Further Integration (Edexcel International A Level (IAL) Maths: Pure 4): Exam Questions

Use a suitable substitution to find

$\int$-15 sin(5x-2) dx

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

${\int }_1^2\frac{x}{x+4}dx=1+4In\frac{5}{6}$

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

${\int }_0^1$(5x - 4)e^3x dx

Show that

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

can be written in the form

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

Hence find

$\int \frac{11}{\left(2x-3\right)\left(x+4\right)}dx$

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

$\int 8x\sin (3x^2+1)dx$

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

$\int \frac{1}{x{(2+lnx)}^3}dx=\frac{-1}{2{(\ln x+2)}^2}+c$

where c is the constant of integration.

Use integration by parts to find

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

Show that

$\int Inxdx=x\ln x-x+c$

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

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

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

$V=\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}$

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 $\frac{311\mathrm{\pi }}{20}$cubic units, find the value of a.

Find an expression for y given that

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

Integrate

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

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

${\int }_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}}\frac{1}{\sqrt{1-x^2}}dx$

Find

$\int x^2\sin 3xdx$

Find

$\int \frac{Inx}{x^3}dx$

Find the integral

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

Use integration by parts to show that

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

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

$V=\frac{4}{3}{\mathrm{\pi r}}^3$

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

Use your answer from part (a) to help find

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