Integration by substitution and by parts: practice worksheet

1. Question 1warm-up · no calculator

   (a)Find $\int x\cos 2x\,\mathrm{d}x$.[3]

   Markscheme

   attempt at integration by parts with $u = x$, $\frac{\mathrm{d}v}{\mathrm{d}x} = \cos 2x$	M1
   $\frac{x}{2}\sin 2x - \int \frac{1}{2}\sin 2x\,\mathrm{d}x$	A1
   $\frac{x}{2}\sin 2x + \frac{1}{4}\cos 2x + c$	A1

   Award M1 for a correct structure $uv - \int v\,u'$ even with a sign or coefficient error in $v$. Condone absence of $+c$ until the final A1.

   Answer: $\frac{x}{2}\sin 2x + \frac{1}{4}\cos 2x + c$

2. Question 2warm-up · no calculator

   (a)Using the substitution $u = 1 + x^2$, find the exact value of $\int_0^{\sqrt{3}} \frac{x}{\sqrt{1+x^2}}\,\mathrm{d}x$.[4]

   Markscheme

   $\frac{\mathrm{d}u}{\mathrm{d}x} = 2x$ (or $\mathrm{d}u = 2x\,\mathrm{d}x$)	A1
   $\frac{1}{2}\int u^{-\frac{1}{2}}\,\mathrm{d}u$	M1
   new limits $u = 1$ and $u = 4$ (or returning to $x$ and using original limits)	A1
   $\left[\sqrt{u}\right]_1^4 = 2 - 1 = 1$	A1

   Condone absence of or incorrect limits up to the M1. The third A1 may be earned by writing $\sqrt{1+x^2}$ and substituting $x = 0$, $x = \sqrt{3}$.

   Answer: $1$

3. Question 3exam-style · no calculator

   (a)Find $\int x^2\sin x\,\mathrm{d}x$.[5]

   Markscheme

   attempt at integration by parts with $u = x^2$, $\frac{\mathrm{d}v}{\mathrm{d}x} = \sin x$	M1
   $-x^2\cos x + \int 2x\cos x\,\mathrm{d}x$	A1
   second application of parts with $u = 2x$ (or $x$), $\frac{\mathrm{d}v}{\mathrm{d}x} = \cos x$	M1
   $2x\sin x - \int 2\sin x\,\mathrm{d}x$	A1
   $-x^2\cos x + 2x\sin x + 2\cos x + c$	A1

   Award the first A1 for both the correct $uv$ term and the correct remaining integral; if only one is correct award M1A0. Second M1 requires the correct direction of differentiation (polynomial degree reduced).

   Answer: $-x^2\cos x + 2x\sin x + 2\cos x + c$

   (b)Hence show that $\int_0^{\pi} x^2\sin x\,\mathrm{d}x = \pi^2 - 4$.[2]

   Markscheme

   substituting limits into their answer to (a): $\left(-\pi^2(-1) + 0 + 2(-1)\right) - \left(0 + 0 + 2\right)$	M1
   $= (\pi^2 - 2) - 2 = \pi^2 - 4$	A1
   $\pi^2 - 4$	AG

   The A1 requires the values $\cos\pi = -1$, $\sin\pi = 0$, $\cos 0 = 1$ seen or clearly used. Do not award A1 if the answer to (a) is incorrect and the given result is reached by manipulation.

   Answer: $\pi^2 - 4$

4. Question 4exam-style · no calculator

   (a)Express $\frac{1}{u(u+1)}$ in partial fractions.[2]

   Markscheme

   $\frac{1}{u(u+1)} \equiv \frac{A}{u} + \frac{B}{u+1}$ and attempt to find $A$, $B$ (e.g. $1 = A(u+1) + Bu$)	M1
   $\frac{1}{u} - \frac{1}{u+1}$	A1

   Accept the cover-up method with $A = 1$, $B = -1$ stated.

   Answer: $\frac{1}{u} - \frac{1}{u+1}$

   (b)Hence, using the substitution $u = \cos x$, find the exact value of $\int_0^{\frac{\pi}{3}} \frac{\sin x}{\cos^2 x + \cos x}\,\mathrm{d}x$.[5]

   Markscheme

   $\frac{\mathrm{d}u}{\mathrm{d}x} = -\sin x$ (or $\mathrm{d}u = -\sin x\,\mathrm{d}x$)	A1
   $-\int \frac{1}{u(u+1)}\,\mathrm{d}u = -\int \left(\frac{1}{u} - \frac{1}{u+1}\right)\mathrm{d}u$	M1
   limits $u = 1$ and $u = \frac{1}{2}$ (or equivalent, e.g. $\int_{\frac{1}{2}}^{1}$ with sign reversed)	A1
   $\left[-\ln u + \ln(u+1)\right]_1^{\frac{1}{2}}$ or $\left[\ln\frac{u+1}{u}\right]_1^{\frac{1}{2}}$	A1
   $\ln 3 - \ln 2 = \ln\frac{3}{2}$	A1

   Condone absence of or incorrect limits up to the second A1. Condone missing modulus signs. Accept $\ln 3 - \ln 2$ or $-\ln\frac{2}{3}$ for the final A1.

   Answer: $\ln\frac{3}{2}$

5. Question 5exam-style · no calculator

   Let $I = \int e^{2x}\cos x\,\mathrm{d}x$.

   (a)Using integration by parts twice, show that $I = \frac{1}{5}e^{2x}\left(\sin x + 2\cos x\right) + c$.[6]

   Markscheme

   attempt at integration by parts with $u = e^{2x}$, $\frac{\mathrm{d}v}{\mathrm{d}x} = \cos x$	M1
   $I = e^{2x}\sin x - 2\int e^{2x}\sin x\,\mathrm{d}x$	A1
   second application of parts on $\int e^{2x}\sin x\,\mathrm{d}x$ with $u = e^{2x}$, $\frac{\mathrm{d}v}{\mathrm{d}x} = \sin x$	M1
   $\int e^{2x}\sin x\,\mathrm{d}x = -e^{2x}\cos x + 2\int e^{2x}\cos x\,\mathrm{d}x$	A1
   $I = e^{2x}\sin x + 2e^{2x}\cos x - 4I$ (recognising the original integral)	M1
   $5I = e^{2x}(\sin x + 2\cos x) + k$ so $I = \frac{1}{5}e^{2x}(\sin x + 2\cos x) + c$	A1
   $I = \frac{1}{5}e^{2x}(\sin x + 2\cos x) + c$	AG

   METHOD 2: taking $u = \cos x$, $\frac{\mathrm{d}v}{\mathrm{d}x} = e^{2x}$ both times gives $I = \frac{1}{2}e^{2x}\cos x + \frac{1}{4}e^{2x}\sin x - \frac{1}{4}I$; award marks equivalently. Candidates must use the same pairing in both applications, otherwise the identity $I = I$ results and at most M1A1M1A0 is available.

   Answer: $I = \frac{1}{5}e^{2x}(\sin x + 2\cos x) + c$

   (b)The region $R$ is bounded by the curve $y = e^{2x}\cos x$, the $x$-axis and the $y$-axis, for $0 \le x \le \frac{\pi}{2}$. Find the exact area of $R$.[2]

   Markscheme

   area $= \left[\frac{1}{5}e^{2x}(\sin x + 2\cos x)\right]_0^{\frac{\pi}{2}}$	M1
   $= \frac{1}{5}\left(e^{\pi}(1 + 0) - (0 + 2)\right) = \frac{e^{\pi} - 2}{5}$	A1

   Accept $\frac{1}{5}e^{\pi} - \frac{2}{5}$. Since the curve is non-negative on the interval no modulus is needed.

   Answer: $\frac{e^{\pi} - 2}{5}$

6. Question 6challenge · no calculator

   A continuous random variable $X$ has probability density function $f$ given by $$f(x) = \begin{cases} kxe^{x}, & 0 \le x \le 1 \\ 0, & \text{otherwise} \end{cases}$$ where $k$ is a positive constant.

   (a)Using integration by parts, show that $k = 1$.[4]

   Markscheme

   attempt at integration by parts on $\int xe^{x}\,\mathrm{d}x$ with $u = x$, $\frac{\mathrm{d}v}{\mathrm{d}x} = e^{x}$	M1
   $\int xe^{x}\,\mathrm{d}x = xe^{x} - e^{x}\;(= (x-1)e^{x})$	A1
   $k\int_0^1 xe^{x}\,\mathrm{d}x = 1$, i.e. $k\left[(x-1)e^{x}\right]_0^1 = 1$	M1
   $k\left(0 - (-1)\right) = 1 \Rightarrow k = 1$	A1
   $k = 1$	AG

   The second M1 requires the total probability condition with limits 0 and 1 to be explicitly stated. The final A1 requires the values at both limits to be shown.

   Answer: $k = 1$

   (b)Find $\mathrm{P}(X > \ln 2)$, giving your answer in the form $a + b\ln 2$ where $a, b \in \mathbb{Z}$.[3]

   Markscheme

   $\mathrm{P}(X > \ln 2) = \int_{\ln 2}^{1} xe^{x}\,\mathrm{d}x$ or $1 - \int_0^{\ln 2} xe^{x}\,\mathrm{d}x$	M1
   use of $e^{\ln 2} = 2$: $\left[(x-1)e^{x}\right]_{\ln 2}^{1} = 0 - 2(\ln 2 - 1)$	A1
   $= 2 - 2\ln 2$	A1

   Award the first A1 for $(\ln 2 - 1)\cdot 2 + 1 = 2\ln 2 - 1$ if the complement route is used. Accept $2(1 - \ln 2)$.

   Answer: $2 - 2\ln 2$

   (c)Show that $\mathrm{E}(X) = e - 2$.[4]

   Markscheme

   $\mathrm{E}(X) = \int_0^1 x^2e^{x}\,\mathrm{d}x$	M1
   attempt at integration by parts with $u = x^2$, $\frac{\mathrm{d}v}{\mathrm{d}x} = e^{x}$	M1
   $x^2e^{x} - 2\int xe^{x}\,\mathrm{d}x = x^2e^{x} - 2(x-1)e^{x}$ (using part (a))	A1
   $\left[x^2e^{x} - 2(x-1)e^{x}\right]_0^1 = (e - 0) - (0 + 2) = e - 2$	A1
   $\mathrm{E}(X) = e - 2$	AG

   Accept a full second application of parts on $\int xe^{x}\,\mathrm{d}x$ in place of quoting part (a). Accept the antiderivative written as $e^{x}(x^2 - 2x + 2)$.

   Answer: $\mathrm{E}(X) = e - 2$

   (d)The median of $X$ is $m$. Show that $m$ satisfies the equation $(1 - m)e^{m} = \frac{1}{2}$.[2]

   Markscheme

   $\int_0^m xe^{x}\,\mathrm{d}x = \frac{1}{2}$, i.e. $\left[(x-1)e^{x}\right]_0^m = \frac{1}{2}$	M1
   $(m-1)e^{m} + 1 = \frac{1}{2} \Rightarrow (1-m)e^{m} = \frac{1}{2}$	A1
   $(1 - m)e^{m} = \frac{1}{2}$	AG

   Accept the equivalent condition $\int_m^1 xe^{x}\,\mathrm{d}x = \frac{1}{2}$ giving $-(m-1)e^{m} = \frac{1}{2}$ directly.

   Answer: $(1 - m)e^{m} = \frac{1}{2}$

Original questions and notes written by Pietro Meloni, PhD, from the IB syllabus. Not endorsed by or affiliated with the International Baccalaureate Organization. Found a mistake? Please tell me privately at pietro@pietromeloni.com. More questions on this topic: Calculus practice.