Integrals with Undefined Endpoints | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Approach an undefined endpoint from inside the domain

A finite interval can also give an improper integral. In $\int_0^1x^{-1/2}\,dx$ , the integrand $1/\sqrt{x}$ is undefined at zero and grows without bound as $x$ approaches zero from above.

Replace zero by a small positive number $\varepsilon$ (epsilon). First integrate where the function is defined; then let $\varepsilon\to0^+$ . The plus sign means approach from positive values.

\begin{gathered}\int_\varepsilon^1x^{-1/2}\,dx\\=[2\sqrt{x}]_\varepsilon^1=2-2\sqrt\varepsilon\\\int_0^1x^{-1/2}\,dx\\=\lim_{\varepsilon\to0^+}(2-2\sqrt\varepsilon)\\=2\end{gathered}

Near x = 0, the shaded regions continue above the plots. The limiting areas, not the heights alone, decide convergence.

The height is unbounded, but the area is finite. For $1/x^2$ instead, the same test gives:

\int_\varepsilon^1x^{-2}\,dx=\frac1\varepsilon-1\ \longrightarrow\ \infty

That integral diverges. An infinite height alone does not decide whether the area is finite.

Check the whole interval before substituting

If an upper endpoint is undefined, approach it from below. Never apply one upper-minus-lower calculation across an undefined point inside the interval. For example, $1/x^2$ is undefined at zero inside $[-1,1]$ ; its two sides must be considered separately, and each diverges.

Your turnChoose the limit

Evaluate $\int_0^4 3/\sqrt{4-x}\,dx$ , or explain why it has no finite value.

Show worked answer

Use an upper bound $b<4$ , because the problem is at the upper endpoint.

\begin{gathered}\int_0^b\frac3{\sqrt{4-x}}\,dx\\=[-6\sqrt{4-x}]_0^b\\=12-6\sqrt{4-b}\end{gathered}

Answer

\begin{gathered}\lim_{b\to4^-}(12-6\sqrt{4-b})\\=12\end{gathered}

Your turnCheck the claim

A student uses $[-1/x]_{-1}^1=-2$ for $\int_{-1}^1x^{-2}\,dx$ . Identify the invalid step.

Show worked answer

The evaluation crosses an undefined point at zero. The positive function cannot have a negative integral with increasing limits; in fact the improper contributions diverge, so there is no finite value.

Approach an undefined endpoint from inside the domain

A finite interval can also give an improper integral. In $\int_0^1x^{-1/2}\,dx$ , the integrand $1/\sqrt{x}$ is undefined at zero and grows without bound as $x$ approaches zero from above.

Replace zero by a small positive number $\varepsilon$ (epsilon). First integrate where the function is defined; then let $\varepsilon\to0^+$ . The plus sign means approach from positive values.

\begin{gathered}\int_\varepsilon^1x^{-1/2}\,dx\\=[2\sqrt{x}]_\varepsilon^1=2-2\sqrt\varepsilon\\\int_0^1x^{-1/2}\,dx\\=\lim_{\varepsilon\to0^+}(2-2\sqrt\varepsilon)\\=2\end{gathered}

Near x = 0, the shaded regions continue above the plots. The limiting areas, not the heights alone, decide convergence.

The height is unbounded, but the area is finite. For $1/x^2$ instead, the same test gives:

\int_\varepsilon^1x^{-2}\,dx=\frac1\varepsilon-1\ \longrightarrow\ \infty

That integral diverges. An infinite height alone does not decide whether the area is finite.

Check the whole interval before substituting

Your turnChoose the limit

Evaluate $\int_0^4 3/\sqrt{4-x}\,dx$ , or explain why it has no finite value.

Show worked answer

Use an upper bound $b<4$ , because the problem is at the upper endpoint.

\begin{gathered}\int_0^b\frac3{\sqrt{4-x}}\,dx\\=[-6\sqrt{4-x}]_0^b\\=12-6\sqrt{4-b}\end{gathered}

Answer

\begin{gathered}\lim_{b\to4^-}(12-6\sqrt{4-b})\\=12\end{gathered}

Your turnCheck the claim

A student uses $[-1/x]_{-1}^1=-2$ for $\int_{-1}^1x^{-2}\,dx$ . Identify the invalid step.

Show worked answer