Further integration (substitution, by parts) and volumes of revolution

1StandardNon-calculatorHL only5 marks

Use integration by parts to find the exact value of $\displaystyle\int_0^1 xe^{2x}\,dx$.

Chooses $u=x,\ \dfrac{dv}{dx}=e^{2x}$ and applies $uv-\int v\,du$M1
$\dfrac{x}{2}e^{2x}-\int\dfrac12e^{2x}dx$A1
$\dfrac{x}{2}e^{2x}-\dfrac14e^{2x}$A1
Substitutes limits $1$ and $0$M1
$\dfrac{e^2}{4}+\dfrac14=\dfrac{e^2+1}{4}$A1
Answer: $\dfrac{e^2+1}{4}$
Examiner tip: Choose $u=x$ so it differentiates away; the other way round makes the integral harder.

2HardNon-calculatorHL only5 marks

Use the substitution $u=x+1$ to find the exact value of $\displaystyle\int_0^3\dfrac{x}{\sqrt{x+1}}\,dx$.

$u=x+1$, $du=dx$, $x=u-1$M1
Limits become $u=1$ to $u=4$A1
Integrand $\dfrac{u-1}{\sqrt u}=u^{1/2}-u^{-1/2}$M1
$\dfrac23u^{3/2}-2u^{1/2}$A1
$\Big(\dfrac{16}{3}-4\Big)-\Big(\dfrac23-2\Big)=\dfrac83$A1
Answer: $\dfrac83$
Examiner tip: Change the limits with the substitution, or convert back to $x$ before substituting; mixing the two loses the final marks.

3StretchNon-calculatorHL only7 marks

(a) By using integration by parts twice, show that $\displaystyle\int e^x\sin x\,dx=\dfrac{e^x(\sin x-\cos x)}{2}+c$. [6]

(b) Hence find the exact value of $\displaystyle\int_0^{\pi}e^x\sin x\,dx$. [1]

Attempts first application of parts with $u=\sin x,\ \dfrac{dv}{dx}=e^x$ (allow $u=e^x$ consistently in both applications)M1
$I=e^x\sin x-\displaystyle\int e^x\cos x\,dx$A1
Attempts second application on $\displaystyle\int e^x\cos x\,dx$ with $u=\cos x,\ \dfrac{dv}{dx}=e^x$M1
$\displaystyle\int e^x\cos x\,dx=e^x\cos x+\int e^x\sin x\,dx$A1
Substitutes back and recognises the original integral: $I=e^x\sin x-e^x\cos x-I$M1
$2I=e^x(\sin x-\cos x)$, leading to $I=\dfrac{e^x(\sin x-\cos x)}{2}+c$ AG (the A1 is for $2I=\ldots$; the AG line earns no extra mark)A1
(b) $\dfrac{e^{\pi}(0+1)-e^0(0-1)}{2}=\dfrac{e^{\pi}+1}{2}$A1
Answer: (a) shown; (b) $\dfrac{e^{\pi}+1}{2}$
Examiner tip: Keep the same choice of $u$ (trigonometric) in both applications, otherwise the integral just undoes itself.

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4HardNon-calculatorHL only4 marks

The region bounded by the curve $y=\sin x$ for $0\le x\le\pi$ and the $x$-axis is rotated through $360^\circ$ about the $x$-axis. Find the exact volume of the solid formed.

$V=\pi\int_0^{\pi}\sin^2x\,dx$ (limits and $\pi$ present)M1
Uses $\sin^2x=\dfrac{1-\cos2x}{2}$M1
$\dfrac{\pi}{2}\Big[x-\dfrac{\sin2x}{2}\Big]_0^{\pi}$A1
$V=\dfrac{\pi^2}{2}$A1
Answer: $V=\dfrac{\pi^2}{2}$
Examiner tip: Square $y$ before integrating: $\int\sin^2x\,dx$ cannot be done without the double-angle identity.

5HardNon-calculatorHL only4 marks

The region enclosed by the curve $y=x^2$ and the line $y=2x$ is rotated through $360^\circ$ about the $x$-axis. Find the exact volume of the solid generated.

Solves $x^2=2x$ to find limits $x=0$ and $x=2$M1
$V=\pi\int_0^2\big[(2x)^2-(x^2)^2\big]dx$ (difference of squares, not square of difference)M1
Antiderivative $\dfrac{4x^3}{3}-\dfrac{x^5}{5}$A1
$V=\dfrac{64\pi}{15}$A1
Answer: $V=\dfrac{64\pi}{15}$
Examiner tip: Subtract the squares of the radii ($\pi\int(R^2-r^2)$); squaring $(2x-x^2)$ is the classic error.

6CoreNon-calculatorHL only4 marks

Find $\displaystyle\int\left(2^{x}+\dfrac{6}{9+x^{2}}\right)dx$.

$\dfrac{2^{x}}{\ln 2}$ (accept $2^x\log_2 e$)A1
Recognises the arctan form: $k\arctan\left(\dfrac{x}{3}\right)$ for any non-zero $k$M1
$2\arctan\left(\dfrac{x}{3}\right)$ (from $6\times\tfrac13$)A1
Fully correct expression including $+\,c$A1
Answer: $\dfrac{2^{x}}{\ln 2}+2\arctan\left(\dfrac{x}{3}\right)+c$
Examiner tip: Use $\int\dfrac{1}{a^2+x^2}dx=\dfrac1a\arctan\dfrac xa$: the factor is $\tfrac13$, so $6$ becomes $2$, not $18$; and $\int 2^x dx$ needs the division by $\ln 2$.

7StandardNon-calculatorHL only7 marks

(a) Find the exact value of $\displaystyle\int_0^{3/2}\dfrac{1}{\sqrt{9-x^{2}}}\,dx$. [2]

(b) Use integration by parts to find the exact value of $\displaystyle\int_1^{e}x^{2}\ln x\,dx$. [5]

(a) $\left[\arcsin\dfrac{x}{3}\right]_0^{3/2}$A1
(a) $\dfrac{\pi}{6}$ (do not accept $30^\circ$ or 0.524)A1
(b) Chooses $u=\ln x$, $\dfrac{dv}{dx}=x^2$ and applies $uv-\int v\dfrac{du}{dx}dx$M1
(b) $\dfrac{x^3}{3}\ln x-\displaystyle\int\dfrac{x^3}{3}\cdot\dfrac1x\,dx$A1
(b) $\dfrac{x^3}{3}\ln x-\dfrac{x^3}{9}$A1
(b) Substitutes limits, using $\ln e=1$ and $\ln 1=0$: $\left(\dfrac{e^3}{3}-\dfrac{e^3}{9}\right)-\left(0-\dfrac19\right)$M1
(b) $\dfrac{2e^{3}+1}{9}$ (accept equivalent, e.g. $\dfrac{2e^3}{9}+\dfrac19$)A1
Answer: (a) $\dfrac{\pi}{6}$ (b) $\dfrac{2e^{3}+1}{9}$
Examiner tip: With $\ln x$ in a product, let $u=\ln x$: you cannot integrate $\ln x$ directly, but differentiating it gives $\tfrac1x$, which cancels a power of $x$.

8StandardNon-calculatorHL only6 marks

The diagram shows part of the curve $y=\ln x$. The shaded region $R$ is enclosed by the curve, the $x$-axis, the $y$-axis and the line $y=2$.

(a) Find the exact area of $R$. [3]

(b) The region $R$ is rotated through $2\pi$ about the $y$-axis. Find the exact volume of the solid formed. [3]

(a) Rearranges to $x=e^{y}$ and sets up area $=\displaystyle\int_0^2 e^{y}\,dy$ (integral with respect to $y$, correct limits)M1
(a) $\left[e^{y}\right]_0^{2}$A1
(a) $e^{2}-1$A1
(b) Uses $V=\pi\displaystyle\int_0^2 x^{2}\,dy$ with $x^2=e^{2y}$ (condone missing $\pi$ here if it appears later)M1
(b) $\pi\left[\dfrac{e^{2y}}{2}\right]_0^{2}$A1
(b) $\dfrac{\pi\left(e^{4}-1\right)}{2}$ (accept $\dfrac{\pi e^4}{2}-\dfrac{\pi}{2}$; do not accept 84.2 alone)A1
Answer: (a) $e^{2}-1$ (b) $\dfrac{\pi\left(e^{4}-1\right)}{2}$
Examiner tip: For a region against the $y$-axis, write $x$ in terms of $y$ and integrate with respect to $y$ using $y$-limits ($0$ to $2$), not $x$-limits ($1$ to $e^2$).

9HardCalculatorHL only6 marks

A glass vase is modelled by rotating the curve $y=3+\sin\left(\dfrac{x}{4}\right)$, for $0\le x\le 12$, through $2\pi$ about the $x$-axis. The $x$-axis is the vertical axis of symmetry of the vase, the base is at $x=0$ and all lengths are in cm.

(a) Write down an integral that gives the volume of the vase, in cm³. [2]

(b) Hence find this volume. [1]

(d) Suggest one reason why the model may overestimate the amount of water the vase can actually hold. [1]

(a) Uses $\pi\displaystyle\int y^{2}\,dx$ (with $\pi$ and squaring)M1
(a) $\pi\displaystyle\int_0^{12}\left(3+\sin\dfrac{x}{4}\right)^{2}dx$ (correct limits and integrand; $dx$ may be missing)A1
(b) $509\ \text{cm}^3$ (509.061…)A1
(c) Sets up $\pi\displaystyle\int_0^{h}\left(3+\sin\dfrac{x}{4}\right)^{2}dx=150$ (variable upper limit; may be seen as a GDC graph/solver)M1
(c) $h=3.97$ cm (3.97443…)A1
(d) The model gives the outer shape, so the thickness of the glass (walls and base) reduces the inside volume (accept: real vase not exactly this curve / a solid of revolution; ignore answers about the vase being open at the top)R1
Answer: (a) $\pi\displaystyle\int_0^{12}\left(3+\sin\frac{x}{4}\right)^{2}dx$ (b) $509\ \text{cm}^3$ (c) $3.97$ cm (d) thickness of the glass
Examiner tip: Set the GDC to radians before integrating a trig function, and in (c) make the upper limit the unknown rather than solving $\pi y^2 h=150$ as if the vase were a cylinder.

10StretchNon-calculatorHL only7 marks

(a) Use the substitution $u=e^{x}$ to show that $$\int_0^{\ln 2}\frac{e^{x}}{e^{2x}+3e^{x}+2}\,dx=\int_1^{2}\frac{1}{(u+1)(u+2)}\,du.$$ [3]

(b) Hence find the exact value of $\displaystyle\int_0^{\ln 2}\frac{e^{x}}{e^{2x}+3e^{x}+2}\,dx$, giving your answer in the form $\ln\left(\dfrac{p}{q}\right)$ where $p,q\in\mathbb{Z}^{+}$. [4]

(a) $\dfrac{du}{dx}=e^{x}$, so $e^{x}\,dx=du$ (accept $dx=\dfrac{du}{u}$)M1
(a) Denominator written as $u^{2}+3u+2=(u+1)(u+2)$A1
(a) Limits changed: $x=0\Rightarrow u=1$, $x=\ln 2\Rightarrow u=2$, leading to the given result (AG)A1
(b) Attempts partial fractions $\dfrac{A}{u+1}+\dfrac{B}{u+2}$M1
(b) $\dfrac{1}{u+1}-\dfrac{1}{u+2}$A1
(b) $\left[\ln(u+1)-\ln(u+2)\right]_1^{2}$ (accept $\left[\ln\dfrac{u+1}{u+2}\right]_1^2$)A1
(b) $\ln\dfrac34-\ln\dfrac23=\ln\left(\dfrac{9}{8}\right)$A1
Answer: $\ln\left(\dfrac{9}{8}\right)$
Examiner tip: In a 'show that', every step must be visible: state $du=e^x dx$, factorise the denominator and change both limits, or the final A1 (AG) is lost.

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