CAIE Further Math A2 2.4 Integration

[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

[Maximum number: 5]

The curve C has parametric equations

x=\frac{2}{3} t^{\frac{3}{2}}-2 t^{\frac{1}{2}}, \quad y=2 t+5, \quad \text { for } 0<t \leqslant 3 .

Find the exact length of C.

$\dot{x}=t^{\frac{1}{2}}-t^{-\frac{1}{2}}, \quad \dot{y}=2$
Differentiates x and y with respect to t.

$\dot{x}^{2}+\dot{y}^{2}=\left(t^{\frac{1}{2}}-t^{-\frac{1}{2}}\right)^{2}+4=t+2+t^{-1}=\left(t^{\frac{1}{2}}+t^{-\frac{1}{2}}\right)^{2}$
Factorises $\dot{x}^{2}+\dot{y}^{2}$ .

$\int_{0}^{3} t^{\frac{1}{2}}+t^{-\frac{1}{2}} \mathrm{~d} t=\left[\frac{2}{3} t^{\frac{3}{2}}+2 t^{\frac{1}{2}}\right]_{0}^{3}=4 \sqrt{3}$
Applies correct formula for arc length.
M1 for their $\int_{0}^{3} \sqrt{\dot{x}^{2}+\dot{y}^{2}} \mathrm{~d} t$ .
Answer must be simplified to $4 \sqrt{3}$ for A1.

5

[Maximum number: 2]

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

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

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

2

[Maximum number: 5]

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

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} .

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

CAIE A-Level Further Math A2 2.4 Integration Questions

Exam points

Covered syllabus points

Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question 2

Question (a)

Question (b)

Question 3

Question 4

Question 5