CAIE A-Level Further Math A2 2.4 Integration Questions

Practise exact integration through substitutions and reduction formulae, then apply the resulting integrals to bounds, arc lengths and surfaces of revolution.

Exam points

complete the square or choose a trigonometric or hyperbolic substitution for radical integrals
derive and apply a reduction formula with its boundary term, index shift and exact base case
form the correct parametric arc-length or surface-area integrand before applying the limits

Covered syllabus points

2.4.1—Hyperbolic functions 2.4.2—And use reduction formulae for the 2.4.3—How the area under a curve 2.4.4—Arc length and surface area

[Maximum number: 11]

The integral $I_{n}$ , where n is an integer, is defined by $I_{n}=\int_{0}^{\frac{4}{3}}\left(1+x^{2}\right)^{\frac{1}{2} n} \mathrm{~d} x$ .

(a)

Find the exact value of $I_{-1}$ giving your answer in the form $\ln a$ , where a is an integer to be determined.

[ 2 ]

$I_{-1}=\int_{0}^{\frac{4}{3}}\left(1+x^{2}\right)^{-\frac{1}{2}} \mathrm{~d} x=\left[\sinh ^{-1} x\right]_{0}^{\frac{4}{3}}=\ln 3$
Recognises integral.

2

(b)

By considering $\frac{\mathrm{d}}{\mathrm{d} x}\left(x\left(1+x^{2}\right)^{\frac{1}{2} n}\right)$ , or otherwise, show that

(n+1) I_{n}=n I_{n-2}+\frac{4}{3}\left(\frac{5}{3}\right)^{n} .

[ 5 ]

$\frac{\mathrm{d}}{\mathrm{d} x}\left(x\left(1+x^{2}\right)^{\frac{1}{2} n}\right)=n x^{2}\left(1+x^{2}\right)^{\frac{1}{2} n-1}+\left(1+x^{2}\right)^{\frac{1}{2} n}$
Uses the product rule to differentiate.

$n\left(1+x^{2}-1\right)\left(1+x^{2}\right)^{\frac{1}{2} n-1}+\left(1+x^{2}\right)^{\frac{1}{2} n}$

M1*

Uses $x^{2}=1+x^{2}-1$ .

$\left[x\left(1+x^{2}\right)^{\frac{1}{2} n}\right]_{0}^{\frac{4}{3}}=n I_{n}-n I_{n-2}+I_{n}$
Integrates both sides using the limits given. Requires
previous method mark.

$\frac{4}{3}\left(\frac{5}{3}\right)^{n}=(n+1) I_{n}-n I_{n-2} \Rightarrow(n+1) I_{n}=n I_{n-2}+\frac{4}{3}\left(\frac{5}{3}\right)^{n}$
Substitutes limits and rearranges. AG.

5

(c)

A curve has equation $y=x^{2}$ , for $0 \leqslant x \leqslant \frac{2}{3}$ . The arc length of the curve is denoted by s. Use the substitution u=2 x to show that $s=\frac{1}{2} I_{1}$ and find the exact value of s.

[ 4 ]

$s=\int_{0}^{\frac{2}{3}} \sqrt{1+4 x^{2}} \mathrm{~d} x$
Forms correct integral with correct limits.

$\frac{1}{2} \int_{0}^{\frac{4}{3}} \sqrt{1+u^{2}} \mathrm{~d} u=\frac{1}{2} I_{1}$
AG.

$2 I_{1}=I_{-1}+\frac{4}{3}\left(\frac{5}{3}\right)$
Applies reduction formula with n=1.

$s=\frac{1}{4} \ln 3+\frac{5}{9}$
4

[Maximum number: 7]

(a)

By considering the sum of the areas of these rectangles, show that

\int_{0}^{1} \cos x \mathrm{~d} x<\frac{1}{2 n}\left(1-\cos 1+\frac{\sin 1 \sin \frac{1}{n}}{1-\cos \frac{1}{n}}\right)

[ 4 ]

$\int_{0}^{1} \cos x \mathrm{~d} x<\frac{1}{n}+\frac{1}{n} \cos \frac{1}{n}+\frac{1}{n} \cos \frac{2}{n}+\ldots \frac{1}{n} \cos \frac{n-1}{n}$
Forms sum of areas of rectangles given in the diagram.
A0 if comparison with integral missing or unclear.

$=\frac{1}{n}\left(1+\cos \frac{1}{n}+\cos \frac{2}{n}+\ldots+\cos \frac{n-1}{n}\right)=\frac{1}{2 n}\left(1-\cos \frac{n}{n}+\frac{\sin \frac{n}{n} \sin \frac{1}{n}}{1-\cos \frac{1}{n}}\right)$
Applies result from part (b) with $\theta=\frac{1}{n}$ . AG.

4

(b)

Use a similar method to find, in terms of n, a lower bound for $\int_{0}^{1} \cos x \mathrm{~d} x$ .

[ 3 ]

$\int_{0}^{1} \cos x \mathrm{~d} x>\frac{1}{n} \cos \frac{1}{n}+\frac{1}{n} \cos \frac{2}{n}+\ldots \frac{1}{n} \cos \frac{n}{n}$
Forms sum of areas of rectangles. A0 if comparison
with integral missing or unclear.

$=\frac{1}{2 n}\left(1-\cos 1+\frac{\sin 1 \sin \frac{1}{n}}{1-\cos \frac{1}{n}}\right)+\frac{1}{n} \cos 1-\frac{1}{n}=\frac{1}{2 n}\left(\cos 1-1+\frac{\sin 1 \sin \frac{1}{n}}{1-\cos \frac{1}{n}}\right)$
3

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question 2

Question (a)

Question (b)