Unit 27 Exam Practice
Integration Techniques
These questions are written in the style of IB examination questions, and they are harder than the review exercises. The marks for each part are shown in square brackets on the right, for example [3]. Paper 1 means no calculator; Paper 2 means a graphic display calculator (GDC) is allowed. Try each question under exam conditions first, then open the markscheme.
| Mark | Meaning |
|---|---|
| M | Method mark: a correct method is used, even with a small slip |
| A | Accuracy mark: a correct answer or line of working (usually needs the M mark before it) |
| R | Reasoning mark: a clear reason or justification is given |
| N | Mark for a correct final answer with no working shown |
| AG | “Answer given” in the question, so no extra mark is awarded for writing it |
The number after the letter is the number of marks, so A2 is worth 2 marks. In the exam, give answers exactly or correct to 3 significant figures (s.f.) unless the question says otherwise.
SL questions
SL 1 Paper 1
A function \(f\) is defined for \(x > 0\). Its derivative is \[ f'(x) = \frac{\left(2\sqrt{x} - 1\right)^{2}}{x^{2}}. \] The graph of \(f\) passes through the point \((1,\ 5)\). Find \(f(x)\). [7]
Total 7 marks
\(\left(2\sqrt{x} - 1\right)^{2} = 4x - 4x^{\frac{1}{2}} + 1\) M1
\(f'(x) = 4x^{-1} - 4x^{-\frac{3}{2}} + x^{-2}\) A1
\(\int 4x^{-1}\,\mathrm{d}x = 4\ln x\) (no modulus needed, as \(x > 0\)) A1
\(\int -4x^{-\frac{3}{2}}\,\mathrm{d}x = 8x^{-\frac{1}{2}}\) A1
\(\int x^{-2}\,\mathrm{d}x = -x^{-1}\) A1
\(f(1) = 0 + 8 - 1 + C = 5\), so \(C = -2\) M1
\(f(x) = 4\ln x + \dfrac{8}{\sqrt{x}} - \dfrac{1}{x} - 2\) A1
Award the M1 for substituting \((1,\ 5)\) only if a constant of integration is present.
[7 marks]
SL 2 Paper 1
(a) Show that \(\cos^{4} x - \sin^{4} x = \cos 2x\). [2]
(b) Show that \(\cos^{4} x + \sin^{4} x = \dfrac{3}{4} + \dfrac{1}{4}\cos 4x\). [3]
(c) Hence find \(\displaystyle\int \sin^{4} x\,\mathrm{d}x\). [3]
Total 8 marks
(a)
\(\cos^{4} x - \sin^{4} x = \left(\cos^{2} x - \sin^{2} x\right)\left(\cos^{2} x + \sin^{2} x\right)\) M1
\(= \cos 2x \times 1 = \cos 2x\) A1 AG
[2 marks]
(b)
\(\cos^{4} x + \sin^{4} x = \left(\cos^{2} x + \sin^{2} x\right)^{2} - 2\sin^{2} x\cos^{2} x\) M1
\(= 1 - \dfrac{1}{2}\left(2\sin x\cos x\right)^{2} = 1 - \dfrac{1}{2}\sin^{2} 2x\) A1
\(= 1 - \dfrac{1}{4}\left(1 - \cos 4x\right) = \dfrac{3}{4} + \dfrac{1}{4}\cos 4x\) A1 AG
[3 marks]
(c)
subtracting (a) from (b): \(2\sin^{4} x = \dfrac{3}{4} + \dfrac{1}{4}\cos 4x - \cos 2x\) M1
\(\sin^{4} x = \dfrac{3}{8} - \dfrac{1}{2}\cos 2x + \dfrac{1}{8}\cos 4x\) A1
\(\int \sin^{4} x\,\mathrm{d}x = \dfrac{3x}{8} - \dfrac{\sin 2x}{4} + \dfrac{\sin 4x}{32} + C\) A1
[3 marks]
See 27.3.
SL 3 Paper 1
(a) Find \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(x^{2}\,\mathrm{e}^{x^{2}}\right)\). [2]
(b) Hence find \(\displaystyle\int x^{3}\,\mathrm{e}^{x^{2}}\,\mathrm{d}x\). [4]
Total 6 marks
(a)
product rule: \(2x\,\mathrm{e}^{x^{2}} + x^{2} \times 2x\,\mathrm{e}^{x^{2}}\) M1
\(= 2x\,\mathrm{e}^{x^{2}} + 2x^{3}\,\mathrm{e}^{x^{2}}\) A1
[2 marks]
(b)
integrating (a): \(x^{2}\,\mathrm{e}^{x^{2}} = \int 2x\,\mathrm{e}^{x^{2}}\,\mathrm{d}x + \int 2x^{3}\,\mathrm{e}^{x^{2}}\,\mathrm{d}x\) M1
\(\int 2x\,\mathrm{e}^{x^{2}}\,\mathrm{d}x = \mathrm{e}^{x^{2}}\) (by inspection, \(f(x) = x^{2}\)) A1
\(\int 2x^{3}\,\mathrm{e}^{x^{2}}\,\mathrm{d}x = x^{2}\,\mathrm{e}^{x^{2}} - \mathrm{e}^{x^{2}}\) A1
\(\int x^{3}\,\mathrm{e}^{x^{2}}\,\mathrm{d}x = \dfrac{1}{2}\mathrm{e}^{x^{2}}\left(x^{2} - 1\right) + C\) A1
[4 marks]
SL 4 Paper 1
A function \(f\) is defined for \(x \in \mathbb{R}\). Its second derivative is \[ f''(x) = 2\,\mathrm{e}^{2x} - 5\,\mathrm{e}^{x}. \] The graph of \(f\) has a horizontal tangent at the point \(\left(0,\ \dfrac{1}{2}\right)\).
(a) Show that \(f'(x) = \mathrm{e}^{2x} - 5\,\mathrm{e}^{x} + 4\). [3]
(b) Find \(f(x)\). [3]
(c) The graph of \(f\) has one other stationary point. Find its exact \(x\)-coordinate. [3]
(d) Determine the nature of each of the two stationary points. [4]
(e) Find the exact \(x\)-coordinate of the point of inflexion of the graph of \(f\). Justify that it is a point of inflexion. [3]
Total 16 marks
(a)
\(f'(x) = \int \left(2\,\mathrm{e}^{2x} - 5\,\mathrm{e}^{x}\right)\mathrm{d}x\) M1
\(= \mathrm{e}^{2x} - 5\,\mathrm{e}^{x} + A\) A1
horizontal tangent: \(f'(0) = 1 - 5 + A = 0\), so \(A = 4\) and \(f'(x) = \mathrm{e}^{2x} - 5\,\mathrm{e}^{x} + 4\) A1 AG
[3 marks]
(b)
\(f(x) = \int \left(\mathrm{e}^{2x} - 5\,\mathrm{e}^{x} + 4\right)\mathrm{d}x\) M1
\(= \dfrac{1}{2}\mathrm{e}^{2x} - 5\,\mathrm{e}^{x} + 4x + B\) A1
\(f(0) = \dfrac{1}{2} - 5 + B = \dfrac{1}{2}\), so \(f(x) = \dfrac{1}{2}\mathrm{e}^{2x} - 5\,\mathrm{e}^{x} + 4x + 5\) A1
[3 marks]
(c)
\(f'(x) = 0\): \(\left(\mathrm{e}^{x}\right)^{2} - 5\,\mathrm{e}^{x} + 4 = 0\) M1
\(\left(\mathrm{e}^{x} - 1\right)\left(\mathrm{e}^{x} - 4\right) = 0\), so \(\mathrm{e}^{x} = 1\) or \(\mathrm{e}^{x} = 4\) A1
\(\mathrm{e}^{x} = 1\) gives \(x = 0\) (the given point), so the other one is at \(x = \ln 4\) A1
[3 marks]
(d)
using the second derivative test M1
\(f''(0) = 2 - 5 = -3\) A1
\(f''(\ln 4) = 2(16) - 5(4) = 12\) A1
\(f''(0) < 0\), so local maximum at \(x = 0\); \(f''(\ln 4) > 0\), so local minimum at \(x = \ln 4\) R1
Accept a correct sign diagram of \(f'(x) = \left(\mathrm{e}^{x} - 1\right)\left(\mathrm{e}^{x} - 4\right)\) instead.
[4 marks]
(e)
\(f''(x) = \mathrm{e}^{x}\left(2\,\mathrm{e}^{x} - 5\right) = 0\) M1
\(\mathrm{e}^{x} = \dfrac{5}{2}\), so \(x = \ln\dfrac{5}{2}\) A1
\(\mathrm{e}^{x} > 0\), and \(2\,\mathrm{e}^{x} - 5\) changes from negative to positive at \(x = \ln\dfrac{5}{2}\), so \(f''\) changes sign R1
Do not award the R1 for “\(f''(x) = 0\)” alone.
[3 marks]
SL 5 Paper 2
Water flows into and out of tank A. The volume of water in the tank is \(V\) litres, \(t\) minutes after the start, where \[ \frac{\mathrm{d}V}{\mathrm{d}t} = \frac{120}{t + 4} - 15, \qquad t \geq 0. \] At \(t = 0\) the tank holds \(100\) litres. The model is valid until the tank becomes empty.
(a) Show that \(V = 100 + 120\ln\left(\dfrac{t + 4}{4}\right) - 15t\). [4]
(b) Find the value of \(t\) at which \(V\) is a maximum. Justify that it is a maximum. [3]
(c) Show that the maximum volume is \(40 + 120\ln 2\) litres. [2]
(d) Find the time at which tank A becomes empty. [2]
Tank B holds \(20\) litres at \(t = 0\). The volume of water in tank B is \(W\) litres, where \[ \frac{\mathrm{d}W}{\mathrm{d}t} = 5 + 4\cos\left(\frac{\pi t}{6}\right). \]
(e) Find an expression for \(W\) in terms of \(t\). [3]
(f) Find the time at which the two tanks hold the same volume of water. [2]
Total 16 marks
(a)
\(V = \int \left(\dfrac{120}{t + 4} - 15\right)\mathrm{d}t\) M1
\(= 120\ln(t + 4) - 15t + C\) (no modulus needed, as \(t + 4 > 0\)) A1
\(t = 0\): \(100 = 120\ln 4 + C\) M1
\(C = 100 - 120\ln 4\), so \(V = 100 + 120\ln\left(\dfrac{t + 4}{4}\right) - 15t\) A1 AG
[4 marks]
(b)
\(\dfrac{120}{t + 4} - 15 = 0\) M1
\(t = 4\) A1
\(\dfrac{\mathrm{d}^{2}V}{\mathrm{d}t^{2}} = -\dfrac{120}{(t + 4)^{2}} < 0\), so it is a maximum R1
Accept a sign change of \(\dfrac{\mathrm{d}V}{\mathrm{d}t}\) from positive to negative at \(t = 4\).
[3 marks]
(c)
\(V(4) = 100 + 120\ln 2 - 60\) M1
\(= 40 + 120\ln 2\) \((123.18\ldots)\) litres A1 AG
[2 marks]
(d)
\(100 + 120\ln\left(\dfrac{t + 4}{4}\right) - 15t = 0\) (for example, using Analyze Graph → Zero on a GDC) M1
\(t = 21.5\) minutes \((21.479\ldots)\) A1
[2 marks]
(e)
\(W = \int \left(5 + 4\cos\left(\dfrac{\pi t}{6}\right)\right)\mathrm{d}t\) M1
\(= 5t + \dfrac{24}{\pi}\sin\left(\dfrac{\pi t}{6}\right) + C\) A1
\(W(0) = 20\) gives \(C = 20\), so \(W = 5t + \dfrac{24}{\pi}\sin\left(\dfrac{\pi t}{6}\right) + 20\) A1
[3 marks]
(f)
\(V = W\) (for example, using the intersection of the two graphs on a GDC) M1
\(t = 12.4\) minutes \((12.383\ldots)\) A1
[2 marks]
HL questions
HL questions on this unit also use the AHL content: integrals of \(\sec^{2} x\) and \(\tan^{2} x\), the \(\arctan\) and \(\arcsin\) forms, partial fractions, general substitutions and integration by parts (including repeated parts).
HL 1 Paper 1
(a) Show that \(\tan^{4} x = \tan^{2} x\sec^{2} x - \sec^{2} x + 1\). [2]
(b) Hence find \(\displaystyle\int \tan^{4} x\,\mathrm{d}x\). [4]
Total 6 marks
(a)
\(\tan^{4} x = \tan^{2} x\left(\sec^{2} x - 1\right) = \tan^{2} x\sec^{2} x - \tan^{2} x\) M1
\(= \tan^{2} x\sec^{2} x - \left(\sec^{2} x - 1\right) = \tan^{2} x\sec^{2} x - \sec^{2} x + 1\) A1 AG
[2 marks]
(b)
\(\int \tan^{2} x\sec^{2} x\,\mathrm{d}x\): \(f(x) = \tan x\), \(f'(x) = \sec^{2} x\) M1
\(= \dfrac{\tan^{3} x}{3}\) A1
\(\int \left(-\sec^{2} x + 1\right)\mathrm{d}x = -\tan x + x\) A1
\(\int \tan^{4} x\,\mathrm{d}x = \dfrac{\tan^{3} x}{3} - \tan x + x + C\) A1
[4 marks]
See 27.3.
HL 2 Paper 1
(a) Express \(\dfrac{x^{3} + 2x^{2} - 5x + 5}{x^{2} + x - 2}\) in the form \(px + q + \dfrac{A}{x - 1} + \dfrac{B}{x + 2}\), where \(p, q, A, B \in \mathbb{Z}\). [4]
(b) Hence find \(\displaystyle\int \frac{x^{3} + 2x^{2} - 5x + 5}{x^{2} + x - 2}\,\mathrm{d}x\). [3]
Total 7 marks
(a)
dividing by \(x^{2} + x - 2\) (the fraction is improper) M1
\(x^{3} + 2x^{2} - 5x + 5 = (x + 1)\left(x^{2} + x - 2\right) + (-4x + 7)\) A1
\(x^{2} + x - 2 = (x - 1)(x + 2)\) and \(-4x + 7 = A(x + 2) + B(x - 1)\) M1
\(x = 1\): \(A = 1\); \(\ x = -2\): \(B = -5\). So \(x + 1 + \dfrac{1}{x - 1} - \dfrac{5}{x + 2}\) A1
[4 marks]
(b)
\(\int (x + 1)\,\mathrm{d}x = \dfrac{x^{2}}{2} + x\) A1
\(\int \dfrac{1}{x - 1}\,\mathrm{d}x = \ln\lvert x - 1\rvert\) A1
\(\int -\dfrac{5}{x + 2}\,\mathrm{d}x = -5\ln\lvert x + 2\rvert\), and \(+\,C\) A1
[3 marks]
See 27.4.
HL 3 Paper 1
(a) Write \(4x + 5\) in the form \(a(2x - 2) + b\), and write \(x^{2} - 2x + 10\) in the form \((x + h)^{2} + k\). [3]
(b) Hence find \(\displaystyle\int \frac{4x + 5}{x^{2} - 2x + 10}\,\mathrm{d}x\). [3]
(c) A function \(f\) has \(f'(x) = \dfrac{4x + 5}{x^{2} - 2x + 10}\) and \(f(1) = 0\). Find the exact value of \(f(4)\). [2]
Total 8 marks
(a)
\(a = 2\) A1
\(b = 9\) A1
\(x^{2} - 2x + 10 = (x - 1)^{2} + 9\) A1
[3 marks]
(b)
\(2\int \dfrac{2x - 2}{x^{2} - 2x + 10}\,\mathrm{d}x = 2\ln\left(x^{2} - 2x + 10\right)\) A1
\(9\int \dfrac{1}{(x - 1)^{2} + 3^{2}}\,\mathrm{d}x\) (arctan form with \(a = 3\)) M1
so the integral is \(2\ln\left(x^{2} - 2x + 10\right) + 3\arctan\left(\dfrac{x - 1}{3}\right) + C\) A1
A modulus in the logarithm is not needed, since \((x - 1)^{2} + 9 > 0\), but do not penalise it.
[3 marks]
(c)
\(f(1) = 2\ln 9 + 0 + C = 0\), so \(C = -4\ln 3\) M1
\(f(4) = 2\ln 18 + 3 \times \dfrac{\pi}{4} - 4\ln 3 = 2\ln 2 + \dfrac{3\pi}{4}\) A1
[2 marks]
See 27.4.
HL 4 Paper 1
By using the substitution \(u = \sqrt{\mathrm{e}^{x} - 1}\), find \(\displaystyle\int \sqrt{\mathrm{e}^{x} - 1}\,\mathrm{d}x\), for \(x > 0\). [6]
Total 6 marks
\(u^{2} = \mathrm{e}^{x} - 1\), so \(2u\,\dfrac{\mathrm{d}u}{\mathrm{d}x} = \mathrm{e}^{x} = u^{2} + 1\) M1
\(\mathrm{d}x = \dfrac{2u}{u^{2} + 1}\,\mathrm{d}u\) A1
\(\int u \times \dfrac{2u}{u^{2} + 1}\,\mathrm{d}u = \int \dfrac{2u^{2}}{u^{2} + 1}\,\mathrm{d}u\) A1
\(= \int \left(2 - \dfrac{2}{u^{2} + 1}\right)\mathrm{d}u\) M1
\(= 2u - 2\arctan u + C\) A1
\(= 2\sqrt{\mathrm{e}^{x} - 1} - 2\arctan\sqrt{\mathrm{e}^{x} - 1} + C\) A1
[6 marks]
HL 5 Paper 2
(a) By using integration by parts twice, find \(\displaystyle\int \sin(\ln x)\,\mathrm{d}x\), for \(x > 0\). [6]
A function \(F\) is defined for \(x > 0\), with \(F'(x) = \sin(\ln x)\) and \(F(1) = 0\).
(b) Find the smallest value of \(x\), \(x > 1\), for which \(F(x) = 0\). [3]
Total 9 marks
(a)
\(u = \sin(\ln x)\), \(\dfrac{\mathrm{d}v}{\mathrm{d}x} = 1\), so \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = \dfrac{\cos(\ln x)}{x}\), \(v = x\) M1
\(I = x\sin(\ln x) - \int \cos(\ln x)\,\mathrm{d}x\) A1
again: \(u = \cos(\ln x)\), \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = -\dfrac{\sin(\ln x)}{x}\), \(v = x\) M1
\(\int \cos(\ln x)\,\mathrm{d}x = x\cos(\ln x) + I\) A1
\(I = x\sin(\ln x) - x\cos(\ln x) - I\) M1
\(I = \dfrac{x}{2}\big(\sin(\ln x) - \cos(\ln x)\big) + C\) A1
[6 marks]
(b)
\(F(1) = \dfrac{1}{2}(0 - 1) + C = 0\), so \(C = \dfrac{1}{2}\) A1
\(\dfrac{x}{2}\big(\sin(\ln x) - \cos(\ln x)\big) + \dfrac{1}{2} = 0\) (for example, using Analyze Graph → Zero on a GDC) M1
\(x = 51.5\) \((51.456\ldots)\) A1
[3 marks]
See 27.6.
HL 6 Paper 1
(a) Show that \[ \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{x}{x^{2} + 4}\right) = \frac{8}{\left(x^{2} + 4\right)^{2}} - \frac{1}{x^{2} + 4}. \] [3]
(b) Hence find \(\displaystyle\int \frac{1}{\left(x^{2} + 4\right)^{2}}\,\mathrm{d}x\). [4]
(c) Use the substitution \(x = 2\tan\theta\), where \(-\dfrac{\pi}{2} < \theta < \dfrac{\pi}{2}\), to show that \[ \int \frac{1}{\left(x^{2} + 4\right)^{2}}\,\mathrm{d}x = \frac{1}{8}\int \cos^{2}\theta\,\mathrm{d}\theta. \] [3]
(d) Hence find \(\displaystyle\int \frac{1}{\left(x^{2} + 4\right)^{2}}\,\mathrm{d}x\) by this second method, and show that your answer agrees with your answer to part (b). [3]
(e) Find \(\displaystyle\int \frac{x^{2}}{\left(x^{2} + 4\right)^{2}}\,\mathrm{d}x\). [3]
Total 16 marks
(a)
quotient rule: \(\dfrac{\left(x^{2} + 4\right) - x(2x)}{\left(x^{2} + 4\right)^{2}}\) M1
\(= \dfrac{4 - x^{2}}{\left(x^{2} + 4\right)^{2}}\) A1
\(= \dfrac{8 - \left(x^{2} + 4\right)}{\left(x^{2} + 4\right)^{2}} = \dfrac{8}{\left(x^{2} + 4\right)^{2}} - \dfrac{1}{x^{2} + 4}\) A1 AG
[3 marks]
(b)
integrating (a): \(\dfrac{x}{x^{2} + 4} = 8\int \dfrac{\mathrm{d}x}{\left(x^{2} + 4\right)^{2}} - \int \dfrac{\mathrm{d}x}{x^{2} + 4}\) M1
\(\int \dfrac{1}{x^{2} + 4}\,\mathrm{d}x = \dfrac{1}{2}\arctan\left(\dfrac{x}{2}\right)\) A1
\(8\int \dfrac{\mathrm{d}x}{\left(x^{2} + 4\right)^{2}} = \dfrac{x}{x^{2} + 4} + \dfrac{1}{2}\arctan\left(\dfrac{x}{2}\right)\) A1
\(\int \dfrac{\mathrm{d}x}{\left(x^{2} + 4\right)^{2}} = \dfrac{x}{8\left(x^{2} + 4\right)} + \dfrac{1}{16}\arctan\left(\dfrac{x}{2}\right) + C\) A1
[4 marks]
(c)
\(\dfrac{\mathrm{d}x}{\mathrm{d}\theta} = 2\sec^{2}\theta\), so \(\mathrm{d}x = 2\sec^{2}\theta\,\mathrm{d}\theta\) M1
\(x^{2} + 4 = 4\tan^{2}\theta + 4 = 4\sec^{2}\theta\) A1
\(\int \dfrac{2\sec^{2}\theta}{16\sec^{4}\theta}\,\mathrm{d}\theta = \dfrac{1}{8}\int \cos^{2}\theta\,\mathrm{d}\theta\) A1 AG
[3 marks]
(d)
\(\dfrac{1}{8}\int \cos^{2}\theta\,\mathrm{d}\theta = \dfrac{1}{16}\int (1 + \cos 2\theta)\,\mathrm{d}\theta\) M1
\(= \dfrac{\theta}{16} + \dfrac{\sin 2\theta}{32} + C\) A1
\(\theta = \arctan\dfrac{x}{2}\) and \(\sin 2\theta = 2\sin\theta\cos\theta = \dfrac{2 \times x \times 2}{x^{2} + 4} = \dfrac{4x}{x^{2} + 4}\),
so \(\dfrac{1}{16}\arctan\left(\dfrac{x}{2}\right) + \dfrac{x}{8\left(x^{2} + 4\right)} + C\), the same as (b) A1
For \(\sin\theta = \dfrac{x}{\sqrt{x^{2} + 4}}\) and \(\cos\theta = \dfrac{2}{\sqrt{x^{2} + 4}}\), accept a right-angled triangle with sides \(x\), \(2\) and \(\sqrt{x^{2} + 4}\).
[3 marks]
(e)
\(x^{2} = \left(x^{2} + 4\right) - 4\), so \(\dfrac{x^{2}}{\left(x^{2} + 4\right)^{2}} = \dfrac{1}{x^{2} + 4} - \dfrac{4}{\left(x^{2} + 4\right)^{2}}\) M1
\(\int = \dfrac{1}{2}\arctan\left(\dfrac{x}{2}\right) - \dfrac{x}{2\left(x^{2} + 4\right)} - \dfrac{1}{4}\arctan\left(\dfrac{x}{2}\right)\) A1
\(= \dfrac{1}{4}\arctan\left(\dfrac{x}{2}\right) - \dfrac{x}{2\left(x^{2} + 4\right)} + C\) A1
[3 marks]
HL 7 Paper 1
Let \(I = \displaystyle\int \mathrm{e}^{2x}\sin 3x\,\mathrm{d}x\) and \(J = \displaystyle\int \mathrm{e}^{2x}\cos 3x\,\mathrm{d}x\).
(a) Use integration by parts to show that \(2I = \mathrm{e}^{2x}\sin 3x - 3J\). (You may ignore constants of integration in parts (a) and (b).) [3]
(b) Show that \(2J = \mathrm{e}^{2x}\cos 3x + 3I\). [2]
(c) Hence find \(I\). [3]
A curve \(y = f(x)\) passes through the origin, and \(f'(x) = \mathrm{e}^{2x}\sin 3x\).
(d) Find the exact value of \(f\left(\dfrac{\pi}{3}\right)\). [3]
(e) Find the \(x\)-coordinates of the stationary points of the curve for \(0 < x < \pi\), and determine their nature. [3]
Total 14 marks
(a)
\(u = \sin 3x\), \(\dfrac{\mathrm{d}v}{\mathrm{d}x} = \mathrm{e}^{2x}\), so \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = 3\cos 3x\), \(v = \dfrac{1}{2}\mathrm{e}^{2x}\) M1
\(I = \dfrac{1}{2}\mathrm{e}^{2x}\sin 3x - \dfrac{3}{2}\int \mathrm{e}^{2x}\cos 3x\,\mathrm{d}x\) A1
\(I = \dfrac{1}{2}\mathrm{e}^{2x}\sin 3x - \dfrac{3}{2}J\), so \(2I = \mathrm{e}^{2x}\sin 3x - 3J\) A1 AG
[3 marks]
(b)
\(u = \cos 3x\), \(\dfrac{\mathrm{d}v}{\mathrm{d}x} = \mathrm{e}^{2x}\): \(\ J = \dfrac{1}{2}\mathrm{e}^{2x}\cos 3x + \dfrac{3}{2}I\) M1
\(2J = \mathrm{e}^{2x}\cos 3x + 3I\) A1 AG
[2 marks]
(c)
from (a), \(3J = \mathrm{e}^{2x}\sin 3x - 2I\)
so \(3 \times\) (b) gives \(2\left(\mathrm{e}^{2x}\sin 3x - 2I\right) = 3\mathrm{e}^{2x}\cos 3x + 9I\) M1
\(13I = \mathrm{e}^{2x}(2\sin 3x - 3\cos 3x)\) A1
\(I = \dfrac{1}{13}\mathrm{e}^{2x}(2\sin 3x - 3\cos 3x) + C\) A1
[3 marks]
(d)
\(f(0) = \dfrac{1}{13}(0 - 3) + C = 0\) M1
\(C = \dfrac{3}{13}\) A1
\(f\left(\dfrac{\pi}{3}\right) = \dfrac{1}{13}\mathrm{e}^{\frac{2\pi}{3}}\big(0 - 3(-1)\big) + \dfrac{3}{13} = \dfrac{3}{13}\left(\mathrm{e}^{\frac{2\pi}{3}} + 1\right)\) A1
[3 marks]
(e)
\(\mathrm{e}^{2x} > 0\), so \(f'(x) = 0\) when \(\sin 3x = 0\): \(\ x = \dfrac{\pi}{3}\) or \(x = \dfrac{2\pi}{3}\) A1
\(\sin 3x\) is positive, then negative, then positive on \(\left(0, \dfrac{\pi}{3}\right)\), \(\left(\dfrac{\pi}{3}, \dfrac{2\pi}{3}\right)\), \(\left(\dfrac{2\pi}{3}, \pi\right)\), so \(f'\) changes sign R1
local maximum at \(x = \dfrac{\pi}{3}\); local minimum at \(x = \dfrac{2\pi}{3}\) A1
[3 marks]
HL 8 Paper 1
Let \(k\) be a positive constant.
(a) Show that the substitution \(u = \mathrm{e}^{x}\) transforms \(\displaystyle\int \frac{1}{\mathrm{e}^{x} + k\,\mathrm{e}^{-x} - 4}\,\mathrm{d}x\) into \(\displaystyle\int \frac{1}{u^{2} - 4u + k}\,\mathrm{d}u\). [3]
(b) In the case \(k = 3\), find \(\displaystyle\int \frac{1}{\mathrm{e}^{x} + 3\,\mathrm{e}^{-x} - 4}\,\mathrm{d}x\), for \(x > \ln 3\). [4]
(c) In the case \(k = 5\), find \(\displaystyle\int \frac{1}{\mathrm{e}^{x} + 5\,\mathrm{e}^{-x} - 4}\,\mathrm{d}x\). [3]
A function \(h\) is defined for \(x \in \mathbb{R}\), with \(h'(x) = \dfrac{1}{\mathrm{e}^{x} + 5\,\mathrm{e}^{-x} - 4}\) and \(h(\ln 2) = 0\).
(d) Find \(h(x)\). [2]
(e) Find the equations of the horizontal asymptotes of the graph of \(h\). Hence find the range of \(h\), justifying your answer. [4]
Total 16 marks
(a)
\(\dfrac{\mathrm{d}u}{\mathrm{d}x} = \mathrm{e}^{x} = u\), so \(\mathrm{d}x = \dfrac{\mathrm{d}u}{u}\) M1
\(\int \dfrac{1}{u + \frac{k}{u} - 4} \times \dfrac{1}{u}\,\mathrm{d}u\) A1
\(= \int \dfrac{1}{u^{2} + k - 4u}\,\mathrm{d}u = \int \dfrac{1}{u^{2} - 4u + k}\,\mathrm{d}u\) A1 AG
[3 marks]
(b)
\(\dfrac{1}{(u - 1)(u - 3)} = \dfrac{A}{u - 1} + \dfrac{B}{u - 3}\), so \(1 = A(u - 3) + B(u - 1)\) M1
\(A = -\dfrac{1}{2}\), \(B = \dfrac{1}{2}\) A1
\(\int = \dfrac{1}{2}\ln\lvert u - 3\rvert - \dfrac{1}{2}\ln\lvert u - 1\rvert + C\) A1
\(= \dfrac{1}{2}\ln\left(\dfrac{\mathrm{e}^{x} - 3}{\mathrm{e}^{x} - 1}\right) + C\) A1
Accept the answer with modulus signs. For \(x > \ln 3\) both \(\mathrm{e}^{x} - 3\) and \(\mathrm{e}^{x} - 1\) are positive.
[4 marks]
(c)
\(u^{2} - 4u + 5 = (u - 2)^{2} + 1\) M1
\(\int \dfrac{1}{(u - 2)^{2} + 1}\,\mathrm{d}u = \arctan(u - 2) + C\) A1
\(= \arctan\left(\mathrm{e}^{x} - 2\right) + C\) A1
[3 marks]
(d)
\(h(\ln 2) = \arctan(2 - 2) + C = 0\) M1
\(C = 0\), so \(h(x) = \arctan\left(\mathrm{e}^{x} - 2\right)\) A1
[2 marks]
(e)
as \(x \to \infty\), \(\mathrm{e}^{x} - 2 \to \infty\), so \(h(x) \to \dfrac{\pi}{2}\): asymptote \(y = \dfrac{\pi}{2}\) A1
as \(x \to -\infty\), \(\mathrm{e}^{x} - 2 \to -2\), so \(h(x) \to \arctan(-2)\): asymptote \(y = -\arctan 2\) A1
\(h'(x) = \dfrac{\mathrm{e}^{x}}{\left(\mathrm{e}^{x} - 2\right)^{2} + 1} > 0\) for all \(x\), so \(h\) is increasing R1
range: \(-\arctan 2 < h(x) < \dfrac{\pi}{2}\) A1
[4 marks]