Further integration (substitution, parts, partial fractions, trapezium rule, differential equations) – A-level Maths exam questions with mark scheme

1CoreCalculator4 marks

Find

(a) $\displaystyle\int(2x+3)^{5}\,\mathrm{d}x$ [2]

(b) $\displaystyle\int\left(4\mathrm{e}^{2x}+\frac{3}{x}\right)\mathrm{d}x,\quad x>0$ [2]

(a) $k(2x+3)^{6}$M1
$\dfrac{(2x+3)^{6}}{12}+c$A1
(b) Either $4\mathrm{e}^{2x}\to2\mathrm{e}^{2x}$ or $\dfrac3x\to3\ln x$M1
$2\mathrm{e}^{2x}+3\ln x+c$A1
Answer: (a) $\dfrac{(2x+3)^{6}}{12}+c$ (b) $2\mathrm{e}^{2x}+3\ln x+c$
Examiner tip: Divide by the derivative of the inside: $6\times2=12$ in (a).

2StandardCalculator5 marks

Use integration by parts to find the exact value of $$\int_{0}^{\frac{\pi}{4}}x\cos2x\,\mathrm{d}x$$ [5]

Parts the right way round: $Ax\sin2x-\displaystyle\int B\sin2x\,\mathrm{d}x$M1
$\tfrac12x\sin2x-\displaystyle\int\tfrac12\sin2x\,\mathrm{d}x$A1
$\tfrac12x\sin2x+\tfrac14\cos2x$A1
Substitutes limits $\tfrac{\pi}{4}$ and $0$ into an expression of the correct form and subtractsM1
$\dfrac{\pi}{8}-\dfrac14$A1
Answer: $\dfrac{\pi}{8}-\dfrac14$
Examiner tip: Don't assume the bottom limit gives zero: at $x=0$, $\tfrac14\cos0=\tfrac14$, which must be subtracted.

3StandardCalculator6 marks

Use the substitution $u=2x-1$ to find the exact value of $$\int_{1}^{5}\frac{x}{\sqrt{2x-1}}\,\mathrm{d}x$$ [6]

$\mathrm{d}x=\tfrac12\,\mathrm{d}u$ and $x=\dfrac{u+1}{2}$B1
Fully substitutes to obtain an integral in $u$ only: $\displaystyle\int\frac{u+1}{2\sqrt u}\cdot\frac12\,\mathrm{d}u$M1
$\displaystyle\tfrac14\int\left(u^{\frac12}+u^{-\frac12}\right)\mathrm{d}u$A1
Integrates: $u^{n}\to u^{n+1}$ for at least one term, giving $\tfrac14\left(\tfrac23u^{\frac32}+2u^{\frac12}\right)$M1
Uses limits $u=9$ and $u=1$ (or reverts to $x$ and uses 5 and 1)M1
$\dfrac{16}{3}$A1
Answer: $\dfrac{16}{3}$
Examiner tip: Change the limits to $u$-values (1 and 9) – using the $x$-limits with a $u$-integral is a very common lost mark.

4StandardCalculator6 marks

(a) Express $\dfrac{x+5}{(x-1)(x+2)}$ in partial fractions. [2]

(b) Hence show that $$\int_{2}^{5}\frac{x+5}{(x-1)(x+2)}\,\mathrm{d}x=\ln\frac{64}{7}$$ [4]

(a) $x+5\equiv A(x+2)+B(x-1)$ and substitutes a value or compares coefficientsM1
$\dfrac{2}{x-1}-\dfrac{1}{x+2}$A1
(b) Integrates to $p\ln(x-1)+q\ln(x+2)$M1
$2\ln(x-1)-\ln(x+2)$A1
Substitutes limits 5 and 2 and subtracts: $(2\ln4-\ln7)-(2\ln1-\ln4)$M1
Uses log laws correctly: $3\ln4-\ln7=\ln64-\ln7=\ln\dfrac{64}{7}$ csoA1*
Answer: (a) $\dfrac{2}{x-1}-\dfrac{1}{x+2}$
Examiner tip: In a 'show that' the log-law steps must be seen: write $3\ln4=\ln64$ before combining.

5StandardCalculator5 marks

Given that $y=0$ when $x=0$, solve the differential equation $$\frac{\mathrm{d}y}{\mathrm{d}x}=(2x+1)\mathrm{e}^{-y}$$ giving your answer in the form $y=\mathrm{f}(x)$. [5]

Separates variables: $\displaystyle\int\mathrm{e}^{y}\,\mathrm{d}y=\int(2x+1)\,\mathrm{d}x$M1
$\mathrm{e}^{y}=x^{2}+x\ (+c)$A1
Uses $x=0,\ y=0$ to find $c$ (must have included a constant)M1
$c=1$A1
$y=\ln(x^{2}+x+1)$A1
Answer: $y=\ln(x^{2}+x+1)$
Examiner tip: Find $c$ before taking logs – writing $y=\ln(x^{2}+x)+c$ is a different (wrong) function.

6HardCalculator8 marks

A cup of tea is placed in a room kept at a constant temperature of $20^\circ$C. The temperature of the tea, $\theta\,^\circ$C, $t$ minutes after it is made, is modelled by $$\frac{\mathrm{d}\theta}{\mathrm{d}t}=-k(\theta-20)$$ where $k$ is a positive constant. Initially the temperature of the tea is $90^\circ$C.

(a) Show that $\theta=20+70\mathrm{e}^{-kt}$. [4]

After 5 minutes the temperature of the tea is $60^\circ$C.

(b) Find the exact value of $k$, giving your answer in the form $a\ln b$ where $a$ and $b$ are rational. [2]

(a) Separates: $\displaystyle\int\frac{1}{\theta-20}\,\mathrm{d}\theta=\int-k\,\mathrm{d}t$M1
$\ln(\theta-20)=-kt+c$A1
Uses $t=0,\ \theta=90$ to find $c=\ln70$ (or $A=70$ in $\theta-20=A\mathrm{e}^{-kt}$)M1
Removes logs correctly to reach $\theta=20+70\mathrm{e}^{-kt}$ csoA1*
(b) $60=20+70\mathrm{e}^{-5k}\Rightarrow\mathrm{e}^{-5k}=\tfrac47$ and takes logsM1
$k=\tfrac15\ln\tfrac74$ (accept $-\tfrac15\ln\tfrac47$)A1
(c) $30=20+70\mathrm{e}^{-kt}\Rightarrow t=\dfrac{\ln7}{k}$ with their $k$M1
$17$ minutes (from $t=17.39\ldots$)A1
Answer: (b) $k=\tfrac15\ln\tfrac74$ (c) 17 minutes
Examiner tip: Include the constant of integration as soon as you integrate – adding it after removing logs gives the wrong form.

7StandardCalculator5 marks

The table shows corresponding values of $x$ and $y$ for $y=\sqrt{2+x^{3}}$, with values of $y$ given to 5 decimal places.

$x$	0	0.5	1	1.5	2
$y$	1.41421	1.45774		2.31840	3.16228

(a) Complete the table, giving the missing value of $y$ to 5 decimal places. [1]

(b) Use the trapezium rule, with all the values of $y$ in the completed table, to find an estimate for $$\int_{0}^{2}\sqrt{2+x^{3}}\,\mathrm{d}x$$ giving your answer to 3 decimal places. [3]

(c) Given that the curve $y=\sqrt{2+x^{3}}$ is convex for $0\leqslant x\leqslant2$, state, with a reason, whether your answer to (b) is an overestimate or an underestimate. [1]

(a) $1.73205$ (accept $\sqrt3$ given as $1.73205$)B1
(b) Strip width $h=0.5$ (seen or used, e.g. as $\tfrac12\times0.5$)B1
Correct structure: $\tfrac{0.5}{2}\left[1.41421+3.16228+2(1.45774+1.73205+2.31840)\right]$ (allow ft on their (a); one slip in copying a value allowed)M1
$3.898$ caoA1
(c) Overestimate, with a correct reason, e.g. the curve is convex so the tops of the trapezia (the chords) lie above the curve. B0 for 'overestimate' with no reason or an incorrect reasonB1
Answer: (a) $1.73205$ (b) $3.898$ (c) overestimate
Examiner tip: The first and last $y$-values are not doubled – put them outside the bracket.

8StretchCalculator6 marks

$$I=\int_{0}^{\pi}\mathrm{e}^{x}\sin x\,\mathrm{d}x$$

By using integration by parts twice, show that $$I=\frac{1}{2}\left(\mathrm{e}^{\pi}+1\right)$$ [6]

First application of parts: $\pm\mathrm{e}^{x}\cos x\pm\displaystyle\int\mathrm{e}^{x}\cos x\,\mathrm{d}x$ (or, taking $u=\sin x$, $\pm\mathrm{e}^{x}\sin x\pm\displaystyle\int\mathrm{e}^{x}\cos x\,\mathrm{d}x$)M1
$-\mathrm{e}^{x}\cos x+\displaystyle\int\mathrm{e}^{x}\cos x\,\mathrm{d}x$ (or $\mathrm{e}^{x}\sin x-\displaystyle\int\mathrm{e}^{x}\cos x\,\mathrm{d}x$ if $u=\sin x$)A1
Second application of parts with the same choice of $u$: $\displaystyle\int\mathrm{e}^{x}\cos x\,\mathrm{d}x=\mathrm{e}^{x}\sin x-\int\mathrm{e}^{x}\sin x\,\mathrm{d}x$ (or, with $u$ trig, $\mathrm{e}^{x}\cos x+\displaystyle\int\mathrm{e}^{x}\sin x\,\mathrm{d}x$)M1
Recognises the original integral and rearranges: $2I=\left[\mathrm{e}^{x}\sin x-\mathrm{e}^{x}\cos x\right]_{0}^{\pi}$M1
Correct limits: $(0+\mathrm{e}^{\pi})-(0-1)$A1
$I=\tfrac12(\mathrm{e}^{\pi}+1)$ csoA1*
Answer: Proof: $2I=\left[\mathrm{e}^{x}\sin x-\mathrm{e}^{x}\cos x\right]_{0}^{\pi}=\mathrm{e}^{\pi}+1$, so $I=\tfrac12(\mathrm{e}^{\pi}+1)$
Examiner tip: Use the same choice of $u$ (or $\mathrm{d}v/\mathrm{d}x$) both times – swapping just undoes the first step.

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