Learning Outcomes
- I can compute improper integrals.
Section 5.8 Improper integrals (TI8)
Subsection 5.8.1 Activities
Activity 5.8.1.
Recall \(\displaystyle \int\frac{1}{x^2}dx=-\frac{1}{x}+C\text{.}\) Compute the following definite integrals.
(a)
\(\displaystyle \int_{1/100}^1 \frac{1}{x^2} dx=\left[-\frac{1}{x}\right]_{1/100}^1\)
(b)
\(\displaystyle \int_{1/10000}^1 \frac{1}{x^2} dx\)
(c)
\(\displaystyle \int_{1/1000000}^1 \frac{1}{x^2} dx\)
Activity 5.8.2.
What do you notice about \(\displaystyle \int_{a}^1 \frac{1}{x^2} dx\) as \(a\) approached 0 in Activity 5.8.1?
- \(\displaystyle \int_{a}^1 \frac{1}{x^2} dx\) approaches \(0\text{.}\)
- \(\displaystyle \int_{a}^1 \frac{1}{x^2} dx\) approaches a finite constant greater than \(0\text{.}\)
- \(\displaystyle \int_{a}^1 \frac{1}{x^2} dx\) approaches \(\infty\text{.}\)
- There is not enough information.
Activity 5.8.3.
Compute the following definite integrals, again using \(\displaystyle \int\frac{1}{x^2}dx=-\frac{1}{x}+C\text{.}\)
(a)
\(\displaystyle \int_{1}^{100} \frac{1}{x^2} dx=\left[-\frac{1}{x}\right]_{1}^{100}\)
(b)
\(\displaystyle \int_{1}^{10000} \frac{1}{x^2} dx\)
(c)
\(\displaystyle \int_{1}^{1000000} \frac{1}{x^2} dx\)
Activity 5.8.4.
What do you notice about \(\displaystyle \int_{1}^b \frac{1}{x^2} dx\) as \(b\) approached \(\infty\) in Activity 5.8.3?
- \(\displaystyle \int_{1}^b \frac{1}{x^2} dx\) approaches \(0\text{.}\)
- \(\displaystyle \int_{1}^b \frac{1}{x^2} dx\) approaches a finite constant greater than \(0\text{.}\)
- \(\displaystyle \int_{1}^b \frac{1}{x^2} dx\) approaches \(\infty\text{.}\)
- There is not enough information.
Activity 5.8.5.
Recall \(\displaystyle \int\frac{1}{\sqrt x}dx=2\sqrt{x}+C\text{.}\) Compute the following definite integrals.
(a)
\(\displaystyle \int_{1/100}^1 \frac{1}{\sqrt{x}} dx=\left[2\sqrt{x}\right]_{1/100}^1\)
(b)
\(\displaystyle \int_{1/10000}^1 \frac{1}{\sqrt{x}} dx\)
(c)
\(\displaystyle \int_{1/1000000}^1 \frac{1}{\sqrt{x}} dx\)
Activity 5.8.6.
(a)
What do you notice about the integral \(\displaystyle \int_{a}^1 \frac{1}{\sqrt{x}} dx\) as \(a\) approached 0 in Activity 5.8.5?
- \(\displaystyle \int_{a}^1 \frac{1}{\sqrt{x}} dx\) approaches \(0\text{.}\)
- \(\displaystyle \int_{a}^1 \frac{1}{\sqrt{x}} dx\) approaches a finite constant greater than \(0\text{.}\)
- \(\displaystyle \int_{a}^1 \frac{1}{\sqrt{x}} dx\) approaches \(\infty\text{.}\)
- There is not enough information.
(b)
How does this compare to what you found in Activity 5.8.1?
Activity 5.8.7.
Compute the following definite integrals using \(\displaystyle \int\frac{1}{\sqrt x}dx=2\sqrt{x}+C\text{.}\)
(a)
\(\displaystyle \int_{1}^{100} \frac{1}{\sqrt{x}} dx = \left[2\sqrt{x}\right]_{1}^{100}\)
(b)
\(\displaystyle \int_{1}^{10000} \frac{1}{\sqrt{x}} dx\)
(c)
\(\displaystyle \int_{1}^{1000000} \frac{1}{\sqrt{x}} dx\)
Activity 5.8.8.
(a)
What do you notice the integral \(\displaystyle \int_{1}^b \frac{1}{\sqrt{x}} dx\) as \(b\) approached \(\infty\) in Activity 5.8.7?
- \(\displaystyle \int_{1}^b \frac{1}{\sqrt{x}} dx\) approaches \(0\text{.}\)
- \(\displaystyle \int_{1}^b \frac{1}{\sqrt{x}} dx\) approaches a finite constant greater than \(0\text{.}\)
- \(\displaystyle \int_{1}^b \frac{1}{\sqrt{x}} dx\) approaches \(\infty\text{.}\)
- There is not enough information.
(b)
How does this compare to what you found in Activity 5.8.3?
Definition 5.8.9.
For a function \(f(x)\) and a constant \(a\text{,}\) we let \(\displaystyle \int_a^\infty f(x) dx\) denote
\begin{equation*}
\int_a^\infty f(x) dx=\lim_{b\to\infty}\left( \int_a^b f(x) dx\right)\text{.}
\end{equation*}
If this limit is a defined real number, then we say \(\displaystyle \int_a^\infty f(x) dx\) is convergent. Otherwise, it is divergent.
Similarly,
\begin{equation*}
\int_{-\infty}^b f(x) dx=\lim_{a\to-\infty}\left( \int_a^b f(x) dx\right).
\end{equation*}
Activity 5.8.10.
Which of these limits is equal to \(\displaystyle\int_1^\infty \frac{1}{x^2} dx\text{?}\)
- \(\displaystyle \displaystyle\lim_{b\to\infty}\int_1^b\frac{1}{x^2}dx\)
- \(\displaystyle \displaystyle\lim_{b\to\infty}\left[-\frac{1}{x}\right]_1^b\)
- \(\displaystyle \displaystyle\lim_{b\to\infty}\left[-\frac{1}{b}+1\right]\)
- All of these.
Activity 5.8.11.
Given the result of Activity 5.8.10, what is \(\displaystyle\int_1^\infty \frac{1}{x^2} dx\text{?}\)
- \(\displaystyle 0\)
- \(\displaystyle 1\)
- \(\displaystyle \infty\)
- \(\displaystyle -\infty\)
Activity 5.8.12.
Does \(\displaystyle\int_1^\infty \frac{1}{\sqrt x} dx\) converge or diverge?
- Converges because \(\displaystyle\lim_{b\to 0^+}\left[2\sqrt b-2\right]\) converges.
- Diverges because \(\displaystyle\lim_{b\to 0^+}\left[2\sqrt b-2\right]\) diverges.
- Converges because \(\displaystyle\lim_{b\to \infty}\left[2\sqrt b-2\right]\) converges.
- Diverges because \(\displaystyle\lim_{b\to \infty}\left[2\sqrt b-2\right]\) diverges.
Definition 5.8.13.
For a function \(f(x)\) with a vertical asymptote at \(x=c>a\text{,}\) we let \(\displaystyle \int_a^c f(x) dx\) denote
\begin{equation*}
\int_a^c f(x) dx=\lim_{b\to c^{-}}\left( \int_a^b f(x) dx\right)\text{.}
\end{equation*}
For a function \(f(x)\) with a vertical asymptote at \(x=c<b\text{,}\) we let \(\displaystyle \int_c^b f(x) dx\) denote
\begin{equation*}
\int_c^b f(x) dx=\lim_{a\to c^{+}}\left( \int_a^b f(x) dx\right)\text{.}
\end{equation*}
Activity 5.8.14.
Which of these limits is equal to \(\displaystyle\int_0^1 \frac{1}{\sqrt x} dx\text{?}\)
- \(\displaystyle \displaystyle\lim_{a\to0^+}\int_a^1\frac{1}{\sqrt x}dx\)
- \(\displaystyle \displaystyle\lim_{a\to0^+}\left[2\sqrt x\right]_a^1\)
- \(\displaystyle \displaystyle\lim_{a\to0^+}\left[2-2\sqrt a\right]\)
- All of these.
Activity 5.8.15.
Given the this result, what is \(\displaystyle\int_0^1 \frac{1}{\sqrt x} dx\text{?}\)
- \(\displaystyle 0\)
- \(\displaystyle 1\)
- \(\displaystyle 2\)
- \(\displaystyle \infty\)
Activity 5.8.16.
Does \(\displaystyle\int_0^1 \frac{1}{x^2} dx\) converge or diverge?
- Converges because \(\displaystyle\lim_{a\to 0^+}\left[-1+\frac{1}{a}\right]\) converges.
- Diverges because \(\displaystyle\lim_{a\to 0^+}\left[-1+\frac{1}{a}\right]\) diverges.
- Converges because \(\displaystyle\lim_{a\to 1^-}\left[-1+\frac{1}{a}\right]\) converges.
- Diverges because \(\displaystyle\lim_{a\to 1^-}\left[-1+\frac{1}{a}\right]\) diverges.
Activity 5.8.17.
Explain and demonstrate how to write each of the following improper integrals as a limit, and why this limit converges or diverges.
(a)
\(\displaystyle\int_{ -2 }^{ +\infty } \frac{1}{\sqrt{x + 6}} dx.\)
(b)
\(\displaystyle\int_{ -4 }^{ -2 } \frac{1}{{\left(x + 4\right)}^{\frac{4}{3}}} dx.\)
(c)
\(\displaystyle\int_{ -5 }^{ 0 } \frac{1}{{\left(x + 5\right)}^{\frac{5}{9}}} dx.\)
(d)
\(\displaystyle\int_{ 10 }^{ +\infty } \frac{1}{{\left(x - 8\right)}^{\frac{4}{3}}} dx.\)
Fact 5.8.18.
Suppose that \(0< p\) and \(p\neq 1\text{.}\) Applying the integration power rule gives us the indefinite integral \(\displaystyle \int \frac{1}{x^p} dx=\frac{1}{(1-p)}x^{1-p}+C\text{.}\)
Activity 5.8.19.
(a)
If \(0< p< 1\text{,}\) which of the following statements must be true? Select all that apply.
- \(\displaystyle 1-p< 0\)
- \(\displaystyle 1-p> 0\)
- \(\displaystyle 1-p< 1\)
- \(\displaystyle \int_1^\infty \frac{1}{x^p} dx\) converges.
- \(\displaystyle \int_1^\infty \frac{1}{x^p} dx\) diverges.
(b)
If \(p>1\text{,}\) which of the following statements must be true? Select all that apply.
- \(\displaystyle 1-p< 0\)
- \(\displaystyle 1-p> 0\)
- \(\displaystyle 1-p< 1\)
- \(\displaystyle \int_1^\infty \frac{1}{x^p} dx\) converges.
- \(\displaystyle \int_1^\infty \frac{1}{x^p} dx\) diverges.
Activity 5.8.20.
(a)
If \(0< p< 1\text{,}\) which of the following statements must be true?
- \(\displaystyle \int_0^1 \frac{1}{x^p} dx\) converges.
- \(\displaystyle \int_0^1 \frac{1}{x^p} dx\) diverges.
(b)
If \(p>1\text{,}\) which of the following statements must be true?
- \(\displaystyle \int_0^1 \frac{1}{x^p} dx\) converges.
- \(\displaystyle \int_0^1 \frac{1}{x^p} dx\) diverges.
Activity 5.8.21.
Consider when \(p=1\text{.}\) Then \(\frac{1}{x^p}=\frac{1}{x}\) and \(\displaystyle \int \frac{1}{x^p}\ dx=\displaystyle \int \frac{1}{x}\ dx=\ln|x|+C\text{.}\)
(a)
What can we conclude about \(\displaystyle \int_1^\infty \frac{1}{x} dx\text{?}\)
- \(\displaystyle \int_1^\infty \frac{1}{x} dx\) converges.
- \(\displaystyle \int_1^\infty \frac{1}{x} dx\) diverges.
- There is not enough information to determine whether this integral converges or diverges.
(b)
What can we conclude about \(\displaystyle \int_0^1 \frac{1}{x} dx\text{?}\)
- \(\displaystyle \int_0^1 \frac{1}{x} dx\) converges.
- \(\displaystyle \int_0^1 \frac{1}{x} dx\) diverges.
- There is not enough information to determine whether this integral converges or diverges.
Fact 5.8.22.
Let \(c, p>0\text{.}\)
- \(\displaystyle \int_0^c \frac{1}{x^p} dx\) converges if and only if \(p< 1\text{.}\)
- \(\displaystyle \int_c^\infty \frac{1}{x^p} dx\) converges if and only if \(p> 1\text{.}\)
Activity 5.8.23.
Consider the plots of \(f(x), g(x), h(x) \) where \(0 < g(x) < f(x) < h(x)\text{.}\)
If \(\displaystyle \int_1^\infty f(x) dx\) is convergent, what can we say about \(g(x), h(x)\text{?}\)
- \(\displaystyle \int_1^\infty g(x) dx\) and \(\displaystyle \int_1^\infty h(x) dx\) are both convergent.
- \(\displaystyle \int_1^\infty g(x) dx\) and \(\displaystyle \int_1^\infty h(x) dx\) are both divergent.
- Whether or not \(\displaystyle \int_1^\infty g(x) dx\) and \(\displaystyle \int_1^\infty h(x) dx\) are convergent or divergent cannot be determined.
- \(\displaystyle \int_1^\infty g(x) dx\) is convergent and \(\displaystyle \int_1^\infty h(x) dx\) is divergent.
- \(\displaystyle \int_1^\infty g(x) dx\) is convergent and \(\displaystyle \int_1^\infty h(x) dx\) could be either convergent or divergent.
Activity 5.8.24.
Consider the plots of \(f(x), g(x), h(x) \) where \(0 < g(x) < f(x) < h(x)\text{.}\) 
Figure 113. Plots of \(f(x), g(x), h(x)\)
- \(\displaystyle \int_1^\infty g(x) dx\) and \(\displaystyle \int_1^\infty h(x) dx\) are both convergent.
- \(\displaystyle \int_1^\infty g(x) dx\) and \(\displaystyle \int_1^\infty h(x) dx\) are both divergent.
- Whether or not \(\displaystyle \int_1^\infty g(x) dx\) and \(\displaystyle \int_1^\infty h(x) dx\) are convergent or divergent cannot be determined.
- \(\displaystyle \int_1^\infty g(x) dx\) could be either convergent or dicergent and \(\displaystyle \int_1^\infty h(x) dx\) is divergent.
- \(\displaystyle \int_1^\infty g(x) dx\) is convergent and \(\displaystyle \int_1^\infty h(x) dx\) is divergent.
Activity 5.8.25.
Consider the plots of \(f(x), g(x), h(x) \) where \(0 < g(x) < f(x) < h(x)\text{.}\) 
Figure 114. Plots of \(f(x), g(x), h(x)\)
- \(\displaystyle \int_0^1 g(x) dx\) and \(\displaystyle \int_0^1 h(x) dx\) are both convergent.
- \(\displaystyle \int_0^1 g(x) dx\) and \(\displaystyle \int_0^1 h(x) dx\) are both divergent.
- Whether or not \(\displaystyle \int_0^1 g(x) dx\) and \(\displaystyle \int_0^1 h(x) dx\) are convergent or divergent cannot be determined.
- \(\displaystyle \int_0^1 g(x) dx\) is convergent and \(\displaystyle \int_0^1 h(x) dx\) is divergent.
- \(\displaystyle \int_0^1 g(x) dx\) is convergent and \(\displaystyle \int_0^1 h(x) dx\) can either be convergent or divergent.
Activity 5.8.26.
Consider the plots of \(f(x), g(x), h(x) \) where \(0 < g(x) < f(x) < h(x)\text{.}\) 
Figure 115. Plots of \(f(x), g(x), h(x)\)
- \(\displaystyle \int_0^1 g(x) dx\) and \(\displaystyle \int_0^1 h(x) dx\) are both convergent.
- \(\displaystyle \int_0^1 g(x) dx\) and \(\displaystyle \int_0^1 h(x) dx\) are both divergent.
- Whether or not \(\displaystyle \int_0^1 g(x) dx\) and \(\displaystyle \int_0^1 h(x) dx\) are convergent or divergent cannot be determined.
- \(\displaystyle \int_0^1 g(x) dx\) can be either convergent or divergent and \(\displaystyle \int_0^1 h(x) dx\) is divergent.
- \(\displaystyle \int_0^1 g(x) dx\) is convergent and \(\displaystyle \int_0^1 h(x) dx\) is divergent.
Fact 5.8.27.
Let \(f(x), g(x)\) be functions such that for \(a<x<b\text{,}\) \(0\leq f(x) \leq g(x)\text{.}\) Then
\begin{equation*}
\displaystyle 0\leq \int_a^b f(x) dx\leq \int_a^b g(x) dx\text{.}
\end{equation*}
In particular:
- If \(\int_a^b g(x)dx\) converges, so does the smaller \(\int_a^b f(x)dx\text{.}\)
- If \(\int_a^b f(x)dx\) diverges, so does the bigger \(\int_a^b g(x)dx\text{.}\)
Activity 5.8.28.
Compare \(\frac{1}{x^3+1}\) to one of the following functions where \(x>2\) and use this to determine if \(\displaystyle \int_2^\infty \frac{1}{x^3+1}dx\) is convergent or divergent.
- \(\displaystyle \displaystyle\frac{1}{x}\)
- \(\displaystyle \displaystyle\frac{1}{\sqrt{x}}\)
- \(\displaystyle \displaystyle\frac{1}{x^2}\)
- \(\displaystyle \displaystyle\frac{1}{x^3}\)
Activity 5.8.29.
Comparing \(\frac{1}{x^3-4}\) to which of the following functions where \(x>3\) allows you to determine that \(\displaystyle\int_3^{\infty} \dfrac{1}{x^3-4}\ dx\) converges?
- \(\displaystyle \displaystyle\frac{1}{x^3+x}\)
- \(\displaystyle \displaystyle\frac{1}{4x^3}\)
- \(\displaystyle \displaystyle\frac{1}{x^3}\)
- \(\displaystyle \displaystyle\frac{1}{x^3-x^3/2}\)
Activity 5.8.30.
(a)
Find \(\displaystyle \int_{\pi/2}^a \cos(x)dx\text{.}\)
(b)
Which of the following is true about \(\int_{\pi/2}^\infty \cos(x)dx\text{?}\)
- \(\displaystyle \int_{\pi/2}^\infty \cos(x)dx\) is convergent.
- \(\displaystyle \int_{\pi/2}^\infty \cos(x)dx\) is divergent.
- More information is needed.
