Secondary 4 Additional Mathematics Calculus Quiz

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Secondary 4 Additional Mathematics Quiz - Calculus

Name: ________________________
Class: ________________________
Date: ________________________
Score: ______ / 60

Duration: 1 hour 15 minutes
Total Marks: 60

Instructions:

• Answer ALL questions in the spaces provided.
• Show all working clearly. Marks are awarded for method.
• Unless otherwise stated, give non-exact answers correct to 3 significant figures.
• You are reminded of the need for clear presentation in your answers.
• The number of marks is given in brackets [ ] at the end of each question or part question.

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Section A: Differentiation Techniques (15 marks)

Answer ALL questions in this section.

1. Differentiate each of the following with respect to $x$:

(a) $y = 3x^4 - 2x^3 + 5x - 7$ [2]

 
 
 
 

(b) $y = (2x + 1)(x^2 - 3)$ [3]

 
 
 
 
 

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2. Find $\frac{dy}{dx}$ for each of the following:

(a) $y = \frac{x^2 + 1}{2x - 1}$, $x \neq \frac{1}{2}$ [3]

 
 
 
 
 

(b) $y = \sqrt{3x^2 + 4}$ [2]

 
 
 
 

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3. A curve has equation $y = \frac{4}{x} + \ln(2x)$, where $x > 0$.

Find the gradient of the curve at the point where $x = 2$. [3]

 
 
 
 
 

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4. Given that $y = e^{2x} \sin x$, find $\frac{dy}{dx}$ in its simplest form. [2]

 
 
 
 

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5. Differentiate $y = \frac{\cos x}{1 + \sin x}$ with respect to $x$. [3]

 
 
 
 
 

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Section B: Applications of Differentiation (15 marks)

Answer ALL questions in this section.

6. A curve has equation $y = x^3 - 6x^2 + 9x + 1$.

(a) Find the coordinates of the stationary points of the curve. [4]

 
 
 
 
 
 

(b) Determine the nature of each stationary point. [3]

 
 
 
 
 
 

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7. The equation of a curve is $y = 2x^3 - 3x^2 - 12x + 7$.

(a) Find the range of values of $x$ for which the curve is increasing. [3]

 
 
 
 
 
 

(b) Find the equation of the tangent to the curve at the point where $x = 1$. [3]

 
 
 
 
 
 

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8. A rectangular box with a square base and an open top is to have a volume of $500 \text{ cm}^3$. The length of each side of the base is $x \text{ cm}$ and the height is $h \text{ cm}$.

(a) Show that the external surface area, $A \text{ cm}^2$, is given by $A = x^2 + \frac{2000}{x}$. [2]

 
 
 
 
 

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9. A curve has equation $y = x^3 - 3x^2 + 2$. Find the coordinates of the point of inflection. [3]

 
 
 
 
 
 

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10. The radius of a circle is increasing at a constant rate of $0.5 \text{ cm/s}$. Find the rate of increase of the area of the circle when the radius is $4 \text{ cm}$. [3]

 
 
 
 
 
 

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Section C: Integration (15 marks)

Answer ALL questions in this section.

11. Find each of the following integrals:

(a) $\displaystyle \int (6x^2 - 4x + 3) \, dx$ [2]

 
 
 
 

(b) $\displaystyle \int \left( \frac{2}{x} + 3e^x \right) dx$, $x > 0$ [2]

 
 
 
 

(c) $\displaystyle \int (2x - 1)^4 \, dx$ [3]

 
 
 
 
 

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12. Evaluate:

(a) $\displaystyle \int_1^4 \left( \sqrt{x} + \frac{1}{x^2} \right) dx$ [4]

 
 
 
 
 
 
 

(b) $\displaystyle \int_0^{\frac{\pi}{6}} \cos 2x \, dx$ [3]

 
 
 
 
 
 

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13. A curve passes through the point $(1, 5)$ and has gradient function $\frac{dy}{dx} = 3x^2 - 2x + 1$.

Find the equation of the curve. [3]

 
 
 
 
 
 

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14. Find $\displaystyle \int \sin^2 x \, dx$. [3]

 
 
 
 
 
 

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15. Evaluate $\displaystyle \int_0^1 \frac{2x}{x^2 + 1} \, dx$. [3]

 
 
 
 
 
 

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Section D: Applications of Integration (15 marks)

Answer ALL questions in this section.

16. The diagram shows the curve $y = 4x - x^2$ and the line $y = 3$.

(a) Find the coordinates of the points $A$ and $B$ where the curve meets the line. [3]

 
 
 
 
 

(b) Find the area of the shaded region bounded by the curve and the line. [5]

 
 
 
 
 
 
 
 

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17. A curve has equation $y = \frac{1}{x^2}$ for $x \geq 1$. Find the volume of revolution when the region bounded by the curve, the $x$-axis, and the lines $x = 1$ and $x = 2$ is rotated completely about the $x$-axis. [4]

 
 
 
 
 
 
 

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18. The velocity $v \text{ m/s}$ of a particle at time $t$ seconds is given by $v = 3t^2 - 4t + 1$. Find the distance travelled by the particle between $t = 1$ and $t = 3$. [3]

 
 
 
 
 
 

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19. Find the area of the region bounded by the curve $y = \sin x$, the $x$-axis, and the lines $x = 0$ and $x = \pi$. [3]

 
 
 
 
 
 

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20. The region bounded by the curve $y = \sqrt{x}$, the $x$-axis, and the line $x = 4$ is rotated completely about the $x$-axis. Find the volume of the solid formed. [3]

 
 
 
 
 
 

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END OF QUIZ

Check your work carefully.

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Secondary 4 Additional Mathematics Quiz - Calculus — ANSWER KEY

Total Marks: 60

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Section A: Differentiation Techniques (15 marks)

1. (a) $y = 3x^4 - 2x^3 + 5x - 7$ [2 marks]

$\frac{dy}{dx} = 12x^3 - 6x^2 + 5$

Marking: M1 for correct differentiation of at least 3 terms; A1 for fully correct answer. Deduct 1 mark for each error.

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1. (b) $y = (2x + 1)(x^2 - 3)$ [3 marks]

Method 1 (Product Rule): Let $u = 2x + 1$, $v = x^2 - 3$ $\frac{du}{dx} = 2$, $\frac{dv}{dx} = 2x$

$\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx} = (2x + 1)(2x) + (x^2 - 3)(2)$ $= 4x^2 + 2x + 2x^2 - 6 = 6x^2 + 2x - 6$

Method 2 (Expand first): $y = 2x^3 + x^2 - 6x - 3$ $\frac{dy}{dx} = 6x^2 + 2x - 6$

Marking: M1 for product rule or expansion; M1 for correct application; A1 for simplified answer.

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2. (a) $y = \frac{x^2 + 1}{2x - 1}$ [3 marks]

Using Quotient Rule: $u = x^2 + 1$, $v = 2x - 1$ $\frac{du}{dx} = 2x$, $\frac{dv}{dx} = 2$

$\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} = \frac{(2x - 1)(2x) - (x^2 + 1)(2)}{(2x - 1)^2}$ $= \frac{4x^2 - 2x - 2x^2 - 2}{(2x - 1)^2} = \frac{2x^2 - 2x - 2}{(2x - 1)^2}$

Marking: M1 for quotient rule formula; M1 for correct substitution; A1 for simplified numerator.

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2. (b) $y = \sqrt{3x^2 + 4} = (3x^2 + 4)^{\frac{1}{2}}$ [2 marks]

Using Chain Rule: $\frac{dy}{dx} = \frac{1}{2}(3x^2 + 4)^{-\frac{1}{2}} \cdot 6x = \frac{3x}{\sqrt{3x^2 + 4}}$

Marking: M1 for chain rule with correct derivative of inside function; A1 for simplified answer.

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3. $y = \frac{4}{x} + \ln(2x) = 4x^{-1} + \ln(2x)$, $x > 0$ [3 marks]

$\frac{dy}{dx} = -4x^{-2} + \frac{1}{x} = -\frac{4}{x^2} + \frac{1}{x}$

At $x = 2$: $\frac{dy}{dx}\bigg|_{x=2} = -\frac{4}{4} + \frac{1}{2} = -1 + \frac{1}{2} = -\frac{1}{2}$

Marking: M1 for derivative of $4/x$; M1 for derivative of $\ln(2x)$; A1 for correct gradient $-\frac{1}{2}$.

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4. $y = e^{2x} \sin x$ [2 marks]

Using Product Rule: $u = e^{2x}$, $v = \sin x$ $\frac{du}{dx} = 2e^{2x}$, $\frac{dv}{dx} = \cos x$

$\frac{dy}{dx} = e^{2x} \cos x + \sin x \cdot 2e^{2x} = e^{2x}(\cos x + 2\sin x)$

Marking: M1 for product rule with correct derivatives; A1 for simplified factored form.

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5. $y = \frac{\cos x}{1 + \sin x}$ [3 marks]

Using Quotient Rule: $u = \cos x$, $v = 1 + \sin x$ $\frac{du}{dx} = -\sin x$, $\frac{dv}{dx} = \cos x$

$\frac{dy}{dx} = \frac{(1 + \sin x)(-\sin x) - \cos x(\cos x)}{(1 + \sin x)^2}$ $= \frac{-\sin x - \sin^2 x - \cos^2 x}{(1 + \sin x)^2}$ $= \frac{-\sin x - (\sin^2 x + \cos^2 x)}{(1 + \sin x)^2}$ $= \frac{-\sin x - 1}{(1 + \sin x)^2} = -\frac{1 + \sin x}{(1 + \sin x)^2} = -\frac{1}{1 + \sin x}$

Marking: M1 for quotient rule; M1 for correct simplification using identity; A1 for final answer.

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Section B: Applications of Differentiation (15 marks)

6. (a) $y = x^3 - 6x^2 + 9x + 1$ [4 marks]

$\frac{dy}{dx} = 3x^2 - 12x + 9$

For stationary points, $\frac{dy}{dx} = 0$: $3x^2 - 12x + 9 = 0$ $x^2 - 4x + 3 = 0$ $(x - 1)(x - 3) = 0$ $x = 1 \text{ or } x = 3$

When $x = 1$: $y = 1 - 6 + 9 + 1 = 5$ → $(1, 5)$ When $x = 3$: $y = 27 - 54 + 27 + 1 = 1$ → $(3, 1)$

Stationary points: $(1, 5)$ and $(3, 1)$

Marking: M1 for differentiation; M1 for setting to zero; M1 for solving quadratic; A1 for both coordinates.

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6. (b) [3 marks]

$\frac{d^2y}{dx^2} = 6x - 12$

At $x = 1$: $\frac{d^2y}{dx^2} = 6(1) - 12 = -6 < 0$ → Maximum point

At $x = 3$: $\frac{d^2y}{dx^2} = 6(3) - 12 = 6 > 0$ → Minimum point

Marking: M1 for second derivative; M1 for evaluating at both points; A1 for correct nature of both points.

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7. (a) $y = 2x^3 - 3x^2 - 12x + 7$ [3 marks]

$\frac{dy}{dx} = 6x^2 - 6x - 12$

Curve is increasing when $\frac{dy}{dx} > 0$: $6x^2 - 6x - 12 > 0$ $x^2 - x - 2 > 0$ $(x - 2)(x + 1) > 0$

Solution: $x < -1$ or $x > 2$

Marking: M1 for differentiation; M1 for factorising and solving inequality; A1 for correct intervals.

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7. (b) [3 marks]

At $x = 1$: $y = 2(1)^3 - 3(1)^2 - 12(1) + 7 = 2 - 3 - 12 + 7 = -6$

Gradient at $x = 1$: $\frac{dy}{dx} = 6(1)^2 - 6(1) - 12 = 6 - 6 - 12 = -12$

Equation of tangent: $y - (-6) = -12(x - 1)$ $y + 6 = -12x + 12$ $y = -12x + 6$

Marking: M1 for finding point; M1 for finding gradient; A1 for correct equation.

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8. (a) [2 marks]

Volume: $V = x^2 h = 500$ → $h = \frac{500}{x^2}$

Surface area (open top): $A = x^2 + 4xh$ $A = x^2 + 4x\left(\frac{500}{x^2}\right) = x^2 + \frac{2000}{x}$

Marking: M1 for expressing $h$ in terms of $x$; A1 for showing $A = x^2 + \frac{2000}{x}$.

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9. $y = x^3 - 3x^2 + 2$ [3 marks]

$\frac{dy}{dx} = 3x^2 - 6x$ $\frac{d^2y}{dx^2} = 6x - 6$

For point of inflection, $\frac{d^2y}{dx^2} = 0$: $6x - 6 = 0 \implies x = 1$

Check sign change of $\frac{d^2y}{dx^2}$: For $x < 1$, e.g. $x = 0$: $\frac{d^2y}{dx^2} = -6 < 0$ For $x > 1$, e.g. $x = 2$: $\frac{d^2y}{dx^2} = 6 > 0$ Sign changes, so point of inflection at $x = 1$.

When $x = 1$: $y = 1 - 3 + 2 = 0$

Point of inflection: $(1, 0)$

Marking: M1 for second derivative; M1 for setting to zero and solving; A1 for coordinates with justification.

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10. [3 marks]

$A = \pi r^2$, $\frac{dr}{dt} = 0.5 \text{ cm/s}$

$\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$

When $r = 4$: $\frac{dA}{dt} = 2\pi (4)(0.5) = 4\pi \text{ cm}^2\text{/s}$

Marking: M1 for $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$; M1 for substitution; A1 for $4\pi \text{ cm}^2\text{/s}$.

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Section C: Integration (15 marks)

11. (a) $\displaystyle \int (6x^2 - 4x + 3) \, dx$ [2 marks]

$= 6 \cdot \frac{x^3}{3} - 4 \cdot \frac{x^2}{2} + 3x + C$ $= 2x^3 - 2x^2 + 3x + C$

Marking: M1 for integrating at least 2 terms correctly; A1 for fully correct answer including constant.

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11. (b) $\displaystyle \int \left( \frac{2}{x} + 3e^x \right) dx$, $x > 0$ [2 marks]

$= 2\ln|x| + 3e^x + C$ $= 2\ln x + 3e^x + C \quad (\text{since } x > 0)$

Marking: M1 for correct integration of one term; A1 for fully correct including constant.

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11. (c) $\displaystyle \int (2x - 1)^4 \, dx$ [3 marks]

Let $u = 2x - 1$, $\frac{du}{dx} = 2$, $dx = \frac{du}{2}$

$\int (2x - 1)^4 \, dx = \int u^4 \cdot \frac{du}{2} = \frac{1}{2} \cdot \frac{u^5}{5} + C$ $= \frac{(2x - 1)^5}{10} + C$

Marking: M1 for substitution or recognising chain rule; M1 for correct integration; A1 for simplified answer.

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12. (a) $\displaystyle \int_1^4 \left( \sqrt{x} + \frac{1}{x^2} \right) dx$ [4 marks]

$= \int_1^4 \left( x^{\frac{1}{2}} + x^{-2} \right) dx$ $= \left[ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} + \frac{x^{-1}}{-1} \right]_1^4$ $= \left[ \frac{2}{3}x^{\frac{3}{2}} - \frac{1}{x} \right]_1^4$ $= \left( \frac{2}{3}(4)^{\frac{3}{2}} - \frac{1}{4} \right) - \left( \frac{2}{3}(1)^{\frac{3}{2}} - 1 \right)$ $= \left( \frac{2}{3} \cdot 8 - \frac{1}{4} \right) - \left( \frac{2}{3} - 1 \right)$ $= \left( \frac{16}{3} - \frac{1}{4} \right) - \left( -\frac{1}{3} \right)$ $= \frac{16}{3} - \frac{1}{4} + \frac{1}{3} = \frac{17}{3} - \frac{1}{4}$ $= \frac{68 - 3}{12} = \frac{65}{12}$

Marking: M1 for correct integration; M1 for correct limits substitution; M1 for correct evaluation; A1 for $\frac{65}{12}$.

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12. (b) $\displaystyle \int_0^{\frac{\pi}{6}} \cos 2x \, dx$ [3 marks]

$= \left[ \frac{\sin 2x}{2} \right]_0^{\frac{\pi}{6}}$ $= \frac{\sin(\frac{\pi}{3})}{2} - \frac{\sin(0)}{2}$ $= \frac{\frac{\sqrt{3}}{2}}{2} - 0 = \frac{\sqrt{3}}{4}$

Marking: M1 for correct integration; M1 for correct substitution of limits; A1 for $\frac{\sqrt{3}}{4}$.

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13. $\frac{dy}{dx} = 3x^2 - 2x + 1$, passes through $(1, 5)$ [3 marks]

$y = \int (3x^2 - 2x + 1) \, dx = x^3 - x^2 + x + C$

Substitute $(1, 5)$: $5 = 1^3 - 1^2 + 1 + C$ $5 = 1 - 1 + 1 + C$ $C = 4$

Equation: $y = x^3 - x^2 + x + 4$

Marking: M1 for integration; M1 for using point to find $C$; A1 for correct equation.

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14. $\displaystyle \int \sin^2 x \, dx$ [3 marks]

Using identity $\sin^2 x = \frac{1 - \cos 2x}{2}$:

$\int \sin^2 x \, dx = \int \frac{1 - \cos 2x}{2} \, dx$ $= \frac{1}{2} \int (1 - \cos 2x) \, dx$ $= \frac{1}{2} \left( x - \frac{\sin 2x}{2} \right) + C$ $= \frac{x}{2} - \frac{\sin 2x}{4} + C$

Marking: M1 for using correct identity; M1 for integration; A1 for final answer with constant.

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15. $\displaystyle \int_0^1 \frac{2x}{x^2 + 1} \, dx$ [3 marks]

Let $u = x^2 + 1$, $\frac{du}{dx} = 2x$, $du = 2x \, dx$

When $x = 0$, $u = 1$; when $x = 1$, $u = 2$

$\int_0^1 \frac{2x}{x^2 + 1} \, dx = \int_1^2 \frac{1}{u} \, du$ $= \left[ \ln|u| \right]_1^2 = \ln 2 - \ln 1 = \ln 2$

Marking: M1 for substitution; M1 for correct limits and integration; A1 for $\ln 2$.

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Section D: Applications of Integration (15 marks)

16. (a) Curve: $y = 4x - x^2$, Line: $y = 3$ [3 marks]

At intersection: $4x - x^2 = 3$ $x^2 - 4x + 3 = 0$ $(x - 1)(x - 3) = 0$ $x = 1$ or $x = 3$

When $x = 1$: $y = 3$ → $A(1, 3)$ When $x = 3$: $y = 3$ → $B(3, 3)$

Marking: M1 for equating; M1 for solving quadratic; A1 for both coordinates.

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16. (b) [5 marks]

Area = $\displaystyle \int_1^3 \left[ (4x - x^2) - 3 \right] dx$

$= \int_1^3 (4x - x^2 - 3) \, dx$ $= \left[ 2x^2 - \frac{x^3}{3} - 3x \right]_1^3$ $= \left( 2(9) - \frac{27}{3} - 9 \right) - \left( 2(1) - \frac{1}{3} - 3 \right)$ $= (18 - 9 - 9) - \left( 2 - \frac{1}{3} - 3 \right)$ $= 0 - \left( -\frac{4}{3} \right) = \frac{4}{3} \text{ square units}$

Marking: M1 for correct integral expression (curve minus line); M1 for integration; M1 for correct limits; M1 for correct evaluation; A1 for $\frac{4}{3}$.

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17. $y = \frac{1}{x^2}$, $x \geq 1$, rotated about $x$-axis from $x = 1$ to $x = 2$ [4 marks]

Volume $V = \pi \int_1^2 y^2 \, dx = \pi \int_1^2 \frac{1}{x^4} \, dx$

$= \pi \int_1^2 x^{-4} \, dx$ $= \pi \left[ \frac{x^{-3}}{-3} \right]_1^2$ $= \pi \left[ -\frac{1}{3x^3} \right]_1^2$ $= \pi \left( -\frac{1}{3(8)} - \left( -\frac{1}{3(1)} \right) \right)$ $= \pi \left( -\frac{1}{24} + \frac{1}{3} \right)$ $= \pi \left( \frac{-1 + 8}{24} \right) = \frac{7\pi}{24} \text{ cubic units}$

Marking: M1 for correct volume formula; M1 for correct integration; M1 for correct limits; A1 for $\frac{7\pi}{24}$.

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18. $v = 3t^2 - 4t + 1$, $t = 1$ to $t = 3$ [3 marks]

Distance = $\displaystyle \int_1^3 |v| \, dt$. First check if $v$ changes sign.

$v = 3t^2 - 4t + 1 = (3t - 1)(t - 1)$ For $1 \leq t \leq 3$, $v \geq 0$ (since $t - 1 \geq 0$ and $3t - 1 > 0$).

Distance = $\displaystyle \int_1^3 (3t^2 - 4t + 1) \, dt$ $= \left[ t^3 - 2t^2 + t \right]_1^3$ $= (27 - 18 + 3) - (1 - 2 + 1)$ $= 12 - 0 = 12 \text{ m}$

Marking: M1 for checking sign or integrating; M1 for correct integration and limits; A1 for 12.

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19. Area bounded by $y = \sin x$, $x$-axis, $x = 0$, $x = \pi$ [3 marks]

Area = $\displaystyle \int_0^\pi \sin x \, dx$ $= \left[ -\cos x \right]_0^\pi$ $= (-\cos \pi) - (-\cos 0)$ $= (-(-1)) - (-1) = 1 + 1 = 2 \text{ square units}$

Marking: M1 for correct integral; M1 for correct integration and limits; A1 for 2.

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20. $y = \sqrt{x}$, rotated about $x$-axis from $x = 0$ to $x = 4$ [3 marks]

Volume $V = \pi \int_0^4 (\sqrt{x})^2 \, dx = \pi \int_0^4 x \, dx$ $= \pi \left[ \frac{x^2}{2} \right]_0^4$ $= \pi \left( \frac{16}{2} - 0 \right) = 8\pi \text{ cubic units}$

Marking: M1 for correct volume formula; M1 for integration and limits; A1 for $8\pi$.

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END OF ANSWER KEY