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

Express $\underset{\delta x\to 0}{\lim }\sum _{x=2.1}^{6.3}\frac{2}{x}\delta x$ as an integral.

Hence show that

$\underset{\delta x\to 0}{\lim }\sum _{x=2.1}^{6.3}\frac{2}{x}\delta x=\ln k$

where $k$ is a constant to be found.

Given that $k\in {\mathrm{\mathbb{Z}}}^{+}$

show that ${\int }_k^{3k}\frac{2}{\left(3x-k\right)}dx$ is independent of $k$.

Given that $k\in {\mathrm{\mathbb{Z}}}^{+}$

show that ${\int }_k^{2k}\frac{2}{{\left(2x-k\right)}^2}dx$ is inversely proportional to $k$.

Given that

$\frac{x^2+8x-3}{x+2}\equiv Ax+B+\frac{C}{x+2}x\in \mathrm{ℝ}x\ne -2$

find the values of the constants $A$, $B$ and $C$

Hence, using algebraic integration, find the exact value of

${\int }_0^6\frac{x^2+8x-3}{x+2}\mathrm{d}x$

giving your answer in the form $a+b\ln 2$ where $a$ and $b$ are integers to be found.

In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable.

Show that

${\int }_1^{{\mathrm{e}}^2}{x}^3\ln x\mathrm{d}x=a{\mathrm{e}}^8+b$

where $a$ and $b$ are rational constants to be found.

Find

$\int \frac{4x}{2x^2+5}\mathrm{d}x$

Show that

${\int }_2^{6}4{\left(4x+1\right)}^{\frac{1}{2}}dx=\frac{196}{3}$

Find

$\int -15\sin \left(5x-2\right)\mathrm{d}x$

Find

$\int -3x\sin x\mathrm{d}x$

Use the substitution $u=x+4$ to show that

${\int }_1^2\frac{x}{x+4}\mathrm{d}x=1+4\ln \left(\frac{5}{6}\right)$

Find

$\int \frac{3x^2+2}{2x^3+4x}\mathrm{d}x$

Find

$\int 2\sin x\cos x\mathrm{d}x$

Use algebraic integration to find the exact value of

${\int }_1^3{\left(4x+1\right)}^5\mathrm{d}x$

By using the substitution $u=\sqrt{x}$, find

$\int \frac{1}{x+4\sqrt{x}}\mathrm{d}x$

giving your answer in terms of $x$.

Hence evaluate

${\int }_1^9\frac{1}{x+4\sqrt{x}}\mathrm{d}x$

giving your answer in the form $2\ln \left(\frac{a}{b}\right)$ where $a$ and $b$ are integers.

Show that

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

can be written in the form

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

where $A$ and $B$ are constants to be found.

Hence find

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

writing your answer in the form

$\ln \left|\mathrm{f}\left(x\right)\right|+c$

where $\mathrm{f}\left(x\right)$ is a function you should find and $c$ is a constant.

The figure below shows a sketch of the curves with equations $y=x^2-3x+4$ and $y=4-x^2+2x$.

The shaded region $R$ is bounded by the two curves.

Use algebraic integration to find the exact area of $R$.

Use the substitution $u=2+\ln x$ to show that

$\int \frac{1}{x{\left(2+\ln x\right)}^3}\mathrm{d}x=\frac{-1}{2{\left(2+\ln x\right)}^2}+c$

where $c$ is a constant.

Find 

$\int \sqrt{1+{\cot }^{2}x}dx$

Figure 3 shows a sketch of the curve with equation $y=\sqrt{x}$.

The point $P\left(x,y\right)$ lies on the curve.

The rectangle, shown shaded on Figure 3, has height $y$ and width $\delta x$.

Calculate

$\underset{\delta x\to 0}{\lim }\sum _{x=4}^9\sqrt{x}\delta x$

The curve $C$ with equation

$\begin{array}{ccc}y=\frac{p-3x}{\left(2x-q\right)\left(x+3\right)}& & x\in \mathrm{ℝ},x\ne -3,x\ne 2\end{array}$

where $p$ and $q$ are constants, passes through the point $\left(3,\frac{1}{2}\right)$ and has two vertical asymptotes with equations $x=2$ and $x=-3$

(i) Explain why you can deduce that $q=4$

(ii) Show that $p=15$

Figure 4 shows a sketch of part of the curve $C$. The region $R$, shown shaded in Figure 4, is bounded by the curve $C$, the $x$-axis and the line with equation $x=3$

Show that the exact value of the area of $R$ is $a\ln 2+b\ln 3$, where $a$ and $b$ are rational constants to be found.

Use the substitution $u=1+\sqrt{x}$ to show that

${\int }_0^{16}\frac{x}{1+\sqrt{x}}dx={\int }_p^q\frac{2{\left(u-1\right)}^3}{u}du$

where $p$ and $q$ are constants to be found.

Hence show that

${\int }_0^{16}\frac{x}{1+\sqrt{x}}dx=A-B\ln 5$

where $A$ and $B$ are constants to be found.

The curve $C$ has equation $y=\frac{2x-3}{x^2-3x},x>3$.

Express $y=\frac{2x-3}{x^2-3x}$​ using partial fractions.

The finite region $R$ is bounded by the curve $C$, the $x$-axis and the lines with equations $x=4$ and $x=a$.

Given that the area of $R$ is $\ln \left(\frac{9}{2}\right)$, and that $a>4$, find the exact value of $a$.

Use the substitution $x=u^2+1$ to show that

${\int }_5^{10}\frac{3\mathrm{d}x}{\left(x-1\right)\left(3+2\sqrt{x-1}\right)}={\int }_p^q\frac{6\mathrm{d}u}{u\left(3+2u\right)}$

where $p$ and $q$ are positive constants to be found.

Hence, using algebraic integration, show that

${\int }_5^{10}\frac{3\mathrm{d}x}{\left(x-1\right)\left(3+2\sqrt{x-1}\right)}=\ln a$

where $a$ is a rational constant to be found.

Use the substitution $u=2x+1$ to find the exact value of

${\int }_0^4\frac{x}{\sqrt{2x+1}}dx$

Hence find the exact value of

${\int }_0^4\frac{x+3}{\sqrt{2x+1}}dx$

Use algebraic integration to show that

${\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}{\cos }^2\theta d\theta =\frac{\pi }{8}-\frac{1}{4}$

The figure below shows a sketch of the curve with equation $y=1+2x-\frac{1}{4}{x}^2$ where

• the point  $P(x,y)$ lies on the curve

• the shaded rectangle shown has width $\delta x$ and height $y$

By expressing the series limit as a suitable integral, show that

$\underset{\delta x\to 0}{\lim }\sum _{x=1}^8\left(1+2x-\frac{1}{4}{x}^2\right)\delta x=\frac{329}{12}$

Use algebraic integration to show that

${\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{2-2{\cos }^2\theta }{\sin 2\theta }\mathrm{d}\theta =\frac{1}{2}\ln 3$

Use algebraic integration to show that

${\int }_0^1\frac{3x{\mathrm{e}}^{-3x^2}}{3-2{\mathrm{e}}^{-3x^2}}\mathrm{d}x=\frac{1}{4}\ln (3-2{\mathrm{e}}^{-3})$

Use algebraic integration to show that

${\int }_{\frac{\pi }{2}}^{\frac{5\pi }{6}}\frac{2\cos x}{1-\cos 2x}\mathrm{d}x=-1$

Use algebraic integration to show that

$\int \left(2x^2-1\right){\mathrm{e}}^x\mathrm{d}x=\left(p{x}^2+qx+r\right){\mathrm{e}}^x+c$

where $p$, $q$ and $r$ are integers to be found and $c$ is a constant.

Show that

$\int \ln x\mathrm{d}x=x\ln x-x+c$

where $c$ is a constant.

Use algebraic integration to find

$\int 5{\cos }^{2}4x\sin 4x\mathrm{d}x$

Use algebraic integration to find

$\int 3x{\left(5x^2+4\right)}^4\mathrm{d}x$

Use algebraic integration to find

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

Use algebraic integration to find

$\int \left(16-32x\right)\sin \left[{\left(4x-2\right)}^2\right]\mathrm{d}x$

Use algebraic integration to find

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

Use algebraic integration to find

$\int \frac{\ln x}{x^3}\mathrm{d}x$

$\mathrm{f}\left(x\right)=\frac{3kx-18}{\left(x+4\right)\left(x-2\right)}$ where $k$ is a positive constant

Express $\mathrm{f}\left(x\right)$ in partial fractions in terms of $k$.

Hence find the exact value of $k$ for which

${\int }_{-3}^1\mathrm{f}\left(x\right)dx=21$

Figure 2 shows a sketch of part of the curve $C$ with equation $y=x\ln x,x>0$

The line $l$ is the normal to $C$ at the point $P\left(\mathrm{e},\mathrm{e}\right)$

The region $R$, shown shaded in Figure 2, is bounded by the curve $C$, the line $l$ and the $x$-axis.

Show that the exact area of $R$ is $A{\mathrm{e}}^2+B$ where $A$ and $B$ are rational numbers to be found.

Figure 2 shows a sketch of part of the curve with equation

$y={\left(\ln x\right)}^{2}x>0$

The finite region $R$, shown shaded in Figure 2, is bounded by the curve, the line with equation $x=2$, the $x$-axis and the line with equation $x=4$.

Use algebraic integration to find the exact area of $R$, giving your answer in the form

$a{\left(\ln 2\right)}^2+b\ln 2+c$

where $a$, $b$ and $c$ are integers to be found.

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

Find the first three terms, in ascending powers of $x$, of the binomial expansion of

${\left(3+x\right)}^{-2}$

writing each term in simplest form.

Using the answer to part (a) and using algebraic integration, estimate the value of

${\int }_{0.2}^{0.4}\frac{6x}{{\left(3+x\right)}^2}dx$

giving your answer to 4 significant figures.

Find, using algebraic integration, the exact value of

${\int }_{0.2}^{0.4}\frac{6x}{{\left(3+x\right)}^2}dx$

giving your answer in the form $a\ln \left(b\right)+c$, where $a$, $b$ and $c$ are constants to

be found.

Show that

${\int }_0^{2}2x\sqrt{x+2}dx=\frac{32}{15}\left(2+\sqrt{2}\right)$

Use the substitution $u=4-\sqrt{h}$ to show that

$\int \frac{dh}{4-\sqrt{h}}=-8\ln \left|4-\sqrt{h}\right|-2\sqrt{h}+k$

where $k$ is a constant.

Use the substitution $x=\cos \theta$ to show that

${\int }_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}}\frac{1}{\sqrt{1-x^2}}\mathrm{d}x=\frac{\pi }{12}$

Prove that

$\int \tan kx\mathrm{d}x=\frac{1}{k}\ln \left|\sec kx\right|+c$

where $k$ and $c$ are constants.

Use algebraic integration to show that

${\int }_{\frac{\pi }{18}}^{\frac{\pi }{9}}\frac{{\mathrm{cosec}}^{2}3\theta }{3\cot 3\theta }\mathrm{d}\theta =a\ln 3$

where $a$ is a rational number to be found.

Show that

$\int \frac{8x^2-8x-1}{(4x^2-1)(x-2)}\mathrm{d}x=\ln \left(A|x-2|\sqrt{|4x^2-1|}\right)$

where $A$ is a constant.

Show that

$\int {\mathrm{e}}^x\sin x\mathrm{d}x=\frac{1}{2}{\mathrm{e}}^x\left(\sin x-\cos x\right)+c$

where $c$ is a constant.

The figure below shows a sketch of the curves with equations $y=\sqrt{25-x^2}$     and $y=6x-x^2-5$.

The finite regions bounded by the two curves are shaded.

Show that the $x$-coordinates of the points of intersection are $x=3$, $x=4$ and $x=5$.

[You do not need to solve an equation in $x$.]

Use the substitution $x=5\sin u$ to show that

$\int \sqrt{25-x^2}\mathrm{d}x=\frac{25\arcsin \left(\frac{x}{5}\right)+x\sqrt{25-x^2}}{2}+c$

where $c$ is a constant.

Hence show that the exact area of the shaded regions is

$\frac{25\pi }{4}-4-\frac{25}{2}\left(2\arcsin \left(\frac{4}{5}\right)-\arcsin \left(\frac{3}{5}\right)\right)$