Edexcel A-Level Mathematics A2 P4.6 Integration Questions

Practise integration techniques for areas, volumes, differential equations and exact values using algebraic methods rather than calculators.

Exam points

set up areas or volumes as definite integrals from curves, parametric forms or diagrams
use substitution, parts or partial fractions to produce exact logarithmic or exponential forms
solve separable differential equations by integrating and applying initial conditions

Covered syllabus points

P4.6.1 - Evaluation of volume of revolution P4.6.2 - Integration by substitution and parts P4.6.3 - Simple cases of integration P4.6.4 - Analytical solution of simple first P4.6.5 - Use integration to find the area

[Maximum number: 10]

Figure 3

Figure 3 shows a sketch of the curve C with parametric equations

x=6 t-3 \sin 2 t \quad y=2 \cos t \quad 0 \leqslant t \leqslant \frac{\pi}{2}

The curve meets the y-axis at 2 and the x-axis at k, where k is a constant.

(a)

Show that the volume of this solid is given by

\int_{0}^{\alpha} \beta(1-\cos 4 t) \mathrm{d} t

where $\alpha$ and $\beta$ are constants to be found.

[ 4 ]

Attempts $y^{2} \frac{\mathrm{~d} x}{\mathrm{~d} t}=4 \cos ^{2} t(6-6 \cos 2 t)$ $=(2 \cos 2 t+2)(6-6 \cos 2 t)=6\left(2-2 \cos ^{2} 2 t\right)=6(2-(1+\cos 4 t))$
Volume $=\int_{0}^{\frac{\pi}{2}} \pi y^{2} \frac{\mathrm{~d} x}{\mathrm{~d} t} \mathrm{~d} t=\int_{0}^{\frac{\pi}{2}} 6 \pi(1-\cos 4 t) \mathrm{d} t$

(b)

Hence, using algebraic integration, find the exact volume of this solid.

[ 6 ]

$\left[6 \pi t-\frac{6 \pi \sin 4 t}{4}\right]_{0}^{\frac{\pi}{2}},=3 \pi^{2}$

M1, A1

(6)

(14 marks)

[Maximum number: 12]

(a)

Using a suitable substitution, find, using calculus, the value of

\int_{1}^{5} \frac{3 x}{\sqrt{2 x-1}} \mathrm{~d} x

(Solutions relying entirely on calculator technology are not acceptable.)

[ 6 ]

Use u=2x-1, so $dx=\frac12\,du$ and $x=\frac{u+1}{2}$ .
$\int_1^5\frac{3x}{\sqrt{2x-1}}\,dx =\int_1^9\left(\frac34u^{1/2}+\frac34u^{-1/2}\right)\,du$ $=\left[\frac12u^{3/2}+\frac32u^{1/2}\right]_1^9$ $=\left(\frac12\cdot9^{3/2}+\frac32\cdot9^{1/2}\right) -\left(\frac12+\frac32\right)=16$

(b)

Find

\int \frac{6 x^{2}-16}{(x+1)(2 x-3)} d x

[ 6 ]

$\frac{6x^2-16}{(x+1)(2x-3)} =3+\frac{2}{x+1}-\frac{1}{2x-3}$ $\int\frac{6x^2-16}{(x+1)(2x-3)}\,dx =\int\left(3+\frac{2}{x+1}-\frac{1}{2x-3}\right)\,dx$ $=3x+2\ln|x+1|-\frac12\ln|2x-3|+c$

[Maximum number: 3]

f(x)=\frac{2 x^{4}+15 x^{3}+35 x^{2}+21 x-4}{(x+3)^{2}} \quad x \in \mathbb{R} \quad x>-3

Hence find,

\int \mathrm{f}(x) \mathrm{d} x

$\int f(x)\,dx=\int\left(2x^2+3x-1+\frac{5}{(x+3)^2}\right)\,dx$ $=\frac{2x^3}{3}+\frac{3x^2}{2}-x-\frac{5}{x+3}+c$
(3)

All question bank results loaded

Covered syllabus points

Question 1

Question (a)

Question (b)

Question 2

Question (a)

Question (b)

Question 3