Improper integrals and the mean value of a functionEdexcel A-Level Further Maths: Revision notes

A definite integral $\int_a^b f(x)\,dx$ is improper if either limit is infinite, or if $f(x)$ is undefined at some point in $[a,b]$ (usually at an end point or inside the range). You cannot substitute the bad value directly. Instead replace it by a variable $t$, integrate, then take a limit. If the limit is a finite number the integral converges to that value; if not, it diverges.

$\int_a^\infty f(x)\,dx=\lim_{t\to\infty}\int_a^t f(x)\,dx.$ Example: $\int_0^\infty e^{-x}\,dx=\lim_{t\to\infty}\left[-e^{-x}\right]_0^t=\lim_{t\to\infty}\left(1-e^{-t}\right)=1$, so it converges to $1$. Example: $\int_1^\infty\frac{1}{x^2}\,dx=\lim_{t\to\infty}\left(1-\frac1t\right)=1$, but $\int_1^\infty\frac1x\,dx=\lim_{t\to\infty}\ln t$ is infinite, so it diverges. In general $\int_1^\infty x^{-p}\,dx$ converges only when $p>1$.

If $f$ is undefined at the lower limit $a$: $\int_a^b f(x)\,dx=\lim_{t\to a^+}\int_t^b f(x)\,dx$. If it is undefined at the upper limit $b$ use $\lim_{t\to b^-}\int_a^t$. Example: $\int_0^2\frac{1}{\sqrt{x}}\,dx=\lim_{t\to0^+}\left[2\sqrt{x}\right]_t^2=2\sqrt2-0=2\sqrt2$. Example: $\int_0^1\frac1x\,dx=\lim_{t\to0^+}\left(-\ln t\right)$ is infinite, so it diverges. If the bad point lies inside the range, split the integral there and take a limit on each side. The integral converges only if both parts converge. For example $\int_{-1}^1\frac{1}{x^2}\,dx$ diverges, even though a careless evaluation gives $-2$.

In exam answers state the limit you are using, then evaluate. The standard results are $e^{-t}\to0$ and $te^{-t}\to0$ as $t\to\infty$, and $t\ln t\to0$ as $t\to0^+$ (exponentials beat powers; powers beat logarithms). Example: $\int_0^1\ln x\,dx=\lim_{t\to0^+}\left[x\ln x-x\right]_t^1=-1-\lim_{t\to0^+}(t\ln t-t)=-1$.

The mean value of $f(x)$ over $a\le x\le b$ is $\frac{1}{b-a}\int_a^b f(x)\,dx.$ It is the height of the rectangle on $[a,b]$ with the same area as under the curve. Example: $f(x)=x^2$ on $[1,4]$: $\frac{1}{3}\left[\frac{x^3}{3}\right]_1^4=\frac13\cdot21=7$. The same idea works with an improper integral, provided it converges. For $f(x)=\frac{1}{\sqrt x}$ on $[0,4]$ the mean value is $\frac14\cdot4=1$. To find where a function equals its mean value, set $f(x)=$ mean and solve.