Calculate a convergent improper integral
The area approaches 4
The curve is \(f(x) = 12/x^2\). The teal region lies under it from \(x = 3\) to the orange edge at \(x = \cy{b}\).
An antiderivative of \(12/x^2\) is \(-12/x\). Evaluated from 3 to \(\cy{b}\), it gives \(-12/b\) \(-\,(-12/3)\), so the area is \(A(b) = 4 - 12/b\).
Drag \(\cy{b}\) to the right. At \(b = 6\), the area is 2, at \(b = 12\) it is 3, at \(b = 120\) it is 3.9, and at \(b = 1200\) it is 3.99. The lower graph plots the area against \(b\) on an axis squeezed so that its right end stands for \(\infty\).
The term \(12/b\) approaches 0 as \(b\) grows, so the area approaches 4. The value of the improper integral \(\int_3^{\infty} 12/x^2\,dx\) is this limit, 4.
The value is the limit
The curve runs on without end to the left and to the right, so the integral from \(-\infty\) to \(\infty\) has two trouble spots. A trouble spot is a place that makes an integral improper. The integral is split at the point \(m\), and each piece has its own orange edge.
Drag each edge away from \(m\). Each piece has its own area graph below, and each graph approaches its own dashed line, at the heights \(L_1\) and \(L_2\).
The value of the integral is the sum \(L_1 + L_2\). This sum exists only when both limits are numbers.
Notation
Write the integral as a limit for each trouble spot. With two trouble spots, as here, split the integral at a point \(m\) first, so that each piece has one.
\[\begin{aligned} \int_a^{\infty} f(x)\,dx &= \lim_{t \to \infty} \\ &\quad \int_a^{\cy{t}} f(x)\,dx \end{aligned}\]Find an antiderivative \(F\). Evaluate it at the two ends of the finite interval, and keep the letter \(t\) in the result.
\[\int_a^{t} f(x)\,dx = F(t) - F(a)\]Take the limit of each result. If every limit is a number, their sum is the value of the integral, here \(4 + 6 = 10\).
\[\int_{-\infty}^{\infty} f(x)\,dx = \cb{L_1} + \cb{L_2}\]If any limit is infinite or does not exist, the integral has no value. In that case, do not report the values of the other pieces.
A piece below the x-axis has a negative value, and the sum adds it with its sign.
When it fails
The left piece still converges to 4. The area of the right piece grows without bound, so its limit is not a number.
The sum needs two numbers. With one limit missing, the integral has no value, and the number 4 is not an answer.
Example 1: an infinite limit
Find \(\displaystyle\int_0^{\infty} \frac{12}{x^2 + 16}\,dx\).
Replace the infinite limit by \(\cy{t}\).
\[\lim_{t \to \infty} \int_0^{\cy{t}} \frac{12}{x^2 + 16}\,dx\]An antiderivative is \(3\arctan(x/4)\). Its derivative is \(\tfrac{3/4}{1 + x^2/16}\), which is the integrand after you multiply the top and the bottom by 16. The antiderivative is 0 at 0, because \(\arctan 0 = 0\).
\[\Big[3\arctan\frac{x}{4}\Big]_0^{t} = \cb{3\arctan\frac{t}{4}}\]As \(t\) grows, \(\arctan(t/4)\) approaches \(\pi/2\), so the value is the exact number \(3\pi/2\).
\[\lim_{t \to \infty} 3\arctan\frac{t}{4} = \cb{\frac{3\pi}{2}}\]As a check, \(3\pi/2 \approx 4.712\). The lower graph approaches this height.
Example 2: a trouble spot inside
Find \(\displaystyle\int_{-1}^{8} \frac{dx}{\sqrt[3]{x}}\).
The curve is unbounded at \(x = 0\), inside the interval from \(-1\) to 8, so split at 0 and write one limit for each piece. The integrand is \(x^{-1/3}\), and an antiderivative is \(\tfrac32 x^{2/3}\), where \(x^{2/3}\) means \((\sqrt[3]{x})^2\).
\[\begin{aligned} &\lim_{s \to 0^+} \int_{\cy{s}}^{8} x^{-1/3}\,dx \\ &\quad + \lim_{t \to 0^-} \int_{-1}^{\cy{t}} x^{-1/3}\,dx \end{aligned}\]The right piece is the teal region. At 8 the antiderivative is \(\tfrac32 \cdot 4\), because \((\sqrt[3]{8})^2 = 4\), and the term at \(s\) approaches 0.
\[\tfrac32 \cdot 4 - \tfrac32 s^{2/3} \;\to\; \cb{6}\]The left piece is the red region, which lies below the x-axis and so counts as negative. At \(-1\) the antiderivative is \(\tfrac32 \cdot 1\), because \((\sqrt[3]{-1})^2 = 1\), and the term at \(t\) approaches 0.
\[\tfrac32 t^{2/3} - \tfrac32 \cdot 1 \;\to\; \cr{-\tfrac32}\]A tempting wrong move: Add the sizes of the two regions.
\[6 + \tfrac32 = \tfrac{15}{2}\]The left region is red because it lies below the x-axis, so its limit is \(-\tfrac32\), not \(\tfrac32\). The integral adds \(6\) and \(-\tfrac32\).
Both limits are numbers, so add them with their signs.
\[6 + \left(-\tfrac32\right) = \cb{\tfrac92}\]
Practice
Find \(\displaystyle\int_0^{\infty} x e^{-2x}\,dx\).
Replace the infinite limit by t and integrate by parts. Then take the limit of each term that contains t. The dashed line in the lower graph shows the number to expect.
Solution
Replace the infinite limit by \(\cy{t}\). Integrate by parts with \(u = x\) and \(dv = e^{-2x}\,dx\), so \(du = dx\) and \(v = -\tfrac12 e^{-2x}\), and \(uv - \int v\,du\) gives the antiderivative below.
\[\int x e^{-2x}\,dx = -\left(\frac{x}{2} + \frac14\right)e^{-2x}\]Evaluate from 0 to \(\cy{t}\). At 0 the antiderivative is \(-\tfrac14\), so subtracting it adds \(\tfrac14\).
\[\begin{aligned} \int_0^{t} x e^{-2x}\,dx &= \frac14 \\ &\quad - \left(\frac{t}{2} + \frac14\right)e^{-2t} \end{aligned}\]The factor \(e^{-2t}\) approaches 0 faster than \(t\) grows, so \(te^{-2t}\) approaches 0. The value is \(\tfrac14\).
\[\begin{aligned} &\lim_{t \to \infty} \left[\frac14 - \left(\frac{t}{2} + \frac14\right)e^{-2t}\right] \\ &= \cb{\tfrac14} \end{aligned}\]
Questions
The graph shows the curve \(f(x) = 6/(x+1)^3\). The lower graph shows the area \(A(b)\) under \(f\) from 0 to \(\cy{b}\).
What number does \(A(b)\) approach as \(b\) grows?
- 3
- 6
- 2.25
Answer
3
The area up to \(b = 1\) is 2.25. What is the area of the part of the region to the right of \(x = 1\)?
- 0.75
- 3
- It has no finite area
Answer
0.75
What is \(\displaystyle\int_0^{\infty} \frac{20}{(x+4)^2}\,dx\)?
- 5
- 1.25
- The integral diverges
Answer
5
Summary
The value of a convergent improper integral is the limit of its finite areas, so you integrate up to \(\cy{b}\) and then take the limit.
You add the pieces with their signs only when every limit is a number, because one missing limit leaves the integral without a value.
You use this when a problem asks for the value, the area, or the total accumulated amount of an improper integral.
This is one of five skills in topic 6.13 of Which & Why: AP Calculus.