int \\frac{(\\color{@sky}{u^2} - 9)^2}{\\color{@sky}{u^2} \\cdot \\color{@pink}{u}} (\\color{@amber}{2u \\, du})$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"t-val"},{"color":"pink","style":"text-color","anchor":"sqrt-val"},{"color":"amber","style":"text-color","anchor":"dt-val"}],"connections":[{"to":"t-sub","from":"t-val","color":"sky","label":"substitute t","style":"solid"},{"to":"sqrt-sub","from":"sqrt-val","color":"pink","label":"substitute root","style":"solid"},{"to":"dt-sub","from":"dt-val","color":"amber","label":"substitute dt","style":"solid"}]},"order":1,"title":"Apply substitution","description":"Substitute $t = u^2$, $\\sqrt{t} = u$, and $dt = 2u \\, du$ into the integral."},{"type":"text","order":2,"title":"Simplify integrand","content":"Simplify the integrand by canceling $u$:\n\n$$\\int \\frac{2(u^2 - 9)^2}{u^2} \\, du = 2 \\int \\frac{u^4 - 18u^2 + 81}{u^2} \\, du = 2 \\int \\left(u^2 - 18 + 81u^{-2}\\right) \\, du$$"},{"type":"text","order":3,"title":"Integrate in terms of u","content":"Using the power rule gives:\n\n$$2 \\left[ \\frac{u^3}{3} - 18u - \\frac{81}{u} \\right] + C = \\frac{2}{3}u^3 - 36u - \\frac{162}{u} + C$$"},{"type":"text","order":4,"title":"Substitute back to t","content":"Substituting $u = t^{\\frac{1}{2}}$ back into the expression yields:\n\n$$g(t) = \\frac{2}{3}t^{\\frac{3}{2}} - 36t^{\\frac{1}{2}} - 162t^{-\\frac{1}{2}} + C$$"},{"type":"text","order":5,"title":"Find the constant C","content":"Since $g(4) = 10$, we have: $10 = \\frac{2}{3}(8) - 36(2) - 162\\left(\\frac{1}{2}\\right) + C = -\\frac{443}{3} + C$, which gives $C = \\frac{473}{3}$. Thus:\n\n$$g(t) = \\frac{2}{3}t^{\\frac{3}{2}} - 36t^{\\frac{1}{2}} - 162t^{-\\frac{1}{2}} + \\frac{473}{3}$$"}]}],"generatedAt":"2026-06-09T07:44:21.820Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"13fb003b-7e32-4f57-8d67-3e1b6206fa4a:point_slope","label":"(a) Point-Slope Formula","steps":[{"type":"text","order":0,"title":"Evaluate the gradient","content":"The gradient of the curve $V = g(t)$ at any point is given by the derivative $V'(t)$. \n\nTo find the gradient $m$ of the curve at $P(4, 10)$, we substitute $t = 4$ into the expression for $V'(t)$:\n\n$$V'(4) = \\frac{(4 - 9)^2}{4\\sqrt{4}}$$\n\nCalculating the numerator:\n$$(4 - 9)^2 = (-5)^2 = 25$$\n\nCalculating the denominator:\n$$4\\sqrt{4} = 4 \\times 2 = 8$$\n\nThus, the gradient of the curve at $P$ is:\n\n$$m = \\frac{25}{8} \\quad \\text{or} \\quad 3.125$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\text{Given: } P(\\htmlData{anchor=t1-val}{\\color{@sky}{4}}, \\; \\htmlData{anchor=V1-val}{\\color{@pink}{10}}) \\quad \\text{and} \\quad m = \\htmlData{anchor=m-val}{\\color{@amber}{\\frac{25}{8}}}$$"},{"latex":"$$$V - \\htmlData{anchor=V1-sub}{\\color{@pink}{V_1}} = \\htmlData{anchor=m-sub}{\\color{@amber}{m}}(t - \\htmlData{anchor=t1-sub}{\\color{@sky}{t_1}})$$"},{"latex":"$$$V - \\color{@pink}{10} = \\color{@amber}{\\frac{25}{8}}(t - \\color{@sky}{4})$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"t1-val"},{"color":"pink","style":"text-color","anchor":"V1-val"},{"color":"amber","style":"text-color","anchor":"m-val"}],"connections":[{"to":"t1-sub","from":"t1-val","color":"sky","label":"t-coordinate","style":"solid"},{"to":"V1-sub","from":"V1-val","color":"pink","label":"V-coordinate","style":"solid"},{"to":"m-sub","from":"m-val","color":"amber","label":"gradient","style":"solid"}]},"order":1,"title":"Use point-slope formula","description":"We can write the equation of the tangent line using the point-slope form: $V - V_1 = m(t - t_1)$, where $(t_1, V_1) = (4, 10)$ and the gradient is $m = \\frac{25}{8}$."},{"type":"text","order":2,"title":"Rearrange to general form","content":"Now we rearrange the equation into the general linear form $at + bV + c = 0$ with integer coefficients.\n\nFirst, multiply both sides of the equation by $8$ to eliminate the fraction:\n\n$$8(V - 10) = 25(t - 4)$$\n\nExpand both sides:\n\n$$8V - 80 = 25t - 100$$\n\nMove all terms to one side to get the form $at + bV + c = 0$:\n\n$$25t - 8V - 100 + 80 = 0$$\n\n$$25t - 8V - 20 = 0$$\n\nThus, the equation of the tangent line is:\n\n$$25t - 8V - 20 = 0$$"}]},{"id":"13fb003b-7e32-4f57-8d67-3e1b6206fa4a:slope_intercept","label":"(a) Slope-Intercept Formula","steps":[{"type":"text","order":0,"title":"Evaluate the gradient at $P$","content":"The gradient of the curve at any point is given by its derivative, $V'(t)$. To find the gradient of the curve at the point $P(4, 10)$, we need to evaluate $V'(t)$ at $t = 4$.\n\nSubstitute $t = 4$ into the given rate equation:\n\n$$V'(4) = \\frac{(4 - 9)^2}{4\\sqrt{4}}$$\n\nCalculating the numerator:\n$$(4 - 9)^2 = (-5)^2 = 25$$\n\nCalculating the denominator:\n$$4\\sqrt{4} = 4 \\times 2 = 8$$\n\nTherefore, the gradient of the curve at $P$ is:\n\n$$m = \\frac{25}{8}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$V = \\htmlData{anchor=grad}{\\color{@sky}{m}}t + \\htmlData{anchor=intercept}{\\color{@amber}{c}}$$"},{"latex":"$$$10 = \\left(\\htmlData{anchor=grad-val}{\\color{@sky}{\\frac{25}{8}}}\\right)(4) + \\color{@amber}{c}$$"}],"highlights":[{"color":"sky","style":"box","anchor":"grad-val"},{"color":"amber","style":"text-color","anchor":"intercept"}],"connections":[{"to":"grad","from":"grad-val","color":"sky","label":"substitute gradient","style":"solid"}]},"order":1,"title":"Set up slope-intercept form","description":"We use the slope-intercept equation of a straight line, $V = mt + c$. By substituting our gradient $m = \\frac{25}{8}$ and the coordinates of the point $P(4, 10)$ (where $t = 4$ and $V = 10$), we can solve for the constant $c$."},{"type":"text","order":2,"title":"Solve for constant $c$","content":"Multiply the terms on the right side:\n\n$$10 = \\frac{25}{2} + c$$\n\nSubtract $\\frac{25}{2}$ (which is $12.5$) from both sides to find $c$:\n\n$$c = 10 - 12.5 = -2.5 = -\\frac{5}{2}$$\n\nNow, write down the full equation in slope-intercept form:\n\n$$V = \\frac{25}{8}t - \\frac{5}{2}$$"},{"type":"text","order":3,"title":"Convert to general form","content":"To write the equation in the requested form $at + bV + c = 0$ with integer coefficients, we multiply the entire equation by $8$ to clear the fractions:\n\n$$8V = 25t - 20$$\n\nRearranging all terms to one side gives:\n\n$$25t - 8V - 20 = 0$$\n\nThus, the integer constants are $a = 25$, $b = -8$, and $c = -20$."}]},{"id":"1b0e7fe1-8390-445f-8bc6-d96e7deb6e74:term_by_term","label":"(b) Indefinite Integration by Expansion","steps":[{"type":"text","order":0,"title":"Expand the numerator","content":"To find $g(t)$, we integrate the rate of change $V'(t) = \\dfrac{(t - 9)^2}{t\\sqrt{t}}$. First, we expand the numerator: $(t - 9)^2 = t^2 - 18t + 81$."},{"type":"text","order":1,"title":"Express as individual powers","content":"We write the denominator as $t\\sqrt{t} = t^{\\frac{3}{2}}$. Dividing each term in the numerator by $t^{\\frac{3}{2}}$, we get: $V'(t) = \\dfrac{t^2 - 18t + 81}{t^{\\frac{3}{2}}} = t^{\\frac{1}{2}} - 18t^{-\\frac{1}{2}} + 81t^{-\\frac{3}{2}}$."},{"type":"text","order":2,"title":"Integrate term-by-term","content":"Integrate $V'(t)$ with respect to $t$ using the power rule:\n\n$$g(t) = \\int \\left( t^{\\frac{1}{2}} - 18t^{-\\frac{1}{2}} + 81t^{-\\frac{3}{2}} \\right) dt = \\frac{2}{3}t^{\\frac{3}{2}} - 36t^{\\frac{1}{2}} - 162t^{-\\frac{1}{2}} + C$$"},{"type":"text","order":3,"title":"Substitute the given point","content":"Using the observation point $P(4, 10)$, we substitute $t = 4$ and $g(4) = 10$ into the expression:\n\n$$10 = \\frac{2}{3}(4)^{\\frac{3}{2}} - 36(4)^{\\frac{1}{2}} - 162(4)^{-\\frac{1}{2}} + C$$"},{"type":"text","order":4,"title":"Solve for C","content":"Evaluating the terms gives $\\frac{2}{3}(8) - 36(2) - 162\\left(\\frac{1}{2}\\right) = \\frac{16}{3} - 72 - 81 = -\\frac{443}{3}$. Setting this equal to $10 - C$, we find $C = 10 + \\frac{443}{3} = \\frac{473}{3}$."},{"type":"text","order":5,"title":"State the final function","content":"Substituting $C$ back gives the final equation:\n\n$$g(t) = \\frac{2}{3}t^{\\frac{3}{2}} - 36t^{\\frac{1}{2}} - 162t^{-\\frac{1}{2}} + \\frac{473}{3}$$"}]},{"id":"1b0e7fe1-8390-445f-8bc6-d96e7deb6e74:definite_integral","label":"(b) Definite Integration with Variable Limit","steps":[{"type":"text","order":0,"title":"Set up the integral","content":"By the Fundamental Theorem of Calculus, we write $g(t)$ directly using the initial condition at $P(4, 10)$:\n\n$$g(t) = 10 + \\int_{4}^{t} \\frac{(x - 9)^2}{x\\sqrt{x}} \\, dx$$"},{"type":"text","order":1,"title":"Simplify the integrand","content":"Expand and divide the integrand by $x^{\\frac{3}{2}}$:\n\n$$V'(x) = \\frac{x^2 - 18x + 81}{x^{\\frac{3}{2}}} = x^{\\frac{1}{2}} - 18x^{-\\frac{1}{2}} + 81x^{-\\frac{3}{2}}$$"},{"type":"text","order":2,"title":"Integrate term-by-term","content":"Apply the power rule to find the definite integral:\n\n$$\\int_{4}^{t} (x^{\\frac{1}{2}} - 18x^{-\\frac{1}{2}} + 81x^{-\\frac{3}{2}}) \\, dx = \\left[ \\frac{2}{3}x^{\\frac{3}{2}} - 36x^{\\frac{1}{2}} - 162x^{-\\frac{1}{2}} \\right]_{4}^{t}$$"},{"type":"text","order":3,"title":"Evaluate at the boundaries","content":"Evaluate at the boundaries $t$ and $4$. The value at the lower limit is:\n\n$$\\frac{2}{3}(4)^{\\frac{3}{2}} - 36(4)^{\\frac{1}{2}} - 162(4)^{-\\frac{1}{2}} = \\frac{16}{3} - 72 - 81 = -\\frac{443}{3}$$"},{"type":"text","order":4,"title":"State the final expression","content":"Combine this with the initial constant $10$:\n\n$$g(t) = 10 + \\left( \\frac{2}{3}t^{\\frac{3}{2}} - 36t^{\\frac{1}{2}} - 162t^{-\\frac{1}{2}} \\right) - \\left( -\\frac{443}{3} \\right) = \\frac{2}{3}t^{\\frac{3}{2}} - 36t^{\\frac{1}{2}} - 162t^{-\\frac{1}{2}} + \\frac{473}{3}$$"}]},{"id":"1b0e7fe1-8390-445f-8bc6-d96e7deb6e74:substitution","label":"(b) Integration by Substitution","steps":[{"type":"text","order":0,"title":"Set up substitution","content":"We evaluate $g(t) = \\int \\frac{(t - 9)^2}{t\\sqrt{t}} \\, dt$ by letting $u = \\sqrt{t}$, which gives $t = u^2$ and $dt = 2u \\, du$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$t = \\htmlData{anchor=t-val}{\\color{@sky}{u^2}}, \\quad \\sqrt{t} = \\htmlData{anchor=sqrt-val}{\\color{@pink}{u}}, \\quad dt = \\htmlData{anchor=dt-val}{\\color{@amber}{2u \\, du}}$"},{"latex":"$$\\int \\frac{(\\htmlData{anchor=t-sub}{\\color{@sky}{t}} - 9)^2}{\\htmlData{anchor=t-sub2}{\\color{@sky}{t}}\\htmlData{anchor=sqrt-sub}{\\color{@pink}{\\sqrt{t}}}} \\, \\htmlData{anchor=dt-sub}{\\color{@amber}{dt}} = \\int \\frac{(\\color{@sky}{u^2} - 9)^2}{\\color{@sky}{u^2} \\cdot \\color{@pink}{u}} (\\color{@amber}{2u \\, du})$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"t-val"},{"color":"pink","style":"text-color","anchor":"sqrt-val"},{"color":"amber","style":"text-color","anchor":"dt-val"}],"connections":[{"to":"t-sub","from":"t-val","color":"sky","label":"substitute t","style":"solid"},{"to":"sqrt-sub","from":"sqrt-val","color":"pink","label":"substitute root","style":"solid"},{"to":"dt-sub","from":"dt-val","color":"amber","label":"substitute dt","style":"solid"}]},"order":1,"title":"Apply substitution","description":"Substitute $t = u^2$, $\\sqrt{t} = u$, and $dt = 2u \\, du$ into the integral."},{"type":"text","order":2,"title":"Simplify integrand","content":"Simplify the integrand by canceling $u$:\n\n$$\\int \\frac{2(u^2 - 9)^2}{u^2} \\, du = 2 \\int \\frac{u^4 - 18u^2 + 81}{u^2} \\, du = 2 \\int \\left(u^2 - 18 + 81u^{-2}\\right) \\, du$$"},{"type":"text","order":3,"title":"Integrate in terms of u","content":"Using the power rule gives:\n\n$$2 \\left[ \\frac{u^3}{3} - 18u - \\frac{81}{u} \\right] + C = \\frac{2}{3}u^3 - 36u - \\frac{162}{u} + C$$"},{"type":"text","order":4,"title":"Substitute back to t","content":"Substituting $u = t^{\\frac{1}{2}}$ back into the expression yields:\n\n$$g(t) = \\frac{2}{3}t^{\\frac{3}{2}} - 36t^{\\frac{1}{2}} - 162t^{-\\frac{1}{2}} + C$$"},{"type":"text","order":5,"title":"Find the constant C","content":"Since $g(4) = 10$, we have: $10 = \\frac{2}{3}(8) - 36(2) - 162\\left(\\frac{1}{2}\\right) + C = -\\frac{443}{3} + C$, which gives $C = \\frac{473}{3}$. Thus:\n\n$$g(t) = \\frac{2}{3}t^{\\frac{3}{2}} - 36t^{\\frac{1}{2}} - 162t^{-\\frac{1}{2}} + \\frac{473}{3}$$"}]}],"generatedAt":"2026-06-09T07:44:32.962Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"9b92f948-1837-43f2-8b18-373ba8d6ec3b","revisionId":"c9615fca-c181-46b1-a8ba-a68b60516829","questionNumber":null,"questionType":"LA","specification":"","difficultyBand":"medium","tier":null,"totalMarks":5,"sourceTitle":null,"calculatorUse":null,"isCalculation":false,"parts":[{"id":"349bfd33-533a-49e8-98d2-f6efbccf47f6","label":null,"content":"Find \\[ \\int \\frac{(2\\sqrt{x} - 3)(x + 4)}{5\\sqrt{x}} \\, \\mathrm{d}x \\] writing each term in simplest form.","metadata":null,"markscheme":"- Expands the numerator to $2x\\sqrt{x} + 8\\sqrt{x} - 3x - 12$ or $2x^{3/2} - 3x + 8x^{1/2} - 12$ M1\n- Divides each term by the denominator $5x^{1/2}$ to obtain the integrand $\\frac{2}{5}x - \\frac{3}{5}x^{1/2} + \\frac{8}{5} - \\frac{12}{5}x^{-1/2}$ M1\n- Integrates the expression term by term using the rule $\\int x^n \\, dx = \\frac{x^{n+1}}{n+1}$ M1\n- Obtains the integrated terms $\\frac{2}{5} \\cdot \\frac{x^2}{2} - \\frac{3}{5} \\cdot \\frac{x^{3/2}}{3/2} + \\frac{8}{5}x - \\frac{12}{5} \\cdot \\frac{x^{1/2}}{1/2} + C$ A1\n- Simplifies coefficients to reach the final answer $\\frac{1}{5}x^2 - \\frac{2}{5}x^{3/2} + \\frac{8}{5}x - \\frac{24}{5}x^{1/2} + C$ A1","marks":5,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"expansion","label":"Expansion and Division","steps":[{"type":"text","order":0,"title":"Expand the numerator","content":"First, we expand the numerator of the integrand by multiplying the two brackets: $$(2\\sqrt{x} - 3)(x + 4)$$ Using the distributive law (FOIL method), we expand term-by-term: $$(2\\sqrt{x} - 3)(x + 4) = 2\\sqrt{x}(x) + 2\\sqrt{x}(4) - 3(x) - 3(4)$$ Converting the square roots to fractional exponents ($$x^{1/2}$$) gives: $$2x^{3/2} + 8x^{1/2} - 3x - 12$$ Reordering the terms, we get: $$2x^{3/2} - 3x + 8x^{1/2} - 12$$ This earns the first mark **M1**."},{"type":"text","order":1,"title":"Divide by the denominator","content":"Next, we divide each term of our expanded numerator by the denominator, $$5\\sqrt{x}$$, which can be written in index form as $$5x^{1/2}$$: $$\\frac{2x^{3/2} - 3x + 8x^{1/2} - 12}{5x^{1/2}} = \\frac{2x^{3/2}}{5x^{1/2}} - \\frac{3x}{5x^{1/2}} + \\frac{8x^{1/2}}{5x^{1/2}} - \\frac{12}{5x^{1/2}}$$ Using the index rule $$\\frac{x^a}{x^b} = x^{a-b}$$, we simplify each fraction: $$\\frac{2}{5}x - \\frac{3}{5}x^{1/2} + \\frac{8}{5} - \\frac{12}{5}x^{-1/2}$$ This simplified integrand earns the second mark **M1**."},{"type":"text","order":2,"title":"Integrate term-by-term","content":"Now we integrate each term individually using the power rule of integration: $$\\int x^n \\, \\mathrm{d}x = \\frac{x^{n+1}}{n+1} \\quad (n \\neq -1)$$ Applying this rule to each term gives: $$\\int \\left( \\frac{2}{5}x - \\frac{3}{5}x^{1/2} + \\frac{8}{5} - \\frac{12}{5}x^{-1/2} \\right) \\mathrm{d}x = \\frac{2}{5} \\cdot \\frac{x^2}{2} - \\frac{3}{5} \\cdot \\frac{x^{3/2}}{3/2} + \\frac{8}{5}x - \\frac{12}{5} \\cdot \\frac{x^{1/2}}{1/2} + C$$ Showing the integration process earns the third mark **M1**, and obtaining these integrated terms correctly earns **A1**."},{"type":"text","order":3,"title":"Simplify the coefficients","content":"Finally, simplify the coefficients of each term: * First term: $$\\frac{2}{5} \\cdot \\frac{x^2}{2} = \\frac{1}{5}x^2$$ * Second term: $$\\frac{3}{5} \\cdot \\frac{2}{3}x^{3/2} = \\frac{2}{5}x^{3/2}$$ * Third term: $$\\frac{8}{5}x$$ * Fourth term: $$\\frac{12}{5} \\cdot 2x^{1/2} = \\frac{24}{5}x^{1/2}$$ Combining these simplified terms and including the constant of integration $$C$$, we get: $$\\frac{1}{5}x^2 - \\frac{2}{5}x^{3/2} + \\frac{8}{5}x - \\frac{24}{5}x^{1/2} + C$$ This final, simplest form earns the final mark **A1**."}]},{"id":"substitution","label":"Integration by Substitution","steps":[{"type":"text","order":0,"title":"Define the substitution","content":"To solve this integral using the substitution method, we let: $$u = \\sqrt{x} = x^{1/2}$$ This means that: $$u^2 = x$$ Differentiating both sides with respect to $$u$$ gives: $$\\mathrm{d}x = 2u \\, \\mathrm{d}u$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\int \\frac{(2\\htmlData{anchor=u-sub1}{\\color{@sky}{\\sqrt{x}}} - 3)(\\htmlData{anchor=x-sub}{\\color{@purple}{x}} + 4)}{5\\htmlData{anchor=u-sub2}{\\color{@sky}{\\sqrt{x}}}} \\, \\htmlData{anchor=dx-sub}{\\color{@amber}{\\mathrm{d}x}}$$"},{"latex":"$$$\\int \\frac{(2\\color{@sky}{u} - 3)(\\color{@purple}{u^2} + 4)}{5\\color{@sky}{u}} \\cdot (\\color{@amber}{2u \\, \\mathrm{d}u})$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-sub1"},{"color":"sky","style":"text-color","anchor":"u-sub2"},{"color":"purple","style":"text-color","anchor":"x-sub"},{"color":"amber","style":"text-color","anchor":"dx-sub"}],"connections":[{"to":"dx-sub","from":"u-sub1","color":"sky","label":"u","style":"solid"}]},"order":1,"title":"Substitute into the integral","description":"Now, let's substitute $$u = \\sqrt{x}$$, $$x = u^2$$, and $$\\mathrm{d}x = 2u \\, \\mathrm{d}u$$ into the given integral expression."},{"type":"text","order":2,"title":"Simplify the integrand","content":"Notice that the $$u$$ in the denominator cancels out with the $$u$$ from $$\\mathrm{d}x = 2u \\, \\mathrm{d}u$$: $$\\int \\frac{(2u - 3)(u^2 + 4)}{5u} \\cdot 2u \\, \\mathrm{d}u = \\frac{2}{5} \\int (2u - 3)(u^2 + 4) \\, \\mathrm{d}u$$ Next, expand the product in the integrand: $$(2u - 3)(u^2 + 4) = 2u^3 - 3u^2 + 8u - 12$$ Multiply this expanded expression by the constant factor $$\\frac{2}{5}$$: $$\\frac{2}{5}(2u^3 - 3u^2 + 8u - 12) = \\frac{4}{5}u^3 - \\frac{6}{5}u^2 + \\frac{16}{5}u - \\frac{24}{5}$$ So our integral becomes: $$\\int \\left( \\frac{4}{5}u^3 - \\frac{6}{5}u^2 + \\frac{16}{5}u - \\frac{24}{5} \\right) \\mathrm{d}u$$"},{"type":"text","order":3,"title":"Integrate term by term","content":"Now we integrate each term with respect to $$u$$ using the power rule $$\\int u^n \\, \\mathrm{d}u = \\frac{u^{n+1}}{n+1}$$: $$\\int \\left( \\frac{4}{5}u^3 - \\frac{6}{5}u^2 + \\frac{16}{5}u - \\frac{24}{5} \\right) \\mathrm{d}u = \\frac{4}{5} \\left( \\frac{u^4}{4} \\right) - \\frac{6}{5} \\left( \\frac{u^3}{3} \\right) + \\frac{16}{5} \\left( \\frac{u^2}{2} \\right) - \\frac{24}{5}u + C$$ Simplify the coefficients: $$\\frac{1}{5}u^4 - \\frac{2}{5}u^3 + \\frac{8}{5}u^2 - \\frac{24}{5}u + C$$"},{"type":"text","order":4,"title":"Substitute back to x","content":"Finally, we substitute $$u = x^{1/2}$$ (and thus $$u^2 = x$$, $$u^3 = x^{3/2}$$, $$u^4 = x^2$$) back into our expression to get the final answer in terms of $$x$$: $$\\frac{1}{5}x^2 - \\frac{2}{5}x^{3/2} + \\frac{8}{5}x - \\frac{24}{5}x^{1/2} + C$$"}]}],"generatedAt":"2026-06-09T07:19:02.567Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"349bfd33-533a-49e8-98d2-f6efbccf47f6:expansion","label":"Expansion and Division","steps":[{"type":"text","order":0,"title":"Expand the numerator","content":"First, we expand the numerator of the integrand by multiplying the two brackets: $$(2\\sqrt{x} - 3)(x + 4)$$ Using the distributive law (FOIL method), we expand term-by-term: $$(2\\sqrt{x} - 3)(x + 4) = 2\\sqrt{x}(x) + 2\\sqrt{x}(4) - 3(x) - 3(4)$$ Converting the square roots to fractional exponents ($$x^{1/2}$$) gives: $$2x^{3/2} + 8x^{1/2} - 3x - 12$$ Reordering the terms, we get: $$2x^{3/2} - 3x + 8x^{1/2} - 12$$ This earns the first mark **M1**."},{"type":"text","order":1,"title":"Divide by the denominator","content":"Next, we divide each term of our expanded numerator by the denominator, $$5\\sqrt{x}$$, which can be written in index form as $$5x^{1/2}$$: $$\\frac{2x^{3/2} - 3x + 8x^{1/2} - 12}{5x^{1/2}} = \\frac{2x^{3/2}}{5x^{1/2}} - \\frac{3x}{5x^{1/2}} + \\frac{8x^{1/2}}{5x^{1/2}} - \\frac{12}{5x^{1/2}}$$ Using the index rule $$\\frac{x^a}{x^b} = x^{a-b}$$, we simplify each fraction: $$\\frac{2}{5}x - \\frac{3}{5}x^{1/2} + \\frac{8}{5} - \\frac{12}{5}x^{-1/2}$$ This simplified integrand earns the second mark **M1**."},{"type":"text","order":2,"title":"Integrate term-by-term","content":"Now we integrate each term individually using the power rule of integration: $$\\int x^n \\, \\mathrm{d}x = \\frac{x^{n+1}}{n+1} \\quad (n \\neq -1)$$ Applying this rule to each term gives: $$\\int \\left( \\frac{2}{5}x - \\frac{3}{5}x^{1/2} + \\frac{8}{5} - \\frac{12}{5}x^{-1/2} \\right) \\mathrm{d}x = \\frac{2}{5} \\cdot \\frac{x^2}{2} - \\frac{3}{5} \\cdot \\frac{x^{3/2}}{3/2} + \\frac{8}{5}x - \\frac{12}{5} \\cdot \\frac{x^{1/2}}{1/2} + C$$ Showing the integration process earns the third mark **M1**, and obtaining these integrated terms correctly earns **A1**."},{"type":"text","order":3,"title":"Simplify the coefficients","content":"Finally, simplify the coefficients of each term: * First term: $$\\frac{2}{5} \\cdot \\frac{x^2}{2} = \\frac{1}{5}x^2$$ * Second term: $$\\frac{3}{5} \\cdot \\frac{2}{3}x^{3/2} = \\frac{2}{5}x^{3/2}$$ * Third term: $$\\frac{8}{5}x$$ * Fourth term: $$\\frac{12}{5} \\cdot 2x^{1/2} = \\frac{24}{5}x^{1/2}$$ Combining these simplified terms and including the constant of integration $$C$$, we get: $$\\frac{1}{5}x^2 - \\frac{2}{5}x^{3/2} + \\frac{8}{5}x - \\frac{24}{5}x^{1/2} + C$$ This final, simplest form earns the final mark **A1**."}]},{"id":"349bfd33-533a-49e8-98d2-f6efbccf47f6:substitution","label":"Integration by Substitution","steps":[{"type":"text","order":0,"title":"Define the substitution","content":"To solve this integral using the substitution method, we let: $$u = \\sqrt{x} = x^{1/2}$$ This means that: $$u^2 = x$$ Differentiating both sides with respect to $$u$$ gives: $$\\mathrm{d}x = 2u \\, \\mathrm{d}u$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\int \\frac{(2\\htmlData{anchor=u-sub1}{\\color{@sky}{\\sqrt{x}}} - 3)(\\htmlData{anchor=x-sub}{\\color{@purple}{x}} + 4)}{5\\htmlData{anchor=u-sub2}{\\color{@sky}{\\sqrt{x}}}} \\, \\htmlData{anchor=dx-sub}{\\color{@amber}{\\mathrm{d}x}}$$"},{"latex":"$$$\\int \\frac{(2\\color{@sky}{u} - 3)(\\color{@purple}{u^2} + 4)}{5\\color{@sky}{u}} \\cdot (\\color{@amber}{2u \\, \\mathrm{d}u})$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-sub1"},{"color":"sky","style":"text-color","anchor":"u-sub2"},{"color":"purple","style":"text-color","anchor":"x-sub"},{"color":"amber","style":"text-color","anchor":"dx-sub"}],"connections":[{"to":"dx-sub","from":"u-sub1","color":"sky","label":"u","style":"solid"}]},"order":1,"title":"Substitute into the integral","description":"Now, let's substitute $$u = \\sqrt{x}$$, $$x = u^2$$, and $$\\mathrm{d}x = 2u \\, \\mathrm{d}u$$ into the given integral expression."},{"type":"text","order":2,"title":"Simplify the integrand","content":"Notice that the $$u$$ in the denominator cancels out with the $$u$$ from $$\\mathrm{d}x = 2u \\, \\mathrm{d}u$$: $$\\int \\frac{(2u - 3)(u^2 + 4)}{5u} \\cdot 2u \\, \\mathrm{d}u = \\frac{2}{5} \\int (2u - 3)(u^2 + 4) \\, \\mathrm{d}u$$ Next, expand the product in the integrand: $$(2u - 3)(u^2 + 4) = 2u^3 - 3u^2 + 8u - 12$$ Multiply this expanded expression by the constant factor $$\\frac{2}{5}$$: $$\\frac{2}{5}(2u^3 - 3u^2 + 8u - 12) = \\frac{4}{5}u^3 - \\frac{6}{5}u^2 + \\frac{16}{5}u - \\frac{24}{5}$$ So our integral becomes: $$\\int \\left( \\frac{4}{5}u^3 - \\frac{6}{5}u^2 + \\frac{16}{5}u - \\frac{24}{5} \\right) \\mathrm{d}u$$"},{"type":"text","order":3,"title":"Integrate term by term","content":"Now we integrate each term with respect to $$u$$ using the power rule $$\\int u^n \\, \\mathrm{d}u = \\frac{u^{n+1}}{n+1}$$: $$\\int \\left( \\frac{4}{5}u^3 - \\frac{6}{5}u^2 + \\frac{16}{5}u - \\frac{24}{5} \\right) \\mathrm{d}u = \\frac{4}{5} \\left( \\frac{u^4}{4} \\right) - \\frac{6}{5} \\left( \\frac{u^3}{3} \\right) + \\frac{16}{5} \\left( \\frac{u^2}{2} \\right) - \\frac{24}{5}u + C$$ Simplify the coefficients: $$\\frac{1}{5}u^4 - \\frac{2}{5}u^3 + \\frac{8}{5}u^2 - \\frac{24}{5}u + C$$"},{"type":"text","order":4,"title":"Substitute back to x","content":"Finally, we substitute $$u = x^{1/2}$$ (and thus $$u^2 = x$$, $$u^3 = x^{3/2}$$, $$u^4 = x^2$$) back into our expression to get the final answer in terms of $$x$$: $$\\frac{1}{5}x^2 - \\frac{2}{5}x^{3/2} + \\frac{8}{5}x - \\frac{24}{5}x^{1/2} + C$$"}]}],"generatedAt":"2026-06-09T07:19:14.539Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"a9b4224f-432a-4549-8b38-ab486616e0c3","revisionId":"4875638f-a868-4632-9273-10441c9a7236","questionNumber":null,"questionType":"LA","specification":"The mass of a biochemical sample \$M\$ (measured in milligrams) varies over time \$t\$ (measured in seconds, \$t > 0\$) according to a growth model. The mass-time curve passes through the point \$P(4, 15)\$.\n\nGiven that\n\n\\[\\frac{\\mathrm{d} M}{\\mathrm{d} t} = \\frac{5t^2 - 40}{2\\sqrt{t}}\\]","difficultyBand":"medium","tier":null,"totalMarks":9,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"931a4b19-86b8-4677-92e5-c23dbfafd26a","label":"a","content":"Determine the equation of the tangent to the curve at the point \$P\$, writing your answer in the form \$M = kt + c\$, where \$k\$ and \$c\$ are integers to be found.","metadata":null,"markscheme":"- Simplifies the derivative to $\\frac{dM}{dt} = 2.5t^{1.5} - 20t^{-0.5}$ or substitutes $t=4$ directly into the given form M1\n- Evaluates the gradient at $t=4$ as $m = \\frac{5(16) - 40}{2(2)} = \\frac{40}{4} = 10$ A1\n- Substitutes $m=10$ and $P(4, 15)$ into $M - 15 = 10(t - 4)$ to obtain $M = 10t - 25$ A1","marks":3,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"point_slope","label":"Point-Slope Formula","steps":[{"type":"text","order":0,"title":"Evaluate the gradient","content":"To find the gradient ($m$) of the tangent at $t = 4$, we evaluate the derivative $\\frac{\\mathrm{d}M}{\\mathrm{d}t}$ at this point. \n\nSubstitute $t = 4$ into the given expression:\n\n$$\\frac{\\mathrm{d}M}{\\mathrm{d}t} = \\frac{5(4)^2 - 40}{2\\sqrt{4}}$$\n\nCalculating the numerator and denominator:\n\n$$\\begin{aligned} 5(4)^2 - 40 &= 5(16) - 40 = 80 - 40 = 40 \\\\ 2\\sqrt{4} &= 2(2) = 4 \\end{aligned}$$\n\nSo, the gradient is:\n\n$$m = \\frac{40}{4} = 10$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$M_1 = \\htmlData{anchor=M-coord}{\\color{@sky}{15}}, \\quad t_1 = \\htmlData{anchor=t-coord}{\\color{@amber}{4}}, \\quad m = \\htmlData{anchor=grad-val}{\\color{@purple}{10}}$"},{"latex":"$$\\htmlData{anchor=pt-slope-eq}{M - M_1} = m(t - t_1)$"},{"latex":"$$M - \\htmlData{anchor=M-sub}{\\color{@sky}{15}} = \\htmlData{anchor=grad-sub}{\\color{@purple}{10}}(t - \\htmlData{anchor=t-sub}{\\color{@amber}{4}})$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"M-coord"},{"color":"amber","style":"text-color","anchor":"t-coord"},{"color":"purple","style":"text-color","anchor":"grad-val"}],"connections":[{"to":"M-sub","from":"M-coord","color":"sky","label":null,"style":"solid"},{"to":"t-sub","from":"t-coord","color":"amber","label":null,"style":"solid"},{"to":"grad-sub","from":"grad-val","color":"purple","label":null,"style":"solid"}]},"order":1,"title":"Apply point-slope formula","description":"We now use the point-slope formula $M - M_1 = m(t - t_1)$ with the point $P(4, 15)$ and gradient $m = 10$."},{"type":"text","order":2,"title":"Rearrange to target form","content":"Expand and simplify the equation to match the form $M = kt + c$:\n\n$$\\begin{aligned} M - 15 &= 10(t - 4) \\\\ M - 15 &= 10t - 40 \\\\ M &= 10t - 40 + 15 \\\\ M &= 10t - 25 \\end{aligned}$$\n\nThis gives $k = 10$ and $c = -25$."}]},{"id":"slope_intercept","label":"Slope-Intercept Formula","steps":[{"type":"text","order":0,"title":"Evaluate the gradient","content":"The gradient of the tangent to the curve at any point is given by the derivative $\\frac{\\mathrm{d} M}{\\mathrm{d} t}$. \n\nWe are given the rate of change of mass with respect to time:\n$$\\frac{\\mathrm{d} M}{\\mathrm{d} t} = \\frac{5t^2 - 40}{2\\sqrt{t}}$$\n\nTo find the gradient $k$ of the tangent at the point $P(4, 15)$, we substitute $t = 4$ into this derivative expression:\n$$\\begin{aligned} k &= \\left.\\frac{\\mathrm{d} M}{\\mathrm{d} t}\\right|_{t=4} \\\\ &= \\frac{5(4)^2 - 40}{2\\sqrt{4}} \\\\ &= \\frac{5(16) - 40}{2(2)} \\\\ &= \\frac{80 - 40}{4} \\\\ &= \\frac{40}{4} = 10 \\end{aligned}$$","highlights":[{"text":"P(4, 15)","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$M = \\htmlData{anchor=k-slope}{\\color{@sky}{k}}t + \\htmlData{anchor=c-intercept}{\\color{@amber}{c}}$"},{"latex":"$$M = \\htmlData{anchor=k-sub}{\\color{@sky}{10}}t + \\htmlData{anchor=c-un}{\\color{@amber}{c}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"k-slope"},{"color":"sky","style":"text-color","anchor":"k-sub"}],"connections":[{"to":"k-sub","from":"k-slope","color":"sky","label":"gradient","style":"solid"}]},"order":1,"title":"Set up slope-intercept form","description":"The general equation of a straight line in slope-intercept form is $M = kt + c$, where $k$ is the gradient and $c$ is the $y$-intercept (or $M$-intercept in this context). Substituting our gradient $k = 10$ gives us this partial equation."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Substitute } t = \\htmlData{anchor=t-val}{\\color{@purple}{4}} \\text{ and } M = \\htmlData{anchor=M-val}{\\color{@green}{15}} \\text{ into the equation:}$"},{"latex":"$$\\htmlData{anchor=M-sub}{\\color{@green}{15}} = 10(\\htmlData{anchor=t-sub}{\\color{@purple}{4}}) + \\htmlData{anchor=c-sol}{\\color{@amber}{c}}$"},{"latex":"$$15 = 40 + \\color{@amber}{c}$"},{"latex":"$$\\htmlData{anchor=c-final}{\\color{@amber}{c}} = -25$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"t-val"},{"color":"green","style":"text-color","anchor":"M-val"},{"color":"amber","style":"text-color","anchor":"c-final"}],"connections":[{"to":"t-sub","from":"t-val","color":"purple","label":"t-coordinate","style":"solid"},{"to":"M-sub","from":"M-val","color":"green","label":"M-coordinate","style":"solid"}]},"order":2,"title":"Find the constant c","description":"Since the tangent passes through $P(4, 15)$, we can substitute these coordinates into our equation to solve for the constant $c$."},{"type":"text","order":3,"title":"State the final equation","content":"Substitute $c = -25$ back into our equation $M = 10t + c$:\n\n$$M = 10t - 25$$\n\nThis is in the required form $M = kt + c$ with integers $k = 10$ and $c = -25$."}]}],"generatedAt":"2026-06-09T07:55:55.933Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"9650897c-f3af-4800-98dc-5eec35d92a90","label":"b","content":"Find the function \$M(t)\$.","metadata":null,"markscheme":"- Attempts to integrate $\\int (2.5t^{3/2} - 20t^{-1/2}) \\, dt$ with at least one power of $t$ increased correctly M1\n- Obtains $\\frac{2.5}{2.5}t^{5/2}$ or $t^{5/2}$ A1\n- Obtains $-\\frac{20}{0.5}t^{1/2}$ or $-40t^{1/2}$ A1\n- Adds a constant of integration $+ C$ and substitutes $t=4, M=15$ M1\n- Calculates $15 = 4^{5/2} - 40(4^{1/2}) + C \\Rightarrow 15 = 32 - 80 + C$, leading to $C = 63$ A1\n- States final expression $M(t) = t^{5/2} - 40t^{1/2} + 63$ A1","marks":6,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"direct_integration","label":"Direct Term-by-Term Integration","steps":[{"type":"text","order":0,"title":"Simplify the derivative","content":"To find $M(t)$ from its rate of change $\\frac{\\mathrm{d} M}{\\mathrm{d} t}$, we first need to simplify the expression by dividing each term in the numerator by the denominator, $2\\sqrt{t}$.\n\nRecall that $\\sqrt{t} = t^{1/2}$. We can rewrite the derivative as:\n\n$$\\frac{\\mathrm{d} M}{\\mathrm{d} t} = \\frac{5t^2 - 40}{2t^{1/2}}$$\n\nNow, split this into two separate terms:\n\n$$\\frac{\\mathrm{d} M}{\\mathrm{d} t} = \\frac{5t^2}{2t^{1/2}} - \\frac{40}{2t^{1/2}}$$\n\nUsing index laws (subtracting exponents: $2 - \\frac{1}{2} = \\frac{3}{2}$ and $-\\frac{1}{2}$):\n\n$$\\frac{\\mathrm{d} M}{\\mathrm{d} t} = 2.5t^{3/2} - 20t^{-1/2}$$"},{"type":"text","order":1,"title":"Integrate term-by-term","content":"To find the mass function $M(t)$, we integrate the simplified expression with respect to $t$:\n\n$$M(t) = \\int \\left(2.5t^{3/2} - 20t^{-1/2}\\right) \\mathrm{d} t$$\n\nUsing the power rule for integration, $\\int t^n \\mathrm{d} t = \\frac{t^{n+1}}{n+1} + C$, we integrate each term individually:\n\n- **For the first term:**\n $$\\int 2.5t^{3/2} \\mathrm{d} t = \\frac{2.5}{\\frac{5}{2}} t^{5/2} = \\frac{2.5}{2.5} t^{5/2} = t^{5/2}$$\n\n- **For the second term:**\n $$\\int -20t^{-1/2} \\mathrm{d} t = -\\frac{20}{\\frac{1}{2}} t^{1/2} = -40t^{1/2}$$\n\nCombining these and adding the constant of integration $C$ gives:\n\n$$M(t) = t^{5/2} - 40t^{1/2} + C$$"},{"type":"text","order":2,"title":"Use the initial condition","content":"We are given that the curve passes through the point $P(4, 15)$. This means when $t = 4$, the mass $M = 15$.\n\nSubstitute these values into our equation to solve for $C$:\n\n$$15 = 4^{5/2} - 40(4^{1/2}) + C$$\n\nNow, calculate the values:\n- $4^{5/2} = (\\sqrt{4})^5 = 2^5 = 32$\n- $4^{1/2} = \\sqrt{4} = 2$\n\nSubstitute these simplified values back in:\n\n$$\\begin{aligned} 15 &= 32 - 40(2) + C \\\\ 15 &= 32 - 80 + C \\\\ 15 &= -48 + C \\\\ C &= 15 + 48 \\\\ C &= 63 \\end{aligned}$$","highlights":[{"text":"P(4, 15)","location":"specification"}]},{"type":"text","order":3,"title":"State the final function","content":"Substitute the constant $C = 63$ back into the integrated formula to obtain the final equation for $M(t)$:\n\n$$M(t) = t^{5/2} - 40t^{1/2} + 63$$"}]},{"id":"substitution","label":"Integration by Substitution","steps":[{"type":"text","order":0,"title":"Set up the integral","content":"To find the mass function $M(t)$, we need to integrate the rate of change of mass with respect to time $t$:\n\n$$M(t) = \\int \\frac{5t^2 - 40}{2\\sqrt{t}} \\, \\mathrm{d}t$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\htmlData{anchor=u-def}{u} = \\sqrt{t} \\implies \\htmlData{anchor=t-def}{t} = u^2"},{"latex":"\\frac{\\mathrm{d}t}{\\mathrm{d}u} = 2u \\implies \\htmlData{anchor=dt-def}{\\mathrm{d}t} = 2u \\, \\mathrm{d}u"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-def"},{"color":"amber","style":"text-color","anchor":"t-def"},{"color":"purple","style":"text-color","anchor":"dt-def"}],"connections":[{"to":"dt-def","from":"u-def","color":"sky","label":"differentiate","style":"solid"}]},"order":1,"title":"Define the substitution","description":"We use the substitution $u = \\sqrt{t}$ to simplify the radical expression. Squaring both sides gives $t = u^2$, and differentiating with respect to $u$ gives $\\mathrm{d}t = 2u \\, \\mathrm{d}u$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"M(t) = \\int \\frac{5(\\htmlData{anchor=t-sub}{u^2})^2 - 40}{2\\htmlData{anchor=sqrt-sub}{u}} \\, (\\htmlData{anchor=dt-sub}{2u \\, \\mathrm{d}u})"},{"latex":"\\phantom{M(t)} = \\int \\frac{5u^4 - 40}{\\color{@red}{2u}} \\cdot \\color{@red}{2u} \\, \\mathrm{d}u"},{"latex":"\\phantom{M(t)} = \\int (5u^4 - 40) \\, \\mathrm{d}u"}],"highlights":[{"color":"amber","style":"text-color","anchor":"t-sub"},{"color":"sky","style":"text-color","anchor":"sqrt-sub"},{"color":"purple","style":"text-color","anchor":"dt-sub"}],"connections":[{"to":"sqrt-sub","from":"t-sub","color":"amber","label":null,"style":"dashed"}]},"order":2,"title":"Substitute into the integral","description":"Now, we substitute $t = u^2$, $\\sqrt{t} = u$, and $\\mathrm{d}t = 2u \\, \\mathrm{d}u$ into the integral. Notice how the $2u$ terms conveniently cancel out!"},{"type":"text","order":3,"title":"Integrate with respect to u","content":"Integrating this simplified polynomial with respect to $u$ gives:\n\n$$M(u) = \\int (5u^4 - 40) \\, \\mathrm{d}u = \\frac{5u^5}{5} - 40u + C$$\n\nSimplifying terms:\n\n$$M(u) = u^5 - 40u + C$$"},{"type":"text","order":4,"title":"Substitute back to t","content":"Now we replace $u$ with $\\sqrt{t} = t^{1/2}$ to return to the original variable $t$:\n\n$$M(t) = (t^{1/2})^5 - 40(t^{1/2}) + C$$\n\n$$M(t) = t^{5/2} - 40t^{1/2} + C$$"},{"type":"text","order":5,"title":"Solve for constant C","content":"We are given that the curve passes through the point $P(4, 15)$, meaning $M(4) = 15$:\n\n$$15 = 4^{5/2} - 40(4^{1/2}) + C$$\n\nCalculate the numerical values:\n- $4^{5/2} = (\\sqrt{4})^5 = 2^5 = 32$\n- $40(4^{1/2}) = 40(2) = 80$\n\nSubstitute these back in:\n\n$$\\begin{aligned} 15 &= 32 - 80 + C \\\\ 15 &= -48 + C \\\\ C &= 63 \\end{aligned}$$\n\nThus, the function is:\n\n$$M(t) = t^{5/2} - 40t^{1/2} + 63$$","highlights":[{"text":"passes through the point \$P(4, 15)\$","location":"specification"}]}]}],"generatedAt":"2026-06-09T07:57:52.694Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"931a4b19-86b8-4677-92e5-c23dbfafd26a:point_slope","label":"(a) Point-Slope Formula","steps":[{"type":"text","order":0,"title":"Evaluate the gradient","content":"To find the gradient ($m$) of the tangent at $t = 4$, we evaluate the derivative $\\frac{\\mathrm{d}M}{\\mathrm{d}t}$ at this point. \n\nSubstitute $t = 4$ into the given expression:\n\n$$\\frac{\\mathrm{d}M}{\\mathrm{d}t} = \\frac{5(4)^2 - 40}{2\\sqrt{4}}$$\n\nCalculating the numerator and denominator:\n\n$$\\begin{aligned} 5(4)^2 - 40 &= 5(16) - 40 = 80 - 40 = 40 \\\\ 2\\sqrt{4} &= 2(2) = 4 \\end{aligned}$$\n\nSo, the gradient is:\n\n$$m = \\frac{40}{4} = 10$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$M_1 = \\htmlData{anchor=M-coord}{\\color{@sky}{15}}, \\quad t_1 = \\htmlData{anchor=t-coord}{\\color{@amber}{4}}, \\quad m = \\htmlData{anchor=grad-val}{\\color{@purple}{10}}$"},{"latex":"$$\\htmlData{anchor=pt-slope-eq}{M - M_1} = m(t - t_1)$"},{"latex":"$$M - \\htmlData{anchor=M-sub}{\\color{@sky}{15}} = \\htmlData{anchor=grad-sub}{\\color{@purple}{10}}(t - \\htmlData{anchor=t-sub}{\\color{@amber}{4}})$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"M-coord"},{"color":"amber","style":"text-color","anchor":"t-coord"},{"color":"purple","style":"text-color","anchor":"grad-val"}],"connections":[{"to":"M-sub","from":"M-coord","color":"sky","label":null,"style":"solid"},{"to":"t-sub","from":"t-coord","color":"amber","label":null,"style":"solid"},{"to":"grad-sub","from":"grad-val","color":"purple","label":null,"style":"solid"}]},"order":1,"title":"Apply point-slope formula","description":"We now use the point-slope formula $M - M_1 = m(t - t_1)$ with the point $P(4, 15)$ and gradient $m = 10$."},{"type":"text","order":2,"title":"Rearrange to target form","content":"Expand and simplify the equation to match the form $M = kt + c$:\n\n$$\\begin{aligned} M - 15 &= 10(t - 4) \\\\ M - 15 &= 10t - 40 \\\\ M &= 10t - 40 + 15 \\\\ M &= 10t - 25 \\end{aligned}$$\n\nThis gives $k = 10$ and $c = -25$."}]},{"id":"931a4b19-86b8-4677-92e5-c23dbfafd26a:slope_intercept","label":"(a) Slope-Intercept Formula","steps":[{"type":"text","order":0,"title":"Evaluate the gradient","content":"The gradient of the tangent to the curve at any point is given by the derivative $\\frac{\\mathrm{d} M}{\\mathrm{d} t}$. \n\nWe are given the rate of change of mass with respect to time:\n$$\\frac{\\mathrm{d} M}{\\mathrm{d} t} = \\frac{5t^2 - 40}{2\\sqrt{t}}$$\n\nTo find the gradient $k$ of the tangent at the point $P(4, 15)$, we substitute $t = 4$ into this derivative expression:\n$$\\begin{aligned} k &= \\left.\\frac{\\mathrm{d} M}{\\mathrm{d} t}\\right|_{t=4} \\\\ &= \\frac{5(4)^2 - 40}{2\\sqrt{4}} \\\\ &= \\frac{5(16) - 40}{2(2)} \\\\ &= \\frac{80 - 40}{4} \\\\ &= \\frac{40}{4} = 10 \\end{aligned}$$","highlights":[{"text":"P(4, 15)","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$M = \\htmlData{anchor=k-slope}{\\color{@sky}{k}}t + \\htmlData{anchor=c-intercept}{\\color{@amber}{c}}$"},{"latex":"$$M = \\htmlData{anchor=k-sub}{\\color{@sky}{10}}t + \\htmlData{anchor=c-un}{\\color{@amber}{c}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"k-slope"},{"color":"sky","style":"text-color","anchor":"k-sub"}],"connections":[{"to":"k-sub","from":"k-slope","color":"sky","label":"gradient","style":"solid"}]},"order":1,"title":"Set up slope-intercept form","description":"The general equation of a straight line in slope-intercept form is $M = kt + c$, where $k$ is the gradient and $c$ is the $y$-intercept (or $M$-intercept in this context). Substituting our gradient $k = 10$ gives us this partial equation."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Substitute } t = \\htmlData{anchor=t-val}{\\color{@purple}{4}} \\text{ and } M = \\htmlData{anchor=M-val}{\\color{@green}{15}} \\text{ into the equation:}$"},{"latex":"$$\\htmlData{anchor=M-sub}{\\color{@green}{15}} = 10(\\htmlData{anchor=t-sub}{\\color{@purple}{4}}) + \\htmlData{anchor=c-sol}{\\color{@amber}{c}}$"},{"latex":"$$15 = 40 + \\color{@amber}{c}$"},{"latex":"$$\\htmlData{anchor=c-final}{\\color{@amber}{c}} = -25$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"t-val"},{"color":"green","style":"text-color","anchor":"M-val"},{"color":"amber","style":"text-color","anchor":"c-final"}],"connections":[{"to":"t-sub","from":"t-val","color":"purple","label":"t-coordinate","style":"solid"},{"to":"M-sub","from":"M-val","color":"green","label":"M-coordinate","style":"solid"}]},"order":2,"title":"Find the constant c","description":"Since the tangent passes through $P(4, 15)$, we can substitute these coordinates into our equation to solve for the constant $c$."},{"type":"text","order":3,"title":"State the final equation","content":"Substitute $c = -25$ back into our equation $M = 10t + c$:\n\n$$M = 10t - 25$$\n\nThis is in the required form $M = kt + c$ with integers $k = 10$ and $c = -25$."}]},{"id":"9650897c-f3af-4800-98dc-5eec35d92a90:direct_integration","label":"(b) Direct Term-by-Term Integration","steps":[{"type":"text","order":0,"title":"Simplify the derivative","content":"To find $M(t)$ from its rate of change $\\frac{\\mathrm{d} M}{\\mathrm{d} t}$, we first need to simplify the expression by dividing each term in the numerator by the denominator, $2\\sqrt{t}$.\n\nRecall that $\\sqrt{t} = t^{1/2}$. We can rewrite the derivative as:\n\n$$\\frac{\\mathrm{d} M}{\\mathrm{d} t} = \\frac{5t^2 - 40}{2t^{1/2}}$$\n\nNow, split this into two separate terms:\n\n$$\\frac{\\mathrm{d} M}{\\mathrm{d} t} = \\frac{5t^2}{2t^{1/2}} - \\frac{40}{2t^{1/2}}$$\n\nUsing index laws (subtracting exponents: $2 - \\frac{1}{2} = \\frac{3}{2}$ and $-\\frac{1}{2}$):\n\n$$\\frac{\\mathrm{d} M}{\\mathrm{d} t} = 2.5t^{3/2} - 20t^{-1/2}$$"},{"type":"text","order":1,"title":"Integrate term-by-term","content":"To find the mass function $M(t)$, we integrate the simplified expression with respect to $t$:\n\n$$M(t) = \\int \\left(2.5t^{3/2} - 20t^{-1/2}\\right) \\mathrm{d} t$$\n\nUsing the power rule for integration, $\\int t^n \\mathrm{d} t = \\frac{t^{n+1}}{n+1} + C$, we integrate each term individually:\n\n- **For the first term:**\n $$\\int 2.5t^{3/2} \\mathrm{d} t = \\frac{2.5}{\\frac{5}{2}} t^{5/2} = \\frac{2.5}{2.5} t^{5/2} = t^{5/2}$$\n\n- **For the second term:**\n $$\\int -20t^{-1/2} \\mathrm{d} t = -\\frac{20}{\\frac{1}{2}} t^{1/2} = -40t^{1/2}$$\n\nCombining these and adding the constant of integration $C$ gives:\n\n$$M(t) = t^{5/2} - 40t^{1/2} + C$$"},{"type":"text","order":2,"title":"Use the initial condition","content":"We are given that the curve passes through the point $P(4, 15)$. This means when $t = 4$, the mass $M = 15$.\n\nSubstitute these values into our equation to solve for $C$:\n\n$$15 = 4^{5/2} - 40(4^{1/2}) + C$$\n\nNow, calculate the values:\n- $4^{5/2} = (\\sqrt{4})^5 = 2^5 = 32$\n- $4^{1/2} = \\sqrt{4} = 2$\n\nSubstitute these simplified values back in:\n\n$$\\begin{aligned} 15 &= 32 - 40(2) + C \\\\ 15 &= 32 - 80 + C \\\\ 15 &= -48 + C \\\\ C &= 15 + 48 \\\\ C &= 63 \\end{aligned}$$","highlights":[{"text":"P(4, 15)","location":"specification"}]},{"type":"text","order":3,"title":"State the final function","content":"Substitute the constant $C = 63$ back into the integrated formula to obtain the final equation for $M(t)$:\n\n$$M(t) = t^{5/2} - 40t^{1/2} + 63$$"}]},{"id":"9650897c-f3af-4800-98dc-5eec35d92a90:substitution","label":"(b) Integration by Substitution","steps":[{"type":"text","order":0,"title":"Set up the integral","content":"To find the mass function $M(t)$, we need to integrate the rate of change of mass with respect to time $t$:\n\n$$M(t) = \\int \\frac{5t^2 - 40}{2\\sqrt{t}} \\, \\mathrm{d}t$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\htmlData{anchor=u-def}{u} = \\sqrt{t} \\implies \\htmlData{anchor=t-def}{t} = u^2"},{"latex":"\\frac{\\mathrm{d}t}{\\mathrm{d}u} = 2u \\implies \\htmlData{anchor=dt-def}{\\mathrm{d}t} = 2u \\, \\mathrm{d}u"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-def"},{"color":"amber","style":"text-color","anchor":"t-def"},{"color":"purple","style":"text-color","anchor":"dt-def"}],"connections":[{"to":"dt-def","from":"u-def","color":"sky","label":"differentiate","style":"solid"}]},"order":1,"title":"Define the substitution","description":"We use the substitution $u = \\sqrt{t}$ to simplify the radical expression. Squaring both sides gives $t = u^2$, and differentiating with respect to $u$ gives $\\mathrm{d}t = 2u \\, \\mathrm{d}u$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"M(t) = \\int \\frac{5(\\htmlData{anchor=t-sub}{u^2})^2 - 40}{2\\htmlData{anchor=sqrt-sub}{u}} \\, (\\htmlData{anchor=dt-sub}{2u \\, \\mathrm{d}u})"},{"latex":"\\phantom{M(t)} = \\int \\frac{5u^4 - 40}{\\color{@red}{2u}} \\cdot \\color{@red}{2u} \\, \\mathrm{d}u"},{"latex":"\\phantom{M(t)} = \\int (5u^4 - 40) \\, \\mathrm{d}u"}],"highlights":[{"color":"amber","style":"text-color","anchor":"t-sub"},{"color":"sky","style":"text-color","anchor":"sqrt-sub"},{"color":"purple","style":"text-color","anchor":"dt-sub"}],"connections":[{"to":"sqrt-sub","from":"t-sub","color":"amber","label":null,"style":"dashed"}]},"order":2,"title":"Substitute into the integral","description":"Now, we substitute $t = u^2$, $\\sqrt{t} = u$, and $\\mathrm{d}t = 2u \\, \\mathrm{d}u$ into the integral. Notice how the $2u$ terms conveniently cancel out!"},{"type":"text","order":3,"title":"Integrate with respect to u","content":"Integrating this simplified polynomial with respect to $u$ gives:\n\n$$M(u) = \\int (5u^4 - 40) \\, \\mathrm{d}u = \\frac{5u^5}{5} - 40u + C$$\n\nSimplifying terms:\n\n$$M(u) = u^5 - 40u + C$$"},{"type":"text","order":4,"title":"Substitute back to t","content":"Now we replace $u$ with $\\sqrt{t} = t^{1/2}$ to return to the original variable $t$:\n\n$$M(t) = (t^{1/2})^5 - 40(t^{1/2}) + C$$\n\n$$M(t) = t^{5/2} - 40t^{1/2} + C$$"},{"type":"text","order":5,"title":"Solve for constant C","content":"We are given that the curve passes through the point $P(4, 15)$, meaning $M(4) = 15$:\n\n$$15 = 4^{5/2} - 40(4^{1/2}) + C$$\n\nCalculate the numerical values:\n- $4^{5/2} = (\\sqrt{4})^5 = 2^5 = 32$\n- $40(4^{1/2}) = 40(2) = 80$\n\nSubstitute these back in:\n\n$$\\begin{aligned} 15 &= 32 - 80 + C \\\\ 15 &= -48 + C \\\\ C &= 63 \\end{aligned}$$\n\nThus, the function is:\n\n$$M(t) = t^{5/2} - 40t^{1/2} + 63$$","highlights":[{"text":"passes through the point \$P(4, 15)\$","location":"specification"}]}]}],"generatedAt":"2026-06-09T07:58:15.423Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null}],"subjectCode":"a_level_maths__ocr_mei","summaries":[{"id":"19b1d184-7b2e-4eab-8e8e-e0f619c14e25","difficultyBand":"easy","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":"easy","calculatorUse":null,"totalMarks":6,"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":"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":"04ebdcb9-9fb6-4d98-a0cf-ce942c92ef1d","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"077f0048-b271-4300-8b0c-c4573895aabb","difficultyBand":"hard","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"09a9bff7-3d5f-4ea7-86e9-e0ebdf77d8b1","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":false},{"id":"09ea0b5c-19b3-487b-b3c8-9d30c7f0cb18","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"0afa7a3c-4e9a-488b-9ca3-5594de5450d2","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"0ebe6bbe-b6d2-4725-b255-a4dfaaf64e12","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"16e688aa-9193-46ee-901b-e45d1ebe0d42","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"1b65118c-6ba7-4b39-92be-338ac0c8e13f","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"1d419b9e-87a0-4fa4-94a4-c3a0d71d171c","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"1fda6d6f-10fe-48e4-a618-c2ee9e39e51a","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"2641fa14-1752-403c-bc7b-fdecb6f1b8d0","difficultyBand":"medium","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":false},{"id":"2a1270c9-5bcf-47de-9be7-dc1979a5e630","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"2a961770-ef5f-4ce4-b128-3eab2bfb3412","difficultyBand":"easy","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"2baf7366-c0ab-431f-9aa4-1b167daaf5c8","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"2f86eff4-299f-4e87-8b33-6124a9ae166d","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":false},{"id":"304561c8-bb51-4f14-8e2f-64bc58dd157a","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":false},{"id":"32831ce7-3931-4f98-8c2e-827897ee25c4","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"36e54636-bda2-4f83-9e20-ac19d5c6a13a","difficultyBand":"hard","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"38252659-3641-4dd0-81e7-9a6c7bffaf85","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":false},{"id":"39740dc4-92d4-4d05-bf4c-11e2d764cf8b","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"3b0aae70-80a9-498c-aaad-6020ae06da80","difficultyBand":"hard","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"3bf1d35e-bb20-4156-b36a-eb0c6ff57690","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":false},{"id":"3c78da78-8789-47c6-811d-8f21988df751","difficultyBand":"easy","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":false},{"id":"3fa00e4a-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Integrate kx^n","topicId":"50eb6c16-3d55-4bb9-bf41-19d5c925affb","topicSidebar":"$L3a","topicSlug":"integrate-kx-n","topicTitle":"1.9.20 Integrate kx^n"}]}],"$L3b","$L3c"]}]

1.9.20 Integrate kx^n

Question 17

1.9.20 Integrate kx^n Questions

1.9.20 Integrate kx^n

Question 17

1.9.20 Integrate kx^n Questions