Infinite Integration Limits | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Replace infinity by a finite boundary first

The region under $y=1/x^2$ from $x=1$ continues indefinitely. Infinity, $\infty$ , means no finite upper boundary; it is not a number to substitute into a formula. An integral with an infinite limit is an improper integral.

Use a finite upper boundary $B$ , calculate normally, then let $B$ increase without bound.

\int_1^B x^{-2}\,dx=\left[-\frac1x\right]_1^B=1-\frac1B

As $B\to\infty$ , $1/B\to0$ . The area approaches 1. The notation $\lim$ means this limiting value:

\begin{gathered}\int_1^\infty x^{-2}\,dx\\=\lim_{B\to\infty}\left(1-\frac1B\right)\\=1\end{gathered}

CurveFinite upper limit B4

Area from 1 to B: ≈ 0.7500
Exact finite expression: 1 − 1/B

Convergent: the full area is 1. The unshaded tail from B to infinity has area 1/B, approaching zero.

For every finite $B$ , some area remains after the boundary. A finite limiting total is called convergent. It does not mean that the graph ends or that its height becomes exactly zero.

A shrinking height does not guarantee finite area

For $y=1/\sqrt{x}$ , the height also tends to zero, but:

\int_1^B x^{-1/2}\,dx=2\sqrt B-2

This grows without bound as $B\to\infty$ . The improper integral is divergent: it has no finite value. Judge the limiting integral, not just the appearance of the curve.

Your turnPractice

Find $\int_2^\infty 8x^{-3}\,dx$ , or explain why it has no finite value. Show the finite-boundary calculation.

Show worked answer

\begin{gathered}\int_2^B8x^{-3}\,dx\\=\left[-\frac4{x^2}\right]_2^B\\=1-\frac4{B^2}\end{gathered}

Answer

1

The last term tends to zero, so the integral converges.

For a lower limit of $-\infty$ , replace that lower limit by $A$ and take $A\to-\infty$ . Keep the upper-minus-lower order.

Your turnPractice

Evaluate $\int_{-\infty}^{-1}x^{-2}\,dx$ .

Show worked answer

\begin{gathered}\lim_{A\to-\infty}\left[-\frac1x\right]_A^{-1}\\=\lim_{A\to-\infty}\left(1+\frac1A\right)\\=1\end{gathered}

Replace infinity by a finite boundary first

Use a finite upper boundary $B$ , calculate normally, then let $B$ increase without bound.

\int_1^B x^{-2}\,dx=\left[-\frac1x\right]_1^B=1-\frac1B

As $B\to\infty$ , $1/B\to0$ . The area approaches 1. The notation $\lim$ means this limiting value:

\begin{gathered}\int_1^\infty x^{-2}\,dx\\=\lim_{B\to\infty}\left(1-\frac1B\right)\\=1\end{gathered}

CurveFinite upper limit B4

Area from 1 to B: ≈ 0.7500
Exact finite expression: 1 − 1/B

Convergent: the full area is 1. The unshaded tail from B to infinity has area 1/B, approaching zero.

For every finite $B$ , some area remains after the boundary. A finite limiting total is called convergent. It does not mean that the graph ends or that its height becomes exactly zero.

A shrinking height does not guarantee finite area

For $y=1/\sqrt{x}$ , the height also tends to zero, but:

\int_1^B x^{-1/2}\,dx=2\sqrt B-2

This grows without bound as $B\to\infty$ . The improper integral is divergent: it has no finite value. Judge the limiting integral, not just the appearance of the curve.

Your turnPractice

Find $\int_2^\infty 8x^{-3}\,dx$ , or explain why it has no finite value. Show the finite-boundary calculation.

Show worked answer

\begin{gathered}\int_2^B8x^{-3}\,dx\\=\left[-\frac4{x^2}\right]_2^B\\=1-\frac4{B^2}\end{gathered}

Answer

1

The last term tends to zero, so the integral converges.

For a lower limit of $-\infty$ , replace that lower limit by $A$ and take $A\to-\infty$ . Keep the upper-minus-lower order.

Your turnPractice

Evaluate $\int_{-\infty}^{-1}x^{-2}\,dx$ .

Show worked answer

\begin{gathered}\lim_{A\to-\infty}\left[-\frac1x\right]_A^{-1}\\=\lim_{A\to-\infty}\left(1+\frac1A\right)\\=1\end{gathered}