Free worked solutions · 2026 Edition

IB Maths AA SL Chapter 11: Integration and kinematics worked solutions

Worked solutions to the practice questions in Chapter 11 of The Ultimate Guide for the IB: Mathematics: Analysis and Approaches SL (2026 Edition). Each solution shows every step with the mark it earns, the way an IB mark scheme does.

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Basics

One idea at a time, in the order the chapter teaches them.

Question 1

[8 marks]

Find

(a) \displaystyle\int (9x^2 + 4x - 1)\,dx

[3]

(b) \displaystyle\int (2x^5 - 5x^4 + 3)\,dx

[3]

(c) \displaystyle\int \left(\dfrac{1}{2}x^3 - \dfrac{3}{4}x\right)dx

[2]

Show solutionHide solution

(a) 3x^3 + 2x^2 - x + C A1A1A1

(b) \dfrac{x^6}{3} - x^5 + 3x + C A1A1A1

(c) \dfrac{x^4}{8} - \dfrac{3x^2}{8} + C A1A1

Note: A1 for each correct term. Do not award the final A1 if +C is omitted; penalise this once in the question.

Final answer

(a) 3x^3+2x^2-x+C
(b) \frac{x^6}{3}-x^5+3x+C
(c) \frac{x^4-3x^2}{8}+C.

Question 2

[7 marks]

By first writing each integrand as a sum of integer powers of x, find the following integrals, where x \ne 0.

(a) \displaystyle\int \dfrac{x^4 - 6}{x^2}\,dx

[3]

(b) \displaystyle\int \dfrac{3x^3 + 2x - 5}{x^3}\,dx

[4]

Show solutionHide solution

(a) \dfrac{x^4 - 6}{x^2} = x^2 - 6x^{-2}

\displaystyle\int (x^2 - 6x^{-2})\,dx = \dfrac{x^3}{3} + 6x^{-1} + C A1A1

(b) \dfrac{3x^3 + 2x - 5}{x^3} = 3 + 2x^{-2} - 5x^{-3}

\displaystyle\int (3 + 2x^{-2} - 5x^{-3})\,dx = 3x - 2x^{-1} + \dfrac{5}{2}x^{-2} + C A1A1A1

Note: A1 for each correct term; accept \dfrac{6}{x}, -\dfrac{2}{x}, \dfrac{5}{2x^2}. Do not award the final A1 if +C is omitted; penalise this once in the question.

Final answer

(a) \frac{x^3}{3}+\frac6x+C
(b) 3x-\frac2x+\frac5{2x^2}+C.

Question 3

[4 marks]

A curve has gradient function \dfrac{dy}{dx} = 6x^2 - 2x + 4 and passes through the point (1,\ 9). Find the equation of the curve.

[4]

Show solutionHide solution

y = \displaystyle\int (6x^2 - 2x + 4)\,dx = 2x^3 - x^2 + 4x + C M1A1

substituting (1,\ 9): 9 = 2 - 1 + 4 + C

C = 4, so y = 2x^3 - x^2 + 4x + 4

Note: M1 for an attempt to integrate (power of at least one term increased by 1); A1 for all three terms correct.

Final answer

y=2x^3-x^2+4x+4.

Question 4

[5 marks]

Each statement below gives a proposed antiderivative. By differentiating the right-hand side, determine whether the statement is correct. If it is not correct, write down a correct antiderivative.

(a) \displaystyle\int \sin x\,dx = \cos x + C

[2]

(b) \displaystyle\int \dfrac{2}{x}\,dx = \ln\left( x^2 \right) + C, x > 0

[2]

(c) \displaystyle\int e^{x + 2}\,dx = e^{x + 2} + C

[1]

Show solutionHide solution

(a) \dfrac{d}{dx}(\cos x) = -\sin x \ne \sin x, so the statement is not correct

correct: -\cos x + C

(b) \dfrac{d}{dx}\left[ \ln\left( x^2 \right) \right] = \dfrac{2x}{x^2} (or \ln(x^2) = 2\ln x, derivative \dfrac{2}{x})

= \dfrac{2}{x}, so the statement is correct

(c) \dfrac{d}{dx}\left( e^{x + 2} \right) = e^{x + 2} \times 1 = e^{x + 2}, so the statement is correct

Note: In (c) accept the reason e^{x + 2} = e^2 e^x, a constant multiple of e^x.

Final answer

(a) Incorrect: -\cos x+C
(b), (c) correct.

Question 5

[5 marks]

For a function f, f'(x) = 2 - 5\sin x and f\left( \dfrac{\pi}{2} \right) = 4.

(a) Find f(x).

[4]

(b) Find the exact value of f\left( \dfrac{\pi}{3} \right).

[1]

Show solutionHide solution

(a) f(x) = 2x + 5\cos x + C A1A1

f\left( \dfrac{\pi}{2} \right) = \pi + 5\cos\dfrac{\pi}{2} + C = 4

C = 4 - \pi, so f(x) = 2x + 5\cos x + 4 - \pi

(b) f\left( \dfrac{\pi}{3} \right) = \dfrac{2\pi}{3} + 5 \times \dfrac{1}{2} + 4 - \pi = \dfrac{13}{2} - \dfrac{\pi}{3}

Note: FT their f from (a).

Final answer

(a) 2x+5\cos x+4-\pi
(b) \frac{13}{2}-\frac\pi3.

Question 6

[7 marks]

(a) Show that \displaystyle\int_{0}^{2} \left( e^{x/2} + 1 \right) dx = 2e.

[3]

(b) (i) Show that \dfrac{x}{x - 2} = 1 + \dfrac{2}{x - 2}, x \ne 2.

[1]

(ii) Hence find \displaystyle\int_{3}^{7} \dfrac{x}{x - 2}\,dx, giving your answer in the form a + b\ln 5, where a, b \in \mathbb{Z}.

[3]

Show solutionHide solution

(a) \left[ 2e^{x/2} + x \right]_{0}^{2} A1A1

= (2e + 2) - (2 + 0) = 2e M1AG

Note: A1 for 2e^{x/2}, A1 for x; M1 for substituting both limits.

(b)(i) 1 + \dfrac{2}{x - 2} = \dfrac{(x - 2) + 2}{x - 2} = \dfrac{x}{x - 2} A1AG

(ii) \left[ x + 2\ln(x - 2) \right]_{3}^{7}

= (7 + 2\ln 5) - (3 + 2\ln 1)

= 4 + 2\ln 5

Note: \ln\lvert x - 2 \rvert = \ln(x - 2) here since x > 2 on the interval.

Final answer

(a) 2e
(b)(ii) 4+2\ln5.

Question 7

[5 marks]

It is given that \displaystyle\int_{0}^{k} 2\sin\left( \dfrac{x}{2} \right) dx = 2, where 0 < k < 2\pi. Find the exact value of k.

[5]

Show solutionHide solution

\displaystyle\int 2\sin\left( \dfrac{x}{2} \right) dx = -4\cos\left( \dfrac{x}{2} \right) + C A1A1

Note: A1 for -\cos\left( \dfrac{x}{2} \right), A1 for the factor 4.

\left[ -4\cos\left( \dfrac{x}{2} \right) \right]_{0}^{k} = 4 - 4\cos\left( \dfrac{k}{2} \right) = 2

\cos\left( \dfrac{k}{2} \right) = \dfrac{1}{2}

0 < \dfrac{k}{2} < \pi, so \dfrac{k}{2} = \dfrac{\pi}{3} and k = \dfrac{2\pi}{3}

Final answer

k=\frac{2\pi}{3}.

Question 8

[5 marks]

A particle has velocity v (in m s^{-1}) on a straight track, with v = 9t^2 - 4t + 5. Let s metres denote its position relative to O, and s = 2 when t = 1.

(a) Find s(t).

[4]

(b) Find the displacement of the particle from O when t = 2.

[1]

Show solutionHide solution

(a) s = \displaystyle\int (9t^2 - 4t + 5)\,dt

= 3t^3 - 2t^2 + 5t + C

3 - 2 + 5 + C = 2

C = -4, so s = 3t^3 - 2t^2 + 5t - 4

(b) s(2) = 24 - 8 + 10 - 4 = 22 m

Final answer

s=3t^3-2t^2+5t-4; s(2)=22 m.

Tutorial

Exam-style questions on each question type, with the marks the IB would give.

Question 9

[7 marks]

(a) Find \displaystyle\int \left(e^{2x}+3\right)^2dx.

[3]

(b) Show that \sin^2\!\left(\dfrac{x}{2}\right) = \dfrac12\left(1 - \cos x\right).

[2]

(c) Hence find \displaystyle\int 6\sin^2\!\left(\dfrac{x}{2}\right)dx.

[2]

Show solutionHide solution

(a) Expand first: \left(e^{2x}+3\right)^2 = e^{4x} + 6e^{2x} + 9.

[1]

\displaystyle\int \left(e^{4x} + 6e^{2x} + 9\right)dx = \dfrac14e^{4x} + 3e^{2x} + 9x + c.

[2]

(b) Using \cos 2\theta = 1 - 2\sin^2\theta with \theta = \dfrac x2: \cos x = 1 - 2\sin^2\!\left(\dfrac x2\right).

[1]

Rearranging: 2\sin^2\!\left(\dfrac x2\right) = 1 - \cos x, so \sin^2\!\left(\dfrac x2\right) = \dfrac12(1-\cos x).

[1]

(c) 6\sin^2\!\left(\dfrac x2\right) = 3(1-\cos x), so

[1]

\displaystyle\int 3(1-\cos x)\,dx = 3x - 3\sin x + c.

[1]

Final answer

(a) \frac14e^{4x}+3e^{2x}+9x+C
(c) 3x-3\sin x+C.

Question 10

[6 marks]

Find

(a) \displaystyle\int \dfrac{9x^2}{x^3+2}\,dx, where x > 0,

[3]

(b) \displaystyle\int \dfrac{\sin x}{2-\cos x}\,dx.

[3]

Show solutionHide solution

(a) Let g = x^3 + 2, so g' = 3x^2 and 9x^2 = 3g'.

[1]

\displaystyle\int \dfrac{9x^2}{x^3+2}\,dx = 3\int\dfrac{g'}{g}\,dx = 3\ln|x^3+2| + c.

[1]

For x>0 the bracket is positive, so = 3\ln(x^3+2) + c.

[1]

(b) Let g = 2 - \cos x, so g' = \sin x exactly.

[1]

\displaystyle\int\dfrac{g'}{g}\,dx = \ln|2-\cos x| + c.

[1]

Since 2 - \cos x \ge 1 > 0 for all x, no modulus is needed: \ln(2-\cos x) + c.

[1]

Final answer

(a) 3\ln(x^3+2)+C
(b) \ln(2-\cos x)+C.

Question 11

[5 marks]

The gradient of a curve is \dfrac{dy}{dx} = 6\sin(3x) + 2x. The curve passes through the point \left(\dfrac{\pi}{3},\ \dfrac{\pi^2}{9} + 1\right).

Find the equation of the curve.

[5]

Show solutionHide solution

y = \displaystyle\int(6\sin 3x + 2x)\,dx = -\dfrac{6}{3}\cos 3x + x^2 + c

[1]

= -2\cos 3x + x^2 + c.

[2]

At x = \dfrac\pi3: -2\cos\pi + \dfrac{\pi^2}{9} + c = \dfrac{\pi^2}{9} + 1, i.e.; 2 + \dfrac{\pi^2}{9} + c = \dfrac{\pi^2}{9} + 1.

[1]

So c = -1 and y = x^2 - 2\cos 3x - 1.

[1]

Final answer

y=x^2-2\cos3x-1.

Question 12

[7 marks]

Grain is poured into a silo. At time t minutes the mass of grain in the silo, M tonnes, changes at the rate

\dfrac{dM}{dt} = 25 + 12e^{-0.15t}.

When t = 0 the silo holds 80 tonnes.

(a) Find an expression for M in terms of t.

[4]

(b) Find the mass of grain in the silo after 10 minutes.

[1]

(c) The silo is full when it holds 500 tonnes. Find the time at which the silo becomes full.

[2]

Show solutionHide solution

(a) M = \displaystyle\int\left(25 + 12e^{-0.15t}\right)dt = 25t + \dfrac{12}{-0.15}e^{-0.15t} + c

[2]

= 25t - 80e^{-0.15t} + c.

M(0) = 80: -80 + c = 80, so c = 160.

[1]

M = 25t - 80e^{-0.15t} + 160.

[1]

(b) M(10) = 250 - 80e^{-1.5} + 160 = 392.15\ldots \approx 392 tonnes.

[1]

(c) Solve 25t - 80e^{-0.15t} + 160 = 500 (GDC: graph y = 25t - 80e^{-0.15t} + 160 and y = 500, then Analyze Graph, Intersection).

[1]

t = 13.99\ldots \approx 14.0 minutes.

[1]

Final answer

(a) M=25t-80e^{-0.15t}+160
(b) 392 tonnes
(c) 14.0 min.

Question 13

[5 marks]

The functions f and g satisfy

\int_0^3 f(x)\,dx = 8,\quad\int_3^7 f(x)\,dx = -5.

\int_0^7 g(x)\,dx = 4.

Find

(a) \displaystyle\int_0^7 f(x)\,dx,

[1]

(b) \displaystyle\int_7^3 f(x)\,dx,

[1]

(c) \displaystyle\int_0^7 \bigl(2f(x) - g(x) + 3\bigr)\,dx.

[3]

Show solutionHide solution

(a) Splitting at 3: \displaystyle\int_0^7 f\,dx = 8 + (-5) = 3.

[1]

(b) Swapping the limits changes the sign: \displaystyle\int_7^3 f\,dx = -\int_3^7 f\,dx = 5.

[1]

(c) \displaystyle\int_0^7 \bigl(2f - g + 3\bigr)dx = 2\int_0^7 f\,dx - \int_0^7 g\,dx + \int_0^7 3\,dx

[1]

= 2(3) - 4 + [3x]_0^7

[1]

= 6 - 4 + 21 = 23.

[1]

Final answer

(a) 3
(b) 5
(c) 23.

Question 14

[6 marks]

(a) Show that \dfrac{\left(\sqrt{x}+1\right)^2}{x} = 1 + 2x^{-1/2} + \dfrac1x.

[2]

(b) Hence find the exact value of \displaystyle\int_1^4\dfrac{\left(\sqrt{x}+1\right)^2}{x}\,dx, expressing the result as a + b\ln 2.

[4]

Show solutionHide solution

(a) \left(\sqrt{x}+1\right)^2 = x + 2\sqrt{x} + 1.

[1]

Dividing each term by x: 1 + \dfrac{2\sqrt{x}}{x} + \dfrac1x = 1 + 2x^{-1/2} + \dfrac1x.

[1]

(b) \displaystyle\int_1^4\left(1 + 2x^{-1/2} + \dfrac1x\right)dx = \Bigl[x + 4\sqrt{x} + \ln x\Bigr]_1^4

[2]

= (4 + 8 + \ln4) - (1 + 4 + 0)

[1]

= 7 + \ln 4 = 7 + 2\ln 2.

[1]

Final answer

(b) 7+2\ln2.

Question 15

[6 marks]

(a) Use your GDC to find \displaystyle\int_0^2 e^{-x^2/2}\,dx.

[2]

(b) Find the value of k > 0 such that \displaystyle\int_0^k e^{-x^2/2}\,dx = 1.

[3]

(c) Hence, or otherwise, write down the value of \displaystyle\int_{-2}^{2} e^{-x^2/2}\,dx.

[1]

Show solutionHide solution

(a) GDC: Menu, Calculus, Numerical Integral with lower limit 0, upper limit 2, integrand e^{-x^2/2}, variable x.

[1]

\displaystyle\int_0^2 e^{-x^2/2}\,dx = 1.1962\ldots \approx 1.20.

[1]

(b) Solve \displaystyle\int_0^k e^{-x^2/2}\,dx = 1.

[1]

GDC: graph y = \displaystyle\int_0^x e^{-t^2/2}\,dt and y = 1 (or use solve with the numerical integral inside) and find the intersection.

[1]

k = 1.28 (3 s.f.).

[1]

(c) The integrand is symmetric about the y-axis (e^{-(-x)^2/2} = e^{-x^2/2}), so the integral is 2\times 1.1962\ldots = 2.39.

[1]

Final answer

(a) 1.20
(b) 1.28
(c) 2.39.

Question 16

[5 marks]

The curves y = 1 - x^2 and y = x^2 - 7 enclose a region R, part of which lies below the x-axis.

(a) Find the x-coordinates where the curves meet.

[2]

(b) Find the area of R.

[3]

Show solutionHide solution

(a) 1 - x^2 = x^2 - 7 \Rightarrow 2x^2 = 8.

[1]

x = -2 and x = 2.

[1]

(b) On [-2,2] the curve y = 1-x^2 is on top, so the integrand is top minus bottom: (1-x^2) - (x^2-7) = 8 - 2x^2.

[1]

Area = \displaystyle\int_{-2}^{2}(8 - 2x^2)\,dx = 2\int_0^2(8-2x^2)\,dx = 2\Bigl[8x - \tfrac23x^3\Bigr]_0^2

[1]

= 2\left(16 - \tfrac{16}{3}\right) = \dfrac{64}{3}.

[1]

(Top minus bottom is valid even though part of the region lies below the x-axis.)

Final answer

(a) -2,2
(b) \frac{64}{3}.

Question 17

[9 marks]

A particle moves in a straight line. Its acceleration is a = 2t - 8 m s^{-2} for 0 \le t \le 6, where t is in seconds. When t = 0 its velocity is 12 m s^{-1} and it is at the origin O.

(a) Find an expression for the velocity v(t).

[2]

(b) Find the times at which the particle is at rest.

[2]

(c) Find an expression for the displacement s(t), and hence show that the particle is back at O when t = 6.

[3]

(d) Calculate the total distance covered during 0 \le t \le 6.

[2]

Show solutionHide solution

(a) v = \displaystyle\int(2t - 8)\,dt = t^2 - 8t + c; v(0) = 12 gives c = 12.

[1]

v(t) = t^2 - 8t + 12.

[1]

(b) t^2 - 8t + 12 = (t-2)(t-6) = 0.

[1]

t = 2 and t = 6.

[1]

(c) s = \displaystyle\int(t^2 - 8t + 12)\,dt = \dfrac{t^3}{3} - 4t^2 + 12t + c; s(0)=0 gives c = 0.

[2]

s(6) = 72 - 144 + 72 = 0, so the particle is back at O.

[1]

(d) s(2) = \dfrac83 - 16 + 24 = \dfrac{32}{3}; the particle moves forward \dfrac{32}{3} m to t=2, then returns the same distance to O by t = 6 (no further change of direction).

[1]

Distance = \dfrac{32}{3} + \dfrac{32}{3} = \dfrac{64}{3} m.

[1]

Final answer

(a) v=t^2-8t+12
(b) 2,6 s
(c) s=\frac{t^3}{3}-4t^2+12t
(d) \frac{64}{3} m.

Question 18

[8 marks]

A robot moves along a straight rail. Its velocity at time t seconds is v=8\sin(0.7t)-0.9t m s^{-1}, for 0\le t\le8.

(a) Find the first time t>0 at which the robot is at rest.

[2]

(b) Find the acceleration of the robot when t=3.

[2]

(c) Find the total distance travelled by the robot in the first 6 seconds.

[2]

(d) Find the greatest speed of the robot during the 8 seconds.

[2]

Show solutionHide solution

(a) Solve 8\sin(0.7t)-0.9t=0 with t>0: graph y=v on the GDC (radians) and find the first zero after t=0.

[1]

t=3.85 s.

[1]

(b) a=\dfrac{\mathrm{d}v}{\mathrm{d}t}=5.6\cos(0.7t)-0.9.

[1]

a(3)=5.6\cos2.1-0.9=-3.73 m s^{-2}.

[1]

(c) Distance =\displaystyle\int_{0}^{6}|v|\,\mathrm{d}t, entered on the GDC with the absolute value (or split at t=3.848).

[1]

Distance =29.3 m.

[1]

(d) The greatest speed is the largest value of |v|. Minimum of v on the GDC: v=-14.16\ldots at t=6.96 (where a=0); the end values are v(0)=0 and v(8)=-12.25\ldots, and the maximum of v is much smaller in size.

[1]

Greatest speed =14.2 m s^{-1}.

[1]

Final answer

(a) 3.85 s
(b) -3.73 m s^{-2}
(c) 29.3 m
(d) 14.2 m s^{-1}.

Mastery

Full exam questions that mix question types, at Paper 1 and Paper 2 standard.

Question 19

[8 marks]

Kai and Lena move along the same straight path for 0 \le t \le 15, where t is the time in seconds. Kai's velocity is v_K = 5 - 0.2t m s^{-1} and Lena's velocity is v_L = 8e^{-0.3t} m s^{-1}. At t = 0 Lena is 10 m ahead of Kai, and both are moving in the same direction.

(a) Find the time at which Kai and Lena have the same speed.

[2]

(b) Find how far Lena is ahead of Kai at this time.

[3]

(c) Find the time at which Kai catches Lena.

[3]

Show solutionHide solution

(a) Solve 5 - 0.2t = 8e^{-0.3t} (GDC: intersection of the two velocity graphs).

[1]

t = 1.82 s (3 s.f.).

[1]

(b) Gap = 10 + \displaystyle\int_0^{t_1}\bigl(v_L - v_K\bigr)dt with t_1 = 1.8184\ldots (Lena is faster until then, so the gap grows).

[1]

GDC: \displaystyle\int_0^{1.8184}\bigl(8e^{-0.3t} - 5 + 0.2t\bigr)dt = 2.451\ldots.

[1]

Lena is 10 + 2.45 = 12.5 m ahead (3 s.f.).

[1]

(c) Kai catches Lena when the extra distance Kai has travelled equals the initial gap: \displaystyle\int_0^{T}\bigl(v_K - v_L\bigr)dt = 10.

[1]

GDC: graph y = \displaystyle\int_0^x(v_K - v_L)\,dt and y = 10, intersection (or solve with the integral inside).

[1]

T = 8.24 s (3 s.f.).

[1]

Final answer

(a) 1.82 s
(b) 12.5 m
(c) 8.24 s.

Question 20

[5 marks]

At each point of a curve, \dfrac{dy}{dx} = \dfrac{12x^2}{(x^3+1)^2} for x > 0. The curve passes through the point (1, 5).

Determine its equation.

[5]

Show solutionHide solution

With g = x^3 + 1, g' = 3x^2 and 12x^2 = 4g'.

[1]

y = 4\displaystyle\int g^{-2}g'\,dx = 4\cdot\dfrac{g^{-1}}{-1} + c

[1]

= -\dfrac{4}{x^3+1} + c.

[1]

At (1,5): -\dfrac42 + c = 5, so c = 7.

[2]

y = 7 - \dfrac{4}{x^3+1}.

Final answer

y=7-\frac4{x^3+1}.

Question 21

[5 marks]

Find the exact value of \displaystyle\int_0^{\pi/3}6\sin x\cos^2x\,dx.

[5]

Show solutionHide solution

Let g = \cos x, so g' = -\sin x and 6\sin x = -6g'.

[1]

\displaystyle\int 6\sin x\cos^2x\,dx = -6\cdot\dfrac{\cos^3x}{3} = -2\cos^3x.

[1]

\Bigl[-2\cos^3x\Bigr]_0^{\pi/3} = -2\left(\tfrac12\right)^3 - \left(-2\cdot1\right)

[2]

= -\dfrac14 + 2 = \dfrac74.

[1]

Final answer

\frac74.

Question 22

[11 marks]

After a volcanic eruption begins, ash is released at the rate r(t) = 6\left(e^{-0.15t} - e^{-0.8t}\right) thousand tonnes per hour, where t \ge 0 is the time in hours after the eruption began.

(a) Find the greatest rate at which ash is released and the time at which it occurs.

[2]

The total mass of ash released in the first T hours is A(T) = \displaystyle\int_0^T r(t)\,dt thousand tonnes.

(b) Find A(24).

[2]

(c) Show that A(T) = 40\left(1 - e^{-0.15T}\right) - 7.5\left(1 - e^{-0.8T}\right).

[3]

(d) Hence find the value that A(T) approaches as T \to \infty, and comment on your answer to part (b).

[2]

(e) Find the value of T for which A(T) is 95\% of the value found in part (d).

[2]

Show solutionHide solution

(a) GDC: graph r(t) and use Analyze Graph, Maximum.

[1]

Greatest rate 3.31 thousand tonnes per hour at t = 2.58 hours.

[1]

(b) A(24) = \displaystyle\int_0^{24}6\left(e^{-0.15t} - e^{-0.8t}\right)dt (GDC Numerical Integral).

[1]

= 31.4 thousand tonnes (3 s.f.).

[1]

(c) \displaystyle\int_0^T6e^{-0.15t}\,dt = \Bigl[-40e^{-0.15t}\Bigr]_0^T = 40\left(1 - e^{-0.15T}\right).

[2]

\displaystyle\int_0^T6e^{-0.8t}\,dt = \Bigl[-7.5e^{-0.8t}\Bigr]_0^T = 7.5\left(1 - e^{-0.8T}\right), and subtracting gives the result.

[1]

(d) As T\to\infty, both exponentials tend to 0, so A(T) \to 40 - 7.5 = 32.5.

[1]

A(24) = 31.4 is about 97\% of this limit, so almost all of the ash has been released within 24 hours.

[1]

(e) Solve 40\left(1 - e^{-0.15T}\right) - 7.5\left(1 - e^{-0.8T}\right) = 0.95\times 32.5 (GDC: intersection).

[1]

T = 21.4 hours (3 s.f.).

[1]

Final answer

(a) 3.31 thousand tonnes per hour at t=2.58 h
(b) 31.4 thousand tonnes
(d) 32.5 thousand tonnes, about 97\% released by 24 h
(e) T=21.4 h.

Question 23

[9 marks]

A particle moves in a straight line from the origin. Its velocity is v(t) = 4\sin(0.9t) + 1.5 m s^{-1} for 0 \le t \le 10.

(a) Find the earliest zero of the velocity.

[2]

(b) Calculate the distance covered during the full 10 seconds.

[2]

(c) Find the first time at which the total distance travelled by the particle is exactly 30 m.

[3]

(d) Find the displacement of the particle from the origin at the time found in part (c).

[2]

Show solutionHide solution

(a) Solve 4\sin(0.9t) + 1.5 = 0 (GDC, radians: Zero of the graph).

[1]

t = 3.92 s (3 s.f.).

[1]

(b) The velocity changes sign at t = 3.918 and t = 6.554, so use \displaystyle\int_0^{10}|v|\,dt.

[1]

= 32.1 m (3 s.f.).

[1]

(c) The distance travelled by time T is D(T) = \displaystyle\int_0^T|v|\,dt; solve D(T) = 30.

[1]

GDC: graph y = \displaystyle\int_0^x|v(t)|\,dt and y = 30, then Analyze Graph, Intersection (the value D(10) = 32.1 > 30 shows the solution lies inside the interval).

[1]

T = 9.47 s (3 s.f.).

[1]

(d) Displacement = \displaystyle\int_0^{9.4748}v\,dt (the signed integral, not the modulus)

[1]

= 21.4 m (3 s.f.), which is less than 30 m because the particle moved backwards during 3.92<t<6.55.

[1]

Final answer

(a) 3.92 s
(b) 32.1 m
(c) 9.47 s
(d) 21.4 m.

Question 24

[15 marks]

Let f(x) = 2xe^{-x^2} for x \ge 0.

(a) Show that f'(x) = 2\left(1 - 2x^2\right)e^{-x^2}.

[3]

(b) Find the coordinates of the maximum point of the graph of f.

[3]

(c) Show that f''(x) = 4x\left(2x^2 - 3\right)e^{-x^2}.

[3]

(d) Hence find the inflexion point's x-coordinate and justify the change in concavity.

[2]

(e) Find \displaystyle\int f(x)\,dx.

[2]

(f) Region R lies under y=f(x) and above y=0, with 0 \le x \le 1. Calculate its exact area.

[2]

Show solutionHide solution

(a) Product rule with u = 2x, v = e^{-x^2}, v' = -2xe^{-x^2} (chain rule).

[1]

f'(x) = 2e^{-x^2} + 2x\left(-2xe^{-x^2}\right)

[1]

= 2e^{-x^2}\left(1 - 2x^2\right).

[1]

(b) f'(x) = 0 \Rightarrow 1 - 2x^2 = 0 (as e^{-x^2}>0), so x = \dfrac{1}{\sqrt2} (as x\ge0).

[2]

f\left(\tfrac{1}{\sqrt2}\right) = \dfrac{2}{\sqrt2}e^{-1/2} = \sqrt2\,e^{-1/2}. The maximum point is \left(\dfrac{1}{\sqrt2},\ \sqrt{\dfrac2e}\right).

[1]

(c) Differentiate f'(x) = (2 - 4x^2)e^{-x^2} with the product rule: f''(x) = -8xe^{-x^2} + (2 - 4x^2)\left(-2xe^{-x^2}\right).

[2]

= e^{-x^2}\left(-8x - 4x + 8x^3\right) = e^{-x^2}\left(8x^3 - 12x\right) = 4x\left(2x^2-3\right)e^{-x^2}.

[1]

(d) f''(x) = 0 with x>0 gives 2x^2 = 3, so x = \sqrt{\dfrac32}.

[1]

For 0<x<\sqrt{3/2}, f''<0, and for x>\sqrt{3/2}, f''>0: the sign of f'' changes, so it is a point of inflexion.

[1]

(e) Let g = -x^2, g' = -2x, so 2x = -g' and \displaystyle\int 2xe^{-x^2}dx = -\int e^{g}g'\,dx.

[1]

= -e^{-x^2} + c.

[1]

(f) Area = \displaystyle\int_0^1 2xe^{-x^2}dx = \Bigl[-e^{-x^2}\Bigr]_0^1

[1]

= -e^{-1} + 1 = 1 - \dfrac1e.

[1]

Final answer

(b) (\frac1{\sqrt2},\sqrt{\frac2e})
(d) \sqrt{\frac32}, with sign change
(e) -e^{-x^2}+C
(f) 1-\frac1e.

Going further

Beyond the syllabus: the worked examples and investigations of the chapter's Going further pages.

G1 Sums squeezed by integrals

G1 Worked example

[10 marks]

For n\in\mathbb{Z}^+, let S_n=\displaystyle\sum_{r=1}^{n}\frac{1}{\sqrt r}.

(a) By comparing areas of rectangles with areas under the curve y=\dfrac{1}{\sqrt x}, explain why \displaystyle\int_r^{r+1}\frac{dx}{\sqrt x}<\frac{1}{\sqrt r}<\int_{r-1}^{r}\frac{dx}{\sqrt x} for every integer r\ge2.

[2]

(b) Hence show that 2\sqrt{n+1}-2<S_n<2\sqrt n-1 for n\ge2.

[4]

(c) Use (b) to find an interval of width less than 1 that contains S_{10\,000}, with bounds to three decimal places, rounded outwards.

[2]

(d) You may use that a quantity trapped between two expressions approaching the same value approaches that value. Find \displaystyle\lim_{n\to\infty}\frac{S_n}{\sqrt n}.

[2]

Show solutionHide solution

(a) \dfrac{1}{\sqrt x} is decreasing, so on [r,r+1] the curve lies below the height \dfrac1{\sqrt r}, and on [r-1,r] it lies above it, except at one end

each rectangle has width 1 and area \dfrac{1}{\sqrt r}; the first area under the curve is smaller and the second is larger

(b) lower: add the left inequality for r=1,2,\dots,n (it also holds for r=1): S_n>\displaystyle\int_1^{n+1}\frac{dx}{\sqrt x}

=\Bigl[2\sqrt x\Bigr]_1^{n+1}=2\sqrt{n+1}-2

upper: keep the first term and add the right inequality for r=2,\dots,n: S_n<1+\displaystyle\int_1^{n}\frac{dx}{\sqrt x}

=1+2\sqrt n-2=2\sqrt n-1

A1AG

(c) n=10\,000: 2\sqrt{10\,001}-2=198.0099\ldots and 2\sqrt{10\,000}-1=199

198.009<S_{10\,000}<199.000, an interval of width 0.991<1

(d) divide (b) by \sqrt n: \dfrac{2\sqrt{n+1}-2}{\sqrt n}<\dfrac{S_n}{\sqrt n}<2-\dfrac{1}{\sqrt n}

\dfrac{2\sqrt{n+1}-2}{\sqrt n}=2\sqrt{1+\dfrac1n}-\dfrac{2}{\sqrt n}\to2 and 2-\dfrac1{\sqrt n}\to2, so \dfrac{S_n}{\sqrt n}\to2

Final answer

(c) 198.009<S_{10\,000}<199.000
(d) 2

G1 Investigation

[12 marks]

For n\in\mathbb{Z}^+, let H_n=1+\dfrac12+\dfrac13+\dots+\dfrac1n. You may use the result \ln(n+1)<H_n<1+\ln n for n\ge2, and you may assume that a decreasing sequence whose terms all stay above some fixed number has a limit.

(a) Show that 4.61<H_{100}<5.61.

[2]

(b) Let D_n=H_n-\ln n. Show that D_n-D_{n+1}=\displaystyle\int_n^{n+1}\left(\frac1x-\frac1{n+1}\right)dx, and explain why this is positive. Show also that D_n>0 for every n. Deduce that D_n has a limit \gamma, and that D_n>\gamma for every n.

[4]

(c) Let E_n=H_n-\ln(n+1). Show that E_n is increasing and that D_n-E_n\to0 as n\to\infty. Deduce that E_n<\gamma for every n.

[3]

(d) It is given that \gamma=0.577216, correct to six decimal places. Use (b) and (c) to show that H_n>10 for every n\ge12\,367 and that H_n<10 for every n\le12\,365. Use your GDC to find the least n for which H_n>10.

[3]

Show solutionHide solution

(a) \ln101<H_{100}<1+\ln100

\ln101=4.615\ldots and 1+\ln100=5.605\ldots, so 4.61<H_{100}<5.61

(b) D_n-D_{n+1}=\ln(n+1)-\ln n-\dfrac1{n+1}=\displaystyle\int_n^{n+1}\frac{dx}x-\int_n^{n+1}\frac{dx}{n+1}=\int_n^{n+1}\left(\frac1x-\frac1{n+1}\right)dx

on n<x<n+1, \dfrac1x>\dfrac1{n+1}, so the integrand is positive and D_n>D_{n+1}: the sequence decreases

H_n>\ln(n+1)>\ln n, so D_n>0 (for n=1, D_1=1)

D_n decreases and stays above 0, so it has a limit \gamma; a decreasing sequence lies above its limit, so D_n>\gamma

(c) E_{n+1}-E_n=\dfrac1{n+1}-\ln\dfrac{n+2}{n+1}=\displaystyle\int_{n+1}^{n+2}\left(\frac1{n+1}-\frac1x\right)dx>0, so E_n increases

D_n-E_n=\ln(n+1)-\ln n=\ln\left(1+\dfrac1n\right)\to0, so E_n\to\gamma as well

an increasing sequence lies below its limit, so E_n<\gamma

(d) n\ge12\,367: H_n=D_n+\ln n>\gamma+\ln12\,367>0.5772155+9.4227869>10

n\le12\,365: H_n=E_n+\ln(n+1)<\gamma+\ln12\,366<0.5772165+9.4227061<10

Note: \gamma to six decimal places means 0.5772155\le\gamma<0.5772165; both ends are needed. The bounds leave only n=12\,366 undecided.

GDC: H_{12\,366}=9.99996\ldots<10, so the least n is 12\,367

Final answer

(a) 4.61<H_{100}<5.61
(b), (c) proved in the solutions
(d) 12\,367.

G2 The average value of a function

G2 Worked example

[4 marks]

The average value of f on [a,b] is defined as \bar f=\frac1{b-a}\int_a^b f(x)\,dx. Let f(x)=12\sin x for 0\le x\le\pi.

(a) Find the exact area under the graph.

[2]

(b) Find \bar f as a fraction of the peak height.

[2]

Show solutionHide solution

(a) A=[-12\cos x]_0^\pi

A=24

(b) \bar f=24/\pi

\bar f/12=2/\pi

Final answer

(a) 24
(b) \bar f=24/\pi, or 2/\pi of the peak.

G2 Investigation

[7 marks]

A pump uses power p(t)=b+H\sin(\pi t/T) kilowatts during 0\le t\le T hours, where b,H,T>0. Its average power is defined as \bar p=\frac1T\int_0^T p(t)\,dt.

(a) Show that \bar p=b+2H/\pi.

[3]

(b) Explain why changing T does not change \bar p.

[1]

(c) Take b=4, H=9 and T=6. Find the average power and the energy used, in kWh, correct to three significant figures.

[3]

Show solutionHide solution

(a) \int_0^T p(t)\,dt=[bt-HT\cos(\pi t/T)/\pi]_0^T

=bT+2HT/\pi A1; divide by T: \bar p=b+2H/\pi A1.

(b) The area and the width both scale by T, so their ratio is unchanged.

(c) \bar p=4+18/\pi M1; \bar p=9.73 kW A1; energy =6\bar p=58.4 kWh A1.

Final answer

(a) b+2H/\pi
(b) the factor T cancels
(c) 9.73 kW, 58.4 kWh.

TI-84 Plus CE keystrokes

The GDC panels in the book give keystrokes for the TI-Nspire CX II and the Casio fx-CG50. These are the TI-84 Plus CE keystrokes for the same calculator jobs. The full list of TI-84 keystrokes is in Chapter 12.

Job	TI-84 Plus CE keystrokes
Definite integral	`MATH, 9:fnInt(expression, X, lower, upper)`
Intersections	`Graph screen: Y1 and Y2, 2nd TRACE (CALC), 5:intersect, three ENTERs per crossing`
Solve an equation	`MATH, Numeric Solver (MATH, 0:Solver on older OS versions): enter the equation, then SOLVE. Or graph both sides and use 2nd TRACE (CALC), 5:intersect`

From the book

The Ultimate Guide for the IB: Mathematics: Analysis and Approaches SL

Chapter 11 teaches integration and kinematics from the basics to full exam questions: how the IB asks it, every question type with a marked example, and these practice questions in three steps.

About the book