Further Integration (Edexcel IAL Maths): Exam Questions

Given that u = 3x + 2 show that

(i) du = 3 dx,

(ii) integral3 cos(3x +2) dx = integralcos u du.

Hence find an expression in terms of x for the integral

integral3 cos(3x +2) dx.

Use the substitution u = 4x + 1 to find

$\int_{2}^{6} 4 ( 4 x + 1 )^{1 \frac{1}{2}} d x$

Use integration by parts to find an expression for

$\int - 3 x \sin x d x.$

Show that

$\frac{2 x - 1}{( x + 1 ) ( x - 2 )}$

can be written in the form

$\frac{A}{x+1}+\frac{B}{x-2}$

Hence find

$\int \frac{2 x - 1}{( x + 1 ) ( x - 2 )} d x x > 2$

writing your answer as a single logarithm.

The diagram below shows the region R, bounded by the straight lines with equations y = 2x + 1, x =2, x =4 and the x-axis.

(i) For y = 2x + 1, show that y² = 4x² + 4x + 1.

(ii) Find the volume of the solid formed when the region R is rotated 360° around the x-axis.

Use the substitution u = sin x to find the value of

$\int_{2}^{ \frac{\pi }{2} } \cos ^{2} 2 x \cos x d x.$

Use a suitable substitution to find

integral-15 sin(5x-2) dx

Use calculus and the substitution u = x + 4 to show that

$\int_{1}^{2} \frac{x}{x + 4} d x = 1 + 4 I n \frac{5}{6}$

Use integration by parts to find, in terms of e, the exact value of

$\int_{0}^{1}$(5x - 4)e^3x dx

Show that

$\frac{11}{( 2 x - 3 ) ( x + 4 )}$

can be written in the form

$\frac{A}{( 2 x - 3 )} + \frac{B}{( x + 4 )}$

Hence find

$\int \frac{11}{( 2 x - 3 ) ( x + 4 )} d x$

writing your answer as a single logarithm.

The diagram below shows the graph of the curve with equation y = 4 - x².

(i) Find the x-coordinates of the points where the graph of y = 4 - x² intercepts the x-axis.

(ii) The shaded region, R, is to be rotated 360° around the x-axis.

Find the volume of the shape generated.

Use a suitable substitution to find the following

$\int 8 x \sin ( 3 x^{2} + 1 ) d x$

Use the substitution u = 2 + In x to show that

$\int \frac{1}{x ( 2 + l n x )^{3}} d x = \frac{- 1}{2 ( \ln x + 2 )^{2}} + c$

where c is the constant of integration.

Use integration by parts to find

$\int ( 2 x^{2} - 1 ) e^{x} d x$

Show that

$\int I n x d x = x \ln x - x + c$

where c is the constant of integration.

Express

$\frac{x^{2}-4x+7}{(x-1)(x-3)^{2}}$

as partial fractions.

Hence, or otherwise, find

$\int \frac{x^{2} - 4 x + 7}{( x - 1 ) ( x - 3 )^{2} d x}$

The diagram below shows a right-angled triangle with vertices at the origin, the point (h,0) and the point (h,r), where r > 0 and h > 0.

Find an equation of the line on which the hypotenuse of the right-angled triangle lies, giving your answer in the form y = f(x).

The triangle is rotated 360° about the x-axis to form a cone.

Thus use calculus to prove that the general formula for the volume, V, of a cone is

$V=\frac{1}{3}\mathrm{πr}^{2}h$

where r is the base radius of the cone and h is its perpendicular height.

The diagram below shows part of the curve C defined by the equation $y=\frac{1}{a}x^{2}$, where a is a positive constant. The shaded region R is bounded by the curve, the x-axis, and the lines x = 1 and x = 6.

Given that the volume of the solid formed when the region R is rotated 360° about the x-axis is $\frac{311π}{20}$cubic units, find the value of a.

Find an expression for y given that

$y = \int 6 x^{2} e^{x^{3}} d x$

Integrate

$\int ( 16 - 32 x ) \sin ( 4 x - 2 )^{2} d x$

Use calculus and the substitution x = cos $θ$ to find the exact value of

$\int_{\frac{1}{ \sqrt{2}}^{\frac{ \sqrt{3} }{2}} \frac{1}{ \sqrt{1} - x^{2}} } d x$

Find

$\int x^{2} \sin 3 x d x$

Find

$\int \frac{I n x}{x^{3}} d x$

Find the integral

$\int \frac{8 x^{2} - 8 x - 1}{( 4 x^{2} - 1 ) ( x - 2 )} d x$

Use integration by parts to show that

$\int e^{x} \sin x d x = 1 \frac{1}{2} e^{x} ( \sin x - \cos x ) + C$

where c is the constant of integration.

Sketch the region described by the following inequalities

x ≥ 2 x≤ 6 2y ≤ x + 4 y≥p, where 0<p< 2

The region described in part (a) is rotated through 360° about the x-axis.

Find the volume of the solid formed, giving your answer in terms of p.

The solid formed in part (b) will have a 'hole' in its centre.

(i) Find the volume of this 'hole', giving your answer in terms of p.

(ii) Hence show that there are no values of p in the given interval that make the volume of the solid equal to the volume of the 'hole'.

Starting with the equation of a semicircle of radius r, y = $\sqrt{r^{2}-x^{2}}$(where r > 0), use calculus to prove that the general formula for the volume, V, of a sphere of radius r is

$V=\frac{4}{3}\mathrm{πr}^{3}$

Express $\frac{7y+13}{12y^{2}+43y+36}$ in partial fractions.

Use your answer from part (a) to help find

$\int \frac{7 x^{2} + 13}{12 x^{4} + 43 x^{2} + 36} d x$