Wemazi

Topic 6.13 · BC

Replace the improper part with a limit

Builds on identifying why an integral is improper. 8 minutes.

The area up to a moving edge

The region under the curve \(f(x) = 1/(x+1)^2\) continues to the right with no end, so you cannot find its area in the usual way.

Drag the orange edge \(\cy{b}\) to the right. For every \(\cy{b}\), the teal region from 0 to \(\cy{b}\) is finite, and its area is \(A(b) = 1 - \dfrac{1}{b+1}\), from the antiderivative \(-\dfrac{1}{x+1}\).

The area is 0.5 at \(b = 1\), 0.9 at \(b = 9\), and 0.99 at \(b = 99\).

The lower graph plots \(A(b)\) against \(\cy{b}\) on a squeezed axis whose right end stands for \(\infty\). The graph rises toward the dashed line at height 1 and does not pass it.

The area of the whole region is the number that A(b) approaches, which is 1.

An edge that approaches the trouble spot

This curve continues to the right from a start \(a\), with no end. The trouble spot is the place that makes the integral improper, here infinity.

Drag the orange edge \(\cy{t}\) to the right. For each position of \(\cy{t}\), the region is finite, and the lower graph shows its area.

The improper integral is the number that these areas approach as the edge approaches the trouble spot.

The finite integral comes first, and the limit comes after it.

Notation

  1. Take first the piece from \(a\) to \(c\). The curve is unbounded at its right end \(c\), so the edge \(\cy{t}\) approaches \(c\) from the left.

    \[\int_a^c f(x)\,dx = \lim_{t \to c^-} \int_a^{\cy{t}} f(x)\,dx\]

  2. On the piece from \(c\) to \(b\), the asymptote is at the left end, so the edge \(\cy{s}\) approaches \(c\) from the right.

    \[\int_c^b f(x)\,dx = \lim_{s \to c^+} \int_{\cy{s}}^{b} f(x)\,dx\]

  3. If the asymptote lies between \(a\) and \(b\), split the integral at \(c\). Each piece then has the asymptote at one end and needs its own limit.

    \[\begin{aligned} \int_a^b f(x)\,dx &= \int_a^{c} f(x)\,dx \\ &\quad + \int_{c}^{b} f(x)\,dx \end{aligned}\]

  4. If a limit of integration is infinite, replace it by \(\cy{t}\). Integrate over the finite interval. Then let \(t\) go to \(\infty\) or \(-\infty\).

    \[\begin{aligned} \int_a^{\infty} f(x)\,dx &= \lim_{t \to \infty} \\ &\quad \int_a^{\cy{t}} f(x)\,dx \\ \int_{-\infty}^{b} f(x)\,dx &= \lim_{t \to -\infty} \\ &\quad \int_{\cy{t}}^{b} f(x)\,dx \end{aligned}\]

  5. If both limits are infinite, split at any number \(c\), for example 0. Each piece needs its own limit.

    \[\begin{aligned} \int_{-\infty}^{\infty} f(x)\,dx &= \int_{-\infty}^{c} f(x)\,dx \\ &\quad + \int_{c}^{\infty} f(x)\,dx \end{aligned}\]

  6. In words, an improper integral is the limit of the integrals over finite pieces as the edge approaches the trouble spot.

When it fails

Each side needs its own variable. To see why, make the curve odd around \(c\). The left region lies below the axis, so it turns red and counts as negative. The right region has the same shape above the axis. Then move both ends toward \(c\) together.

With one variable for both ends, the red and teal areas cancel, and the sum is 0 at every position. One at a time, the left area goes to \(-\infty\) and the right to \(\infty\), so the integral has no value.

Example 1: an asymptote at the right end

Write \(\displaystyle\int_0^1 \frac{dx}{\sqrt{1-x}}\) as a limit, then find its value.

  1. The curve is unbounded at the right end \(x = 1\). Replace 1 by \(\cy{t}\), and let \(t\) approach 1 from the left.

    \[\begin{aligned} \int_0^1 \frac{dx}{\sqrt{1-x}} &= \lim_{t \to 1^-} \\ &\quad \int_0^{\cy{t}} (1-x)^{-1/2}\,dx \end{aligned}\]

  2. An antiderivative is \(-2\sqrt{1-x}\). Evaluate it from 0 to \(\cy{t}\).

    \[\begin{aligned} &\Big[-2\sqrt{1-x}\Big]_0^{t} \\ &\quad = -2\sqrt{1-t} - (-2) \\ &\quad = \cb{2 - 2\sqrt{1-t}} \end{aligned}\]

  3. As \(t\) approaches 1 from the left, \(\sqrt{1-t}\) approaches 0. The teal region has a finite area, although the curve is unbounded.

    \[\lim_{t \to 1^-} \left(2 - 2\sqrt{1-t}\right) = \cb{2}\]

Example 2: an asymptote inside

Write \(\displaystyle\int_0^5 \frac{dx}{(x-2)^{2/3}}\) as limits, then find its value.

  1. The curve is unbounded at \(x = 2\), which lies between 0 and 5.

  2. A tempting wrong move: Replace the trouble spot 2 by \(t\) and take one limit.

    \[\lim_{t \to 2} \int_0^{t} (x-2)^{-2/3}\,dx\]

    The single orange edge stops at the asymptote, so the region from 2 to 5 is never counted. One limit cannot reach both sides of the asymptote.

  3. Split at 2. Each piece has the trouble spot at one end.

    \[\int_0^5 = \int_0^{2} + \int_{2}^{5}\]

  4. Write each piece as a limit. The edge \(\cy{t}\) approaches 2 from the left, and the edge \(\cy{s}\) approaches 2 from the right.

    \[\begin{aligned} &\lim_{t \to 2^-} \int_0^{\cy{t}} (x-2)^{-2/3}\,dx \\ &\quad + \lim_{s \to 2^+} \int_{\cy{s}}^5 (x-2)^{-2/3}\,dx \end{aligned}\]

  5. An antiderivative is \(3\sqrt[3]{x-2}\). At \(x = 0\) it is \(-3\sqrt[3]{2}\), and at the trouble spot it is 0, so the terms with \(t\) and \(s\) go to 0.

    \[\begin{aligned} &\left(3\sqrt[3]{t-2} + 3\sqrt[3]{2}\right) \\ &\quad + \left(3\sqrt[3]{3} - 3\sqrt[3]{s-2}\right) \\ &\to \cb{3\sqrt[3]{2} + 3\sqrt[3]{3}} \end{aligned}\]

  6. As a check, the sum is about 8.107.

Practice

Write \(\displaystyle\int_{-\infty}^{0} 5e^{2x}\,dx\) as a limit, then find its value.

The orange edge \(\cy{t}\) sits at the end that causes the trouble. Decide where \(t\) must go.

Solution

  1. The lower limit is infinite. Replace it by \(\cy{t}\), and let \(t\) go to \(-\infty\).

    \[\lim_{t \to -\infty} \int_{\cy{t}}^{0} 5e^{2x}\,dx\]

  2. An antiderivative is \(\tfrac{5}{2}e^{2x}\). Evaluate it from \(\cy{t}\) to 0.

    \[\Big[\tfrac{5}{2}e^{2x}\Big]_t^0 = \tfrac{5}{2} - \tfrac{5}{2}e^{2t}\]

  3. As \(t\) goes to \(-\infty\), \(e^{2t}\) goes to 0, so the limit is \(\tfrac{5}{2}\).

    \[\lim_{t \to -\infty} \left(\tfrac{5}{2} - \tfrac{5}{2}e^{2t}\right) = \cb{\tfrac{5}{2}}\]

Questions

The graph shows \(f(x) = 1/\sqrt{x}\), which is unbounded at \(x = 0\). The orange edge \(t\) cuts off the teal region from \(t\) to 4.

  1. Which limit gives the area of the region under the curve from 0 to 4?

    • \(\displaystyle\lim_{t \to 0^+} \int_t^4 f(x)\,dx\)
    • \(\displaystyle\lim_{t \to 4^-} \int_0^t f(x)\,dx\)
    • \(\displaystyle\lim_{t \to 0^-} \int_t^4 f(x)\,dx\)
    Answer

    \(\displaystyle\lim_{t \to 0^+} \int_t^4 f(x)\,dx\)

  2. With \(t = 1\) the area is 2. It is 3 for \(t = 0.25\), and 3.8 for \(t = 0.01\). What happens to the area as \(t\) moves toward 0?

    • It approaches 4
    • It grows without bound
    • It stays at 2
    Answer

    It approaches 4

  3. What is \(\displaystyle\lim_{t \to \infty} \int_0^{t} 6e^{-2x}\,dx\)?

    • 3
    • 6
    • The limit is infinite
    Answer

    3

Summary

You replace the trouble spot by t, integrate over the finite part, and let t approach that spot from inside the interval.

This works because every finite part has an area, and the integral is the number that these areas approach.

You use this form for every improper integral, and split at an asymptote inside the interval.

This is one of five skills in topic 6.13 of Which & Why: AP Calculus.