Free worked solutions · 2026 Edition

Chapter 15: Integration worked solutions

Worked solutions to the practice questions in Chapter 15 of The Ultimate Guide for the IB: Mathematics: Analysis and Approaches HL (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

[6 marks]

A curve has gradient function \dfrac{dy}{dx} = \dfrac{8}{2x - 1}, defined for x > \dfrac{1}{2}, and the point (1,\ 5) lies on it.

(a) Find y in terms of x.

[4]

(b) Find the exact value of x for which y = 9.

[2]

Show solutionHide solution

(a) y = \displaystyle\int \dfrac{8}{2x - 1}\,dx = 4\ln(2x - 1) + C (M1)A1

Note: (M1) for k\ln(2x - 1).

5 = 4\ln 1 + C

C = 5, so y = 4\ln(2x - 1) + 5

(b) 4\ln(2x - 1) + 5 = 9 \Rightarrow \ln(2x - 1) = 1 \Rightarrow 2x - 1 = e

x = \dfrac{e + 1}{2}

Final answer

(a) y=4\ln(2x-1)+5
(b) x=\frac{e+1}{2}

Question 2

[9 marks]

By recognising each integrand as a multiple of g'(x) f(g(x)), find

(a) \displaystyle\int (6x + 3)\,e^{x^2 + x}\,dx

[2]

(b) \displaystyle\int 8\sin^3 x \cos x\,dx

[2]

(c) \displaystyle\int \dfrac{x^2}{x^3 + 4}\,dx, x > 0

[2]

(d) \displaystyle\int \dfrac{\sin\sqrt{x}}{\sqrt{x}}\,dx, x > 0

[2]

(e) \displaystyle\int \dfrac{e^x}{e^x + 5}\,dx

[1]

Show solutionHide solution

(a) g(x) = x^2 + x, g'(x) = 2x + 1, 6x + 3 = 3(2x + 1) (M1)

3e^{x^2 + x} + C

(b) g(x) = \sin x, g'(x) = \cos x: 8 \times \dfrac{\sin^4 x}{4} (M1)

= 2\sin^4 x + C

(c) g(x) = x^3 + 4, g'(x) = 3x^2: \dfrac{1}{3}\displaystyle\int \dfrac{3x^2}{x^3 + 4}\,dx (M1)

= \dfrac{1}{3}\ln(x^3 + 4) + C

(d) g(x) = \sqrt{x}, g'(x) = \dfrac{1}{2\sqrt{x}}, so the integrand is 2g'(x)\sin(g(x)) (M1)

-2\cos\sqrt{x} + C

(e) \ln\left( e^x + 5 \right) + C

Note: No modulus is needed in (c) and (e), as x^3 + 4 > 0 and e^x + 5 > 0. Note: Do not award the final A1 if +C is omitted; penalise this once in the question.

Final answer

(a) 3e^{x^2+x}+C
(b) 2\sin^4x+C
(c) \frac13\ln(x^3+4)+C
(d) -2\cos\sqrt x+C
(e) \ln(e^x+5)+C

Question 3

[8 marks]

(a) (i) Express \dfrac{5x + 6}{(x + 6)(x - 2)} in partial fractions.

[2]

(ii) Hence find \displaystyle\int \dfrac{5x + 6}{(x + 6)(x - 2)}\,dx.

[2]

(b) Use partial fractions to find \displaystyle\int \dfrac{9}{x^2 + x - 20}\,dx, giving your answer as a single logarithm plus a constant.

[4]

Show solutionHide solution

(a)(i) 5x + 6 = A(x - 2) + B(x + 6)

x = -6: -24 = -8A, A = 3; x = 2: 16 = 8B, B = 2, so \dfrac{3}{x + 6} + \dfrac{2}{x - 2}

(ii) 3\ln\lvert x + 6 \rvert + 2\ln\lvert x - 2 \rvert + C A1A1

(b) x^2 + x - 20 = (x - 4)(x + 5)

\dfrac{9}{(x - 4)(x + 5)} = \dfrac{1}{x - 4} - \dfrac{1}{x + 5} M1A1

\displaystyle\int \ldots\,dx = \ln\left\lvert \dfrac{x - 4}{x + 5} \right\rvert + C

Note: The modulus is needed: no interval for x is given. Note: Do not award the final A1 if +C is omitted; penalise this once in the question.

Final answer

(a)(i) \frac{3}{x+6}+\frac{2}{x-2}
(ii) 3\ln|x+6|+2\ln|x-2|+C
(b) \ln\left|\frac{x-4}{x+5}\right|+C

Question 4

[12 marks]

(a) Given that \displaystyle\int_{1}^{k} (2x - 1)\,dx = 12, where k > 1, find the value of k.

[4]

(b) Given that \displaystyle\int_{0}^{a} e^{2x}\,dx = 4, find the exact value of a.

[4]

(c) Given that \displaystyle\int_{0}^{m} \dfrac{3}{3x + 2}\,dx = \ln 4, where m > 0, find the value of m.

[4]

Show solutionHide solution

(a) \left[ x^2 - x \right]_{1}^{k} = k^2 - k - 0

k^2 - k = 12

(k - 4)(k + 3) = 0 (M1)

k = 4 (reject k = -3 since k > 1)

(b) \left[ \dfrac{1}{2}e^{2x} \right]_{0}^{a} = \dfrac{1}{2}\left( e^{2a} - 1 \right)

\dfrac{1}{2}\left( e^{2a} - 1 \right) = 4

e^{2a} = 9

a = \dfrac{1}{2}\ln 9 = \ln 3

(c) \left[ \ln(3x + 2) \right]_{0}^{m} = \ln(3m + 2) - \ln 2

\ln\left( \dfrac{3m + 2}{2} \right) = \ln 4

3m + 2 = 8

m = 2

Final answer

(a) k=4
(b) a=\ln3
(c) m=2

Question 5

[7 marks]

Let f(x) = x^2 - 2x.

(a) Show that \displaystyle\int_{0}^{k} f(x)\,dx = \dfrac{k^2(k - 3)}{3}.

[3]

(b) Hence find the value of k > 0 for which \displaystyle\int_{0}^{k} f(x)\,dx = 0.

[1]

(c) Find the area of the region enclosed by the graph of f and the x-axis.

[2]

(d) Using your answers to parts (b) and (c), write down the area of the region enclosed by the graph of f, the x-axis and the line x = 3, for 2 \le x \le 3.

[1]

Show solutionHide solution

(a) \left[ \dfrac{x^3}{3} - x^2 \right]_{0}^{k} A1A1

= \dfrac{k^3}{3} - k^2 - 0 = \dfrac{k^2(k - 3)}{3} M1AG

(b) k = 3

(c) f(x) = x(x - 2), so the region is 0 \le x \le 2, below the axis: \displaystyle\int_{0}^{2} f(x)\,dx = \dfrac{4(-1)}{3} = -\dfrac{4}{3} (M1)

area = \dfrac{4}{3}

(d) \displaystyle\int_{0}^{3} f(x)\,dx = 0, so \displaystyle\int_{2}^{3} f(x)\,dx = +\dfrac{4}{3}: the area between x = 2 and x = 3 (above the axis) is \dfrac{4}{3}

Note: award A1 for \dfrac{4}{3} with a reason based on the zero integral; direct integration \displaystyle\int_{2}^{3} f(x)\,dx = \dfrac{4}{3} is also accepted.

Final answer

(a) shown
(b) k=3
(c) \frac43
(d) \frac43

Question 6

[11 marks]

A particle moves along a straight line so that its velocity, v\ \text{m s}^{-1}, at time t seconds is

v = t^2 - 7t + 10, \quad 0 \le t \le 6.

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

[2]

(b) Find the displacement of the particle at t = 6 relative to its position at t = 0.

[3]

(c) Find the total distance travelled by the particle for 0 \le t \le 6.

[5]

(d) Explain why the answers to (b) and (c) are different.

[1]

Show solutionHide solution

(a) (t - 2)(t - 5) = 0

t = 2 and t = 5

(b) \displaystyle\int_{0}^{6} (t^2 - 7t + 10)\,dt

= \left[ \dfrac{t^3}{3} - \dfrac{7t^2}{2} + 10t \right]_{0}^{6}

= 72 - 126 + 60 = 6 m

(c) split at the zeros of v: distance = \displaystyle\int_{0}^{2} v\,dt + \left\lvert \int_{2}^{5} v\,dt \right\rvert + \int_{5}^{6} v\,dt

\displaystyle\int_{0}^{2} v\,dt = \dfrac{26}{3}

\displaystyle\int_{2}^{5} v\,dt = -\dfrac{9}{2}

\displaystyle\int_{5}^{6} v\,dt = \dfrac{11}{6}

distance = \dfrac{26}{3} + \dfrac{9}{2} + \dfrac{11}{6} = 15 m

(d) for 2 < t < 5, v < 0 and the particle moves backwards: displacement subtracts these \dfrac{9}{2} m, the distance adds them

Final answer

(a) t=2, 5
(b) 6 m
(c) 15 m
(d) the particle moves backwards for 2<t<5

Question 7

[8 marks]

The line segment y = 1 + \dfrac{x}{2}, for 0 \le x \le 4, is rotated through 2\pi about the x-axis. Together with its two circular ends, the surface formed encloses a frustum (a cone with its top cut off).

(a) Write down the radii of the two circular ends and the distance between them.

[2]

(b) Use integration to find the exact volume of the frustum.

[3]

(c) The line y = 1 + \dfrac{x}{2} meets the x-axis where x = -2. By considering the frustum as the difference of two cones, confirm your answer to part (b).

[3]

Show solutionHide solution

(a) radii 1 (at x = 0) and 3 (at x = 4)

distance 4

(b) V = \pi\displaystyle\int_{0}^{4} \left( 1 + \dfrac{x}{2} \right)^2 dx = \pi\int_{0}^{4} \left( 1 + x + \dfrac{x^2}{4} \right) dx

= \pi\left[ x + \dfrac{x^2}{2} + \dfrac{x^3}{12} \right]_{0}^{4}

= \pi\left( 4 + 8 + \dfrac{16}{3} \right) = \dfrac{52\pi}{3}

(c) large cone: apex at x = -2, height 6, radius 3: \dfrac{1}{3}\pi \times 9 \times 6 = 18\pi

small cone: height 2, radius 1: \dfrac{1}{3}\pi \times 1 \times 2 = \dfrac{2\pi}{3}

18\pi - \dfrac{2\pi}{3} = \dfrac{52\pi}{3}, the same as part (b)

Note: Accept \pi\left[ \dfrac{2}{3}\left( 1 + \dfrac{x}{2} \right)^3 \right]_{0}^{4} in (b). The final M1 is for subtracting the two cone volumes.

Final answer

(a) radii 1 and 3, distance 4
(b) \frac{52\pi}{3}
(c) 18\pi-\frac{2\pi}{3}=\frac{52\pi}{3}

Tutorial

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

Question 8

[6 marks]

The depth of water in a harbour, h metres, t hours after midnight, changes at the rate

\frac{dh}{dt} = 0.8\cos(0.5t).

At midnight the depth is 4 m.

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

[4]

(b) Find the first time after midnight at which the depth is 5 m. Give your answer in hours, correct to 3 significant figures.

[2]

Show solutionHide solution

(a) h = \int 0.8\cos(0.5t)\,dt = \dfrac{0.8}{0.5}\sin(0.5t) + C = 1.6\sin(0.5t) + C.

[2]

h(0) = 4 gives C = 4.

[1]

h = 1.6\sin(0.5t) + 4.

[1]

(b) Solve 1.6\sin(0.5t) + 4 = 5, i.e.; \sin(0.5t) = 0.625 (GDC in radian mode, or graph intersection of y = 1.6\sin(0.5x)+4 with y=5).

[1]

0.5t = 0.6751\ldots so t = 1.3503\ldots \approx 1.35 hours.

[1]

Final answer

(a) h=1.6\sin(0.5t)+4
(b) t=1.35 hours

Question 9

[6 marks]

(a) Find \displaystyle\int \dfrac{1}{4x^2+9}\,dx.

[3]

(b) By first completing the square in the denominator, find \displaystyle\int \dfrac{1}{x^2+2x+5}\,dx.

[3]

Show solutionHide solution

; (a) Write 4x^2 + 9 = 9 + (2x)^2, which is a^2 + u^2 with a = 3 and the linear inside u = 2x.

[1]

Dividing by the coefficient 2 of x in the inside: \dfrac12

[1]

\int \dfrac{1}{9+(2x)^2}\,dx = \dfrac12\cdot\dfrac13\arctan\dfrac{2x}{3} = \dfrac16\arctan\dfrac{2x}{3} + C.

[1]

(b) x^2 + 2x + 5 = (x+1)^2 + 4.

[1]

This is a^2 + u^2 with a=2 and u = x+1, so \int \dfrac{dx}{(x+1)^2 + 4} = \dfrac12\arctan\dfrac{x+1}{2}.

[1]

= \dfrac12\arctan\dfrac{x+1}{2} + C.

[1]

Final answer

(a) \frac16\arctan\frac{2x}{3}+C
(b) \frac12\arctan\frac{x+1}{2}+C

Question 10

[8 marks]

Let J=\displaystyle\int_0^3\frac{1}{(49+x^2)^2}\,dx.

(a) Use the substitution x=7\tan\theta to write J as an integral in \theta, with new limits.

[3]

(b) Hence find J exactly, using a double-angle identity.

[3]

(c) Deduce the exact value of K=\displaystyle\int_0^3\frac{x^2}{(49+x^2)^2}\,dx.

[2]

Show solutionHide solution

(a) The differential is dx=7\sec^2\theta\,d\theta and 49+x^2=49\sec^2\theta. [1] The new limits are 0 and \alpha=\arctan(3/7). [1] Hence J=\frac{1}{343}\int_0^\alpha\cos^2\theta\,d\theta.

[1]

(b) Apply \cos^2\theta=(1+\cos2\theta)/2. [1] This gives J=\frac{1}{343}(\alpha/2+\sin(2\alpha)/4). [1] Since \tan\alpha=3/7, \sin(2\alpha)=21/29, and J=\frac{\arctan(3/7)}{686}+\frac{3}{5684}.

[1]

(c) Use \frac{x^2}{(49+x^2)^2}=\frac{1}{49+x^2}-\frac{49}{(49+x^2)^2}. [1] Thus K=\frac17\arctan(3/7)-49J=\frac{\arctan(3/7)}{14}-\frac{3}{116}.

[1]

Final answer

(a) J=\frac{1}{343}\int_0^{\arctan(3/7)}\cos^2\theta\,d\theta
(b) J=\frac{\arctan(3/7)}{686}+\frac{3}{5684}
(c) K=\frac{\arctan(3/7)}{14}-\frac{3}{116}

Question 11

[5 marks]

Use integration by parts to find the exact value of \displaystyle\int_0^2 5\arctan(3t)\,dt.

[5]

Show solutionHide solution

Select u=\arctan(3t) and dv=5\,dt, so du=\frac{3}{1+9t^2}\,dt and v=5t.

[1]

The parts calculation is H=[5t\arctan(3t)]_0^2-\int_0^2\frac{15t}{1+9t^2}\,dt.

[1]

Using z=1+9t^2 in the remaining integral gives the primitive \frac56\ln(1+9t^2).

[1]

Consequently H=[5t\arctan(3t)-\frac56\ln(1+9t^2)]_0^2.

[1]

Therefore H=10\arctan6-\frac56\ln37.

[1]

Final answer

10\arctan6-\frac56\ln37

Question 12

[5 marks]

Let f(x) = x^3 - 5x + 1.

(a) Find \displaystyle\int_{-3}^{3} f(x)\,dx.

[1]

(b) Find the total area of the regions enclosed between the graph of f and the x-axis for -3 \le x \le 3.

[3]

(c) Explain why your answers to (a) and (b) are different.

[1]

Give your answers to 3 significant figures where appropriate.

Show solutionHide solution

(a) \int_{-3}^3 (x^3 - 5x + 1)\,dx = \left[\dfrac{x^4}{4} - \dfrac{5x^2}{2} + x\right]_{-3}^{3} = 6 (GDC or by hand).

[1]

(b) The graph meets the x-axis at x = -2.330, 0.2016 and 2.128 (GDC solve).

[1]

Total area = \displaystyle\int_{-3}^{3} |f(x)|\,dx (GDC, using the absolute-value function, or integrating between the roots and adding the moduli).

[1]

= 20.90\ldots \approx 20.9.

[1]

(c) In (a) the parts of the graph below the axis count as negative and cancel against the parts above; in (b) every region counts as positive.

[1]

Final answer

(a) 6
(b) 20.9
(c) regions below the axis count as negative in (a)

Question 13

[7 marks]

Let f(x) = \dfrac{12}{(x+1)(x+4)} for x \ge 0.

(a) Express f(x) in partial fractions.

[3]

(b) Find the exact area of the region bounded by the graph of f, the coordinate axes and the line x = 5, giving your answer as a single logarithm.

[3]

(c) Use your GDC to check your answer to part (b), giving the area to 3 significant figures.

[1]

Show solutionHide solution

; (a) 12 = A(x+4) + B(x+1).

[1]

x = -1: 12 = 3A, A = 4; x = -4: 12 = -3B, B = -4.

[2]

f(x) = \dfrac{4}{x+1} - \dfrac{4}{x+4}.

(b) \int_0^5 f\,dx = \Bigl[4\ln(x+1) - 4\ln(x+4)\Bigr]_0^5

[1]

= (4\ln6 - 4\ln9) - (0 - 4\ln4)

[1]

= 4\ln\dfrac{6\times4}{9} = 4\ln\dfrac83.

[1]

(c) GDC: \int_0^5 \dfrac{12}{(x+1)(x+4)}\,dx = 3.923\ldots; and 4\ln\frac83 = 3.923\ldots. The area is 3.92 (3 s.f.).

[1]

Final answer

(a) \frac{4}{x+1}-\frac{4}{x+4}
(b) 4\ln\frac83
(c) 3.92

Question 14

[7 marks]

The region R lies between y=5x^2 and y=11x. The line y=mx, with 0<m<11, cuts R into two pieces.

(a) Find, in terms of m, the area between y=5x^2 and the line.

[3]

(b) Find the exact value of m for which the two pieces have equal areas.

[4]

Show solutionHide solution

(a) Intersection requires 5x^2=mx, giving x=0 and x=m/5. [1] The area is \int_0^{m/5}(mx-5x^2)\,dx=[mx^2/2-5x^3/3]_0^{m/5}. [1] Its value is m^3/50-m^3/75=m^3/150.

[1]

(b) The entire stencil has area 11^3/150=1331/150. [1] Equality of the pieces requires m^3/150=1331/300. [1] Therefore m^3=1331/2. [1] The unique admissible slope is m=11/\sqrt[3]{2}.

[1]

Final answer

(a) \frac{m^3}{150}
(b) m=\frac{11}{\sqrt[3]2}

Question 15

[5 marks]

The curve y = x^{1.5} + 1 for x \ge 0, the y-axis and the lines y = 2 and y = 5 enclose a region R.

(a) Express x in terms of y on the curve.

[1]

(b) Hence find the area of R by integrating with respect to y. Give your answer to 3 significant figures.

[4]

Show solutionHide solution

; (a) x^{1.5} = y - 1, and x \ge 0, so x = (y-1)^{1/1.5} = (y-1)^{2/3}.

[1]

(b) Area = \displaystyle\int_2^5 x\,dy = \int_2^5 (y-1)^{2/3}\,dy.

[1]

= \left[\dfrac35 (y-1)^{5/3}\right]_2^5

[1]

= \dfrac35\bigl(4^{5/3} - 1^{5/3}\bigr) (evaluated on the calculator; the integral template gives the same value)

[1]

= 5.447\ldots \approx 5.45.

[1]

Final answer

(a) x=(y-1)^{2/3}
(b) 5.45

Question 16

[7 marks]

A paperweight is made by rotating the region bounded by the curve y = x^{2.5} (for x \ge 0), the y-axis and the line y = h through 2\pi about the y-axis. All lengths are in centimetres.

(a) Show that x^2 = y^{0.8} on the curve.

[2]

(b) Find the volume of the paperweight when h = 6.

[2]

(c) Find the value of h for which the volume is 100\ \text{cm}^3.

[3]

Give your answers to 3 significant figures.

Show solutionHide solution

; (a) y = x^{2.5} gives x = y^{1/2.5} = y^{0.4}.

[1]

x^2 = y^{0.8}.

[1]

(b) V = \pi\displaystyle\int_0^6 y^{0.8}\,dy.

[1]

= \pi\left[\dfrac{y^{1.8}}{1.8}\right]_0^6 = 43.9\ldots \approx 43.9\ \text{cm}^3.

[1]

(c) V = \pi\displaystyle\int_0^h y^{0.8}\,dy = \dfrac{\pi h^{1.8}}{1.8}.

[1]

\dfrac{\pi h^{1.8}}{1.8} = 100 gives h^{1.8} = \dfrac{180}{\pi}.

[1]

h = 9.478\ldots \approx 9.48 cm.

[1]

Final answer

(a) shown
(b) 43.9 cm^3
(c) h=9.48 cm

Question 17

[8 marks]

A remote-controlled boat moves along a straight canal. Its velocity, v\ \text{m\,s}^{-1}, at time t seconds is v = 4\cos(0.5t) - 1 for 0 \le t \le 12.

(a) Find the times at which the boat is instantaneously at rest.

[2]

(b) Find the total distance the boat travels in the 12 seconds.

[2]

(c) Find the displacement of the boat from its starting point at t = 12.

[1]

(d) Find the acceleration at t = 4, and state, with a reason, whether the boat is speeding up or slowing down at that time.

[3]

Give your answers to 3 significant figures.

Show solutionHide solution

(a) Solve 4\cos(0.5t) - 1 = 0 on the GDC (radians): \cos(0.5t) = 0.25.

[1]

t = 2.64 s and t = 9.93 s.

[1]

(b) v changes sign at both times, so distance = \displaystyle\int_0^{12}|v|\,dt.

[1]

= 31.3\ldots \approx 31.3 m.

[1]

(c) Displacement = \displaystyle\int_0^{12} v\,dt = -14.2\ldots \approx -14.2 m (the boat ends 14.2 m behind its start).

[1]

(d) a = \dfrac{dv}{dt} = -2\sin(0.5t), so a(4) = -2\sin2 = -1.82\ \text{m\,s}^{-2}.

[2]

v(4) = 4\cos2 - 1 = -2.66 < 0 and a(4) < 0: velocity and acceleration have the same sign, so the boat is speeding up.

[1]

Final answer

(a) t=2.64, 9.93
(b) 31.3 m
(c) -14.2 m
(d) a(4)=-1.82 m s^{-2}; speeding up, since v(4)<0 and a(4)<0

Mastery

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

Question 18

[12 marks]

A curve C is defined for x > 3. Its gradient function is

\dfrac{dy}{dx} = \dfrac{18}{(x - 3)(x + 6)},

and C passes through the point (4,\ 0).

(a) Express \dfrac{18}{(x - 3)(x + 6)} in partial fractions.

[2]

(b) Show that the equation of C is y = 2\ln\left( \dfrac{10(x - 3)}{x + 6} \right).

[5]

(c) Explain why y increases as x increases.

[1]

(d) Find the exact value of x for which y = 2\ln 4.

[2]

(e) Find the equation of the horizontal asymptote of C.

[2]

Show solutionHide solution

(a) 18 = A(x + 6) + B(x - 3); x = 3: A = 2; x = -6: B = -2

\dfrac{2}{x - 3} - \dfrac{2}{x + 6}

(b) y = 2\ln(x - 3) - 2\ln(x + 6) + D (no modulus needed since x > 3) A1A1

(4,\ 0): 0 = 2\ln 1 - 2\ln 10 + D

D = 2\ln 10

y = 2\ln\left( \dfrac{x - 3}{x + 6} \right) + 2\ln 10 = 2\ln\left( \dfrac{10(x - 3)}{x + 6} \right) M1AG

Note: The second M1 is for combining the logarithms with the laws of logarithms.

(c) for x > 3 both x - 3 and x + 6 are positive, so \dfrac{dy}{dx} > 0

(d) \dfrac{10(x - 3)}{x + 6} = 4 \Rightarrow 10x - 30 = 4x + 24

x = 9

(e) \dfrac{x - 3}{x + 6} = \dfrac{1 - \frac{3}{x}}{1 + \frac{6}{x}} \to 1 as x \to \infty

y \to 2\ln 10: the asymptote is y = 2\ln 10 (or y = \ln 100)

Final answer

(a) \frac{2}{x-3}-\frac{2}{x+6}
(b) shown
(c) \frac{dy}{dx}>0 for x>3
(d) x=9
(e) y=2\ln10

Question 19

[13 marks]

Let f(x) = 2\sin x and g(x) = \tan x, for 0 \le x < \dfrac{\pi}{2}. The graphs of f and g meet at the origin and at the point P. The shaded region R is enclosed by the two graphs.

(a) Find the exact x-coordinate of P.

[3]

(b) Show that the area of R is 1 - \ln 2.

[4]

(c) The region R is rotated through 2\pi about the x-axis. Show that the volume of the solid formed is

\pi\int_{0}^{\pi/3} \left( 3 - 2\cos 2x - \sec^2 x \right) dx.

[3]

(d) Hence find the exact volume of the solid.

[3]

Show solutionHide solution

(a) 2\sin x = \dfrac{\sin x}{\cos x}

\sin x \ne 0 at P, so \cos x = \dfrac{1}{2}

x = \dfrac{\pi}{3}

(b) 2\sin x \ge \tan x for 0 \le x \le \dfrac{\pi}{3}, so A = \displaystyle\int_{0}^{\pi/3} \left( 2\sin x - \tan x \right) dx

= \left[ -2\cos x + \ln(\cos x) \right]_{0}^{\pi/3} A1A1

= \left( -1 + \ln\dfrac{1}{2} \right) - (-2 + \ln 1)

= 1 - \ln 2

(c) V = \pi\displaystyle\int_{0}^{\pi/3} \left( 4\sin^2 x - \tan^2 x \right) dx

4\sin^2 x = 2(1 - \cos 2x)

\tan^2 x = \sec^2 x - 1

so 4\sin^2 x - \tan^2 x = 3 - 2\cos 2x - \sec^2 x

(d) \pi\left[ 3x - \sin 2x - \tan x \right]_{0}^{\pi/3} A1A1

= \pi\left( \pi - \dfrac{\sqrt{3}}{2} - \sqrt{3} \right) = \pi^2 - \dfrac{3\sqrt{3}\pi}{2}

Note: In (b) A1 for -2\cos x, A1 for \ln(\cos x) (no modulus needed: \cos x > 0). In (c) award M0 for \pi\displaystyle\int (2\sin x - \tan x)^2\,dx. In (d) A1 for 3x - \sin 2x, A1 for -\tan x.

Final answer

(a) x=\frac\pi3
(b) shown
(c) shown
(d) \pi^2-\frac{3\sqrt3\,\pi}{2}

Question 20

[10 marks]

(a) The region R is enclosed by the curve y = 2\ln(x + 1), for x \ge 0, the y-axis and the line y = 2\ln 3.

(i) Show that, for points on the curve, x^2 = e^y - 2e^{y/2} + 1.

[2]

(ii) R is rotated through 2\pi about the y-axis. Show that the volume of the solid formed is 2\pi\ln 3.

[4]

(b) The region S is enclosed by the curve y = \arccos\left( \dfrac{x}{3} \right), for 0 \le x \le 3, and the coordinate axes. Find the exact volume of the solid formed when S is rotated through 2\pi about the y-axis.

[4]

Show solutionHide solution

(a)(i) \dfrac{y}{2} = \ln(x + 1) \Rightarrow x = e^{y/2} - 1

x^2 = \left( e^{y/2} - 1 \right)^2 = e^y - 2e^{y/2} + 1 A1AG

(ii) the curve passes through O, so y runs from 0 to 2\ln 3

V = \pi\displaystyle\int_{0}^{2\ln 3} \left( e^y - 2e^{y/2} + 1 \right) dy

= \pi\left[ e^y - 4e^{y/2} + y \right]_{0}^{2\ln 3}

= \pi\left( (9 - 12 + 2\ln 3) - (1 - 4 + 0) \right)

= 2\pi\ln 3

(b) x = 3\cos y, and y runs from 0 (at x = 3) to \dfrac{\pi}{2} (at x = 0)

V = \pi\displaystyle\int_{0}^{\pi/2} 9\cos^2 y\,dy

= \dfrac{9\pi}{2}\displaystyle\int_{0}^{\pi/2} (1 + \cos 2y)\,dy = \dfrac{9\pi}{2}\left[ y + \dfrac{1}{2}\sin 2y \right]_{0}^{\pi/2}

= \dfrac{9\pi^2}{4}

Note: In (a)(ii) use e^{2\ln 3} = 9 and e^{\ln 3} = 3.

Final answer

(a)(i) shown
(ii) shown
(b) \frac{9\pi^2}{4}

Question 21

[7 marks]

(a) Use integration by parts to find \displaystyle\int \ln\left( x^2 + 16 \right) dx.

[5]

(b) Hence find the exact value of \displaystyle\int_{0}^{4} \ln\left( x^2 + 16 \right) dx, giving your answer in the form a\ln 2 + b + c\pi, where a, b, c \in \mathbb{Z}.

[2]

Show solutionHide solution

(a) u = \ln\left( x^2 + 16 \right), \dfrac{du}{dx} = \dfrac{2x}{x^2 + 16}, v = x

x\ln\left( x^2 + 16 \right) - \displaystyle\int \dfrac{2x^2}{x^2 + 16}\,dx

\dfrac{2x^2}{x^2 + 16} = 2 - \dfrac{32}{x^2 + 16}

\displaystyle\int \left( 2 - \dfrac{32}{x^2 + 16} \right) dx = 2x - 8\arctan\left( \dfrac{x}{4} \right)

x\ln\left( x^2 + 16 \right) - 2x + 8\arctan\left( \dfrac{x}{4} \right) + C

(b) 4\ln 32 - 8 + 8\arctan 1 - 0 = 4\ln 32 - 8 + 2\pi

= 20\ln 2 - 8 + 2\pi

Note: The second M1 in (a) is for dividing (or splitting) the improper fraction. Note: Do not award the final A1 if +C is omitted; penalise this once in the question.

Final answer

(a) x\ln(x^2+16)-2x+8\arctan\frac x4+C
(b) 20\ln2-8+2\pi

Question 22

[14 marks]

Let f(t)=5te^{-3t}, for t\ge0.

(a) Find the coordinates of the maximum point of the graph of f, and justify that it is a maximum.

[3]

(b) Use integration by parts to find \displaystyle\int f(t)\,dt.

[3]

(c) Find the exact area of the region under the graph of f from t=0 to the maximum point.

[2]

(d) This region is rotated through 2\pi about the t-axis. Use integration by parts twice to find the exact volume of the solid formed.

[6]

Show solutionHide solution

(a) The derivative is f'(t)=5e^{-3t}(1-3t). [1] It vanishes at t=1/3, where f=5/(3e). [1] It is positive before this point and negative afterwards, so (1/3,5/(3e)) is the maximum.

[1]

(b) Use u=5t, dv=e^{-3t}\,dt, giving du=5\,dt and v=-e^{-3t}/3. [1] Parts gives -5te^{-3t}/3+(5/3)\int e^{-3t}\,dt. [1] A primitive is F(t)=-\frac59(3t+1)e^{-3t}.

[1]

(c) The area is F(1/3)-F(0). [1] It equals \frac59(1-2/e).

[1]

(d) The volume is V=25\pi\int_0^{1/3}t^2e^{-6t}\,dt.

[1]

First set u=t^2, dv=e^{-6t}\,dt to obtain \int t^2e^{-6t}\,dt=-t^2e^{-6t}/6+\frac13\int te^{-6t}\,dt.

[2]

A second pass with u=t, dv=e^{-6t}\,dt gives \int te^{-6t}\,dt=-te^{-6t}/6-e^{-6t}/36.

[2]

Hence V=25\pi[-e^{-6t}(18t^2+6t+1)/108]_0^{1/3}=\frac{25\pi}{108}(1-5/e^2).

[1]

Final answer

(a) maximum at \left(\frac13,\frac{5}{3e}\right)
(b) -\frac59(3t+1)e^{-3t}+C
(c) \frac59\left(1-\frac2e\right)
(d) \frac{25\pi}{108}\left(1-\frac{5}{e^2}\right)

Question 23

[14 marks]

Let f(x) = \dfrac{18}{(x+1)(x+4)} for x \ge 0.

(a) Express f(x) in partial fractions.

[3]

(b) Hence show that \displaystyle\int_0^k f(x)\,dx = 6\ln\dfrac{4(k+1)}{k+4}, where k > 0.

[3]

(c) Show that, as k increases without limit, the area under the graph of f from x = 0 to x = k approaches 12\ln2.

[2]

(d) Find the value of k for which the area in part (b) is equal to 7.

[3]

(e) The region bounded by the graph of f, the coordinate axes and the line x = 3 is rotated through 2\pi about the x-axis. Find the volume of the solid formed.

[3]

Give answers to parts (d) and (e) to 3 significant figures.

Show solutionHide solution

; (a) 18 = A(x+4) + B(x+1).

[1]

x = -1: 18 = 3A, A = 6; x = -4: 18 = -3B, B = -6.

[2]

f(x) = \dfrac{6}{x+1} - \dfrac{6}{x+4}.

(b) \int_0^k f\,dx = \Bigl[6\ln(x+1) - 6\ln(x+4)\Bigr]_0^k.

[1]

= 6\ln(k+1) - 6\ln(k+4) - \bigl(0 - 6\ln4\bigr)

[1]

= 6\ln\dfrac{4(k+1)}{k+4}.

[1]

(c) \dfrac{4(k+1)}{k+4} = \dfrac{4(1 + 1/k)}{1 + 4/k} \to 4 as k \to \infty.

[1]

The area approaches 6\ln4 = 6\cdot2\ln2 = 12\ln2.

[1]

(d) 6\ln\dfrac{4(k+1)}{k+4} = 7 gives \dfrac{4(k+1)}{k+4} = e^{7/6}.

[1]

Solve (algebraically or on the GDC): 4k + 4 = e^{7/6}(k+4).

[1]

k = 11.2 (from 11.214\ldots).

[1]

(e) V = \pi\displaystyle\int_0^3 \left(\dfrac{18}{(x+1)(x+4)}\right)^2dx.

[2]

GDC: V = 34.61\ldots \approx 34.6.

[1]

Final answer

(a) \frac{6}{x+1}-\frac{6}{x+4}
(b) shown
(c) shown
(d) k=11.2
(e) 34.6

Question 24

[16 marks]

The cross-section of a stone arch is the region between two curves, as shown. The upper edge has equation y = 4 - \dfrac{x^2}{4} and the lower edge has equation y = 2\cos\!\left(\dfrac{\pi x}{8}\right), for -4 \le x \le 4. Both x and y are measured in metres.

(a) Show that both edges pass through the points (-4,\,0) and (4,\,0).

[2]

(b) Find the area of the cross-section.

[3]

The arch is 6.5 m long, and the stone has density 2400\ \text{kg\,m}^{-3}.

(c) Find the volume of stone in the arch and its mass.

[3]

(d) Find the greatest vertical thickness of the arch.

[2]

(e) The region bounded by the upper edge, for x \ge 0, and the coordinate axes is rotated through 2\pi about the y-axis to form a model hill. Find the exact volume of the hill.

[4]

(f) The arch is thinner than 1 m vertically wherever |x| > k. Find the value of k.

[2]

Give non-exact answers to 3 significant figures.

Show solutionHide solution

; (a) Upper edge at x = \pm4: 4 - \dfrac{16}{4} = 0.

[1]

Lower edge at x = \pm4: 2\cos\dfrac{\pi}{2} = 0.

[1]

(b) Area = \displaystyle\int_{-4}^{4}\left(4 - \dfrac{x^2}{4} - 2\cos\dfrac{\pi x}{8}\right)dx (upper - lower).

[2]

GDC: 11.147\ldots \approx 11.1\ \text{m}^2.

[1]

(c) Volume = 11.147\times6.5.

[1]

= 72.46 \approx 72.5\ \text{m}^3.

[1]

Mass = 72.46\times2400 = 1.739\ldots\times10^5 \approx 1.74\times10^5 kg.

[1]

(d) The vertical thickness is g(x) = 4 - \dfrac{x^2}{4} - 2\cos\dfrac{\pi x}{8}. Plot g on the GDC and find its maximum.

[1]

The maximum is at x = 0: thickness = 4 - 2 = 2.00 m.

[1]

(e) y = 4 - \dfrac{x^2}{4} gives x^2 = 16 - 4y, and y runs from 0 to 4.

[1]

V = \pi\displaystyle\int_0^4 (16 - 4y)\,dy

[1]

= \pi\bigl[16y - 2y^2\bigr]_0^4

[1]

= \pi(64 - 32) = 32\pi\ \text{m}^3.

[1]

(f) Solve g(x) = 1 for 0 < x < 4 on the GDC (graph intersection).

[1]

x = 2.98, so k = 2.98 (the thickness decreases from 2 at x=0 to 0 at x=4).

[1]

Final answer

(a) shown
(b) 11.1 m^2
(c) 72.5 m^3 and 1.74\times10^5 kg
(d) 2.00 m
(e) 32\pi m^3
(f) k=2.98

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}.

[2]

(d) 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, an interval of width 0.991<1

Note: the GDC gives S_{10\,000}=198.54\ldots.

(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
(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) 4.615<H_{100}<5.605
(b), (c) proved in the solutions
(d) 12\,367

G2 The reflection trick

G2 Worked example

[9 marks]

Let I=\displaystyle\int_0^{\pi}\frac{x\sin x}{1+\cos^2x}\,dx.

(a) Use the substitution u=\pi-x to show that I=\displaystyle\int_0^{\pi}\frac{(\pi-x)\sin x}{1+\cos^2x}\,dx.

[2]

(b) Hence show that I=\dfrac{\pi}{2}\displaystyle\int_0^{\pi}\frac{\sin x}{1+\cos^2x}\,dx.

[2]

(c) Use the substitution v=\cos x to find the exact value of I.

[4]

(d) Use your GDC to check your answer to (c).

[1]

Show solutionHide solution

(a) x=\pi-u, dx=-du; x=0\Rightarrow u=\pi, x=\pi\Rightarrow u=0; \sin(\pi-u)=\sin u and \cos^2(\pi-u)=\cos^2u

I=\displaystyle\int_{\pi}^{0}\frac{(\pi-u)\sin u}{1+\cos^2u}(-du)=\int_0^{\pi}\frac{(\pi-u)\sin u}{1+\cos^2u}\,du, and the letter does not matter

A1AG

(b) add the two forms of I: 2I=\displaystyle\int_0^{\pi}\frac{x\sin x+(\pi-x)\sin x}{1+\cos^2x}\,dx

=\pi\displaystyle\int_0^{\pi}\frac{\sin x}{1+\cos^2x}\,dx, so I=\dfrac{\pi}{2}\displaystyle\int_0^{\pi}\frac{\sin x}{1+\cos^2x}\,dx

A1AG

(c) dv=-\sin x\,dx; x=0\Rightarrow v=1, x=\pi\Rightarrow v=-1

\displaystyle\int_0^{\pi}\frac{\sin x}{1+\cos^2x}\,dx=\int_1^{-1}\frac{-dv}{1+v^2}=\int_{-1}^{1}\frac{dv}{1+v^2}

=\Bigl[\arctan v\Bigr]_{-1}^{1}=\dfrac\pi4-\left(-\dfrac\pi4\right)=\dfrac\pi2

I=\dfrac\pi2\times\dfrac\pi2=\dfrac{\pi^2}{4}

(d) GDC: I=2.4674\ldots and \dfrac{\pi^2}4=2.4674\ldots

Final answer

(c) \frac{\pi^2}{4}
(d) 2.467

G2 Investigation

[11 marks]

For n\in\mathbb{Z}^+, let I_n=\displaystyle\int_0^{\pi/2}\frac{\sin^nx}{\sin^nx+\cos^nx}\,dx.

(a) Use the substitution u=\dfrac\pi2-x to show that I_1=\displaystyle\int_0^{\pi/2}\frac{\cos x}{\sin x+\cos x}\,dx. Hence find the exact value of I_1.

[3]

(b) Use your GDC to find I_2, I_3 and I_4. Make a conjecture about I_n.

[2]

(c) Prove your conjecture.

[3]

(d) Let g be a function that is continuous and positive on [b,c]. Use the substitution u=b+c-x to show that \displaystyle\int_b^c\frac{g(x)}{g(x)+g(b+c-x)}\,dx=\frac{c-b}{2}. Hence find \displaystyle\int_1^3\frac{x^2}{x^2+(4-x)^2}\,dx.

[3]

Show solutionHide solution

(a) dx=-du, the limits swap, and \sin\left(\frac\pi2-u\right)=\cos u, \cos\left(\frac\pi2-u\right)=\sin u

I_1=\displaystyle\int_{\pi/2}^{0}\frac{\cos u}{\cos u+\sin u}(-du)=\int_0^{\pi/2}\frac{\cos x}{\sin x+\cos x}\,dx

adding the two forms, 2I_1=\displaystyle\int_0^{\pi/2}1\,dx=\frac\pi2, so I_1=\dfrac\pi4

(b) I_2=I_3=I_4=0.785398\ldots

conjecture: I_n=\dfrac\pi4 for every n

(c) the same substitution turns \sin^n into \cos^n and \cos^n into \sin^n: I_n=\displaystyle\int_0^{\pi/2}\frac{\cos^nx}{\cos^nx+\sin^nx}\,dx

adding, 2I_n=\displaystyle\int_0^{\pi/2}\frac{\sin^nx+\cos^nx}{\sin^nx+\cos^nx}\,dx=\int_0^{\pi/2}1\,dx

I_n=\dfrac\pi4 for every n\in\mathbb{Z}^+

Note: the denominator is positive on \left(0,\frac\pi2\right), so the integrand is defined; nothing in the proof uses that n is a whole number.

(d) with u=b+c-x the limits swap and J=\displaystyle\int_b^c\frac{g(x)}{g(x)+g(b+c-x)}\,dx=\int_b^c\frac{g(b+c-u)}{g(b+c-u)+g(u)}\,du

adding, 2J=\displaystyle\int_b^c1\,dx=c-b, so J=\dfrac{c-b}{2}

A1AG

g(x)=x^2, b=1, c=3, b+c-x=4-x: the integral is \dfrac{3-1}{2}=1

Final answer

(a) I_1=\frac\pi4
(b) 0.785398 each; I_n=\frac\pi4
(c) proved in the solutions
(d) 1

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 19.

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 HL

Chapter 15 teaches integration 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