Integration (Cambridge (CIE) IGCSE Additional Maths): Exam Questions

Find the exact value of ${\int }_2^4\frac{{\left(x+1\right)}^2}{x^2}\mathrm{d}x$.

Given

$\frac{\mathrm{d}}{\mathrm{d}x}\left(x\cos x\right)=-x\sin x+\cos x$

find the exact value of ${\int }_0^{\frac{\mathrm{\pi }}{6}}x\sin x\mathrm{d}x$.

Given that ${\int }_1^a\left(\frac{2}{2x+3}+\frac{3}{3x-1}-\frac{1}{x}\right)\mathrm{d}x=\ln 2.4$ where $a>1$ , find the value of $a$.

(i) Find $\frac{\mathrm{d}}{\mathrm{d}x}\left(6{\sin }^{3}kx\right)$, where $k$ is a constant.

(ii) Hence find $\int \left({\sin }^{2}2x\cos 2x\right)dx.$

Find  ${\int }_3^5\left(\frac{1}{x-1}-\frac{1}{{\left(x-1\right)}^2}\right)\mathrm{d}x$, giving your answer in the form $a+\ln b$, where $a$ and $b$ are rational numbers.

A curve is such that $\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}=\sin \left(6x-\frac{\pi }{2}\right)$. Given that $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{2}$ at the point $\left(\frac{\pi }{4},\frac{13\pi }{12}\right)$ on the curve, find the equation of the curve.

The diagram shows part of the graphs of $y=4x^{\frac{2}{3}}$ and $y={(x-3)}^2$ .The graph of $y={(x-3)}^2$ meets the $x$-axis at the point $A(a,0)$ and the two graphs intersect at the point $B(b,4)$.

Find the value of $a$ and of $b$.

Find the area of the shaded region.

The diagram shows the straight line $2x+y=-5$ and part of the curve $xy+3=0$. The straight line intersects the $x$-axis at the point $A$ and intersects the curve at the point $B$. The point $C$ lies on the curve. The point $D$ has coordinates (1, 0). The line $CD$ is parallel to the $y$-axis.

Find the coordinates of each of the points $A$ and $B$.

Find the area of the shaded region, giving your answer in the form $p+\ln q$, where $p$ and $q$ are positive integers.

The diagram shows part of the curve $xy=2$ intersecting the straight line $y=5x-3$ at the point $A$.
The straight line meets the $x$-axis at the point $B$. The point $C$ lies on the $x$-axis and the point $D$ lies on the curve such that the line $CD$ has equation $x=3$. Find the exact area of the shaded region, giving your answer in the form $p+\ln q$, where $p$ and $q$ are constants.

(i) Given that $\mathrm{f}\left(x\right)=\frac{1}{\cos x}$, show that $\mathrm{f}\text{'}\left(x\right)=\tan x\sec x$.

[3]

(ii) Hence find $\int \left(3\tan x\sec x-\sqrt[4]{e^{3x}}\right)\mathrm{d}x$.

[3]

Given that ${\int }_2^5\frac{p}{px+10}\mathrm{d}x=\ln 2$, find the value of the positive constant $p$.

Given that ${\int }_1^a\left(\frac{1}{x}-\frac{1}{2x+3}\right)\mathrm{d}x=\ln 3$, where $a>0$, find the exact value of $a$, giving your answer in simplest surd form.

Find the exact value of ${\int }_0^{\frac{\mathrm{\pi }}{3}}\left(\sin \left(2x+\frac{\mathrm{\pi }}{3}\right)-1+\cos 2x\right)\mathrm{d}x$.

Throughout this question, $x$ is measured in radians.

The curves $y={\sec }^{2}x$ and $y=2+\cos \left(2x\right)$ intersect at the points $\left(-\frac{\pi }{4},2\right)$ and $\left(\frac{\pi }{4},2\right)$, as shown.

Find the exact area of the shaded region enclosed.

Show that $\frac{1}{x+1}+\frac{2}{3x+10}$ can be written as $\frac{5x+12}{3x^2+13x+10}$

The diagram shows part of the curve $y=\frac{5x+12}{3x^2+13x+10}$, the line $x=2$ and a straight line of gradient 1. The curve intersects the $y$-axis at the point $P$. The line of gradient 1 passes through $P$ and intersects the $x$-axis at the point $Q$. Find the area of the shaded region, giving your answer in the form $a+\frac{2}{3}\ln \left(b\sqrt{3}\right)$, where $a$ and $b$ are constants.

The diagram shows part of the curve $y=(9–x)(x–3)$ and the line $y=k–3$, where $k>3$.
The line through the maximum point of the curve, parallel to the $y$-axis, meets the $x$-axis at $A$.
The curve meets the $x$-axis at $B$, and the line $y=k–3$ meets the curve at the point $C(k,k–3)$ .

Find the area of the shaded region.

Show that $\frac{3}{2x-3}+\frac{3}{2x+3}$ can be written as $\frac{12x}{4x^2-9}$.

Hence find $\int \frac{12x}{4x^2-9}\mathrm{d}x$, giving your answer as a single logarithm and an arbitrary constant.

Given that ${\int }_2^a\frac{12x}{4x^2-9}\mathrm{d}x=\ln 5\sqrt{5}$, where $a>2$, find the exact value of $a$.

A curve is such that $\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}=5\cos 2x$. This curve has a gradient of $\frac{3}{4}$ at the point $\left(-\frac{\mathrm{\pi }}{12},\frac{5\mathrm{\pi }}{4}\right)$. Find the equation of this curve.

The gradient of the normal to a curve at the point $(x,y)$ is given by $\frac{x}{x+1}$

Given that the curve passes through the point $(1,4)$, show that its equation is $y=5-\ln x-x$.

Find, in the form $y=mx+c$, the equation of the tangent to the curve at the point where $x=3$.