A-level Mathematics/CIE/Pure Mathematics 1/Integration

The Antiderivative

Integration is defined as the reverse process of differentiation. Thus, it is the process of finding the antiderivative of an expression.

The antiderivative is also called the integral of an expression, and is represented using the symbol $\int$.

e.g. $\int 2xdx=x^2$

The Constant of Integration

A problem with integration is that many different expressions have the same derivative, such as $\frac{d}{dx}(x^2+3)=2x$ and $\frac{d}{dx}(x^2-5)=2x$. Expressions with different constant terms may have the same derivative, so when we integrate an expression, we need to add an arbitrary constant $+c$ to the end, which represents this unknown value.

Therefore, $\int 2xdx=x^2+c$

In some scenarios, we have a point on a curve and an expression for its derivative. From that, we need to find the equation of the curve, which will require us to find the constant of integration by substituting the values from the point.

e.g. The point $(3,2)$ is on a curve with gradient $\frac{dy}{dx}=x^2$. Find the equation of the curve.

$\begin{array}{cc}& y=\int x^{2}dx=\frac{x^3}{3}+c\\ (3,2)⟹& 2=\frac{3^3}{3}+c\\ & 2=9+c\\ & c=-7\\ & y=\frac{x^3}{3}-7\end{array}$

Definite Integrals

A definite integral is an integral between two given bounds $a$ and $b$. These bounds are written $\underset{a}{\overset{b}{\int }}$.

For a function $f(x)$ with integral $F(x)$, the definite integral $\underset{a}{\overset{b}{\int }}f(x)dx=F(b)-F(a)$

e.g. Find $\underset{0}{\overset{1}{\int }}\sqrt{x}dx$

$\begin{array}{cc}\int \sqrt{x}dx& =\int x^{\frac{1}{2}}dx\\ & =\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+c\\ & =\frac{2}{3}{x}^{\frac{3}{2}}+c\\ \underset{0}{\overset{1}{\int }}\sqrt{x}dx& ={[\frac{2}{3}{x}^{\frac{3}{2}}+c]}_0^1\\ & =\frac{2}{3}{1}^{\frac{3}{2}}+c-(\frac{2}{3}{0}^{\frac{3}{2}}+c)\\ & =\frac{2}{3}+c-0-c=\frac{2}{3}\end{array}$

Note that with definite integrals, the arbitrary constants cancel out. This means we don't actually need to write them when working with definite integrals.

Improper Integrals

An improper integral is a definite integral where one of the bounds is invalid.

e.g. $\underset{-1}{\overset{1}{\int }}{x}^{-2}dx$ is invalid at $x=0$

To evaluate an improper integral, we need to find the limit of the integral as one of the bounds approaches the value we are looking for.

$\begin{array}{cc}\underset{-1}{\overset{1}{\int }}{x}^{-2}dx& =\underset{a\to 1}{\lim }\underset{-1}{\overset{a}{\int }}{x}^{-2}dx\\ & =\underset{a\to 1}{\lim }{[-\frac{x^{-3}}{3}]}_{-1}^a\\ & =\underset{a\to 1}{\lim }-\frac{a^{-3}}{3}-(-\frac{-1^{-3}}{3})\\ & =\underset{a\to 1}{\lim }-\frac{a^{-3}}{3}+\frac{-1}{3}\\ & =-\frac{1^{-3}}{3}+\frac{-1}{3}\\ & =-\frac{1}{3}+\frac{-1}{3}\\ & =-\frac{2}{3}\end{array}$

Area under a Curve

A definite integral provides the signed area under a curve.

A definite integral can be used to find the area under a curve.

e.g. Find the area bounded by $y=x^2+2$, the x-axis, the line $x=3$ and the line $x=6$

$\begin{array}{cc}A& =\underset{3}{\overset{6}{\int }}{x}^2+2dx\\ & ={[\frac{x^3}{3}+2x]}_3^6\\ & =\frac{6^3}{3}+2(6)-(\frac{3^3}{3}+2(3))\\ & =\frac{216}{3}+12-\frac{27}{3}-6\\ & =72+12-9-6\\ & =69\end{array}$

Solids of Revolution

A solid of revolution can be made by rotating a curve about an axis. (The animation shows a surface of revolution which encloses the solid)

A solid of revolution is a volume which is obtained by rotating a curve about an axis between two bounds.

The volume can be calculated as the sum of a series of tiny cylinders. If we're rotating about the x-axis, this sum is equal to $\sum _{x=a}^b\pi (f(x){)}^2\delta x$ where $\delta x$ is the width of each cylinder. As $\delta x$ approaches zero, the sum becomes $\underset{a}{\overset{b}{\int }}\pi (f(x){)}^{2}dx$.