TBIL-Cal Improper Integrals (TI8)

\(\displaystyle \int_{a}^1 \frac{1}{x^2}\, dx\) approaches \(0\text{.}\)
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\(\displaystyle \int_{a}^1 \frac{1}{x^2}\, dx\) approaches a finite constant greater than \(0\text{.}\)
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\(\displaystyle \int_{a}^1 \frac{1}{x^2}\, dx\) approaches \(\infty\text{.}\)
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There is not enough information.
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Activity 5.8.3.

Compute the following definite integrals, again using \(\displaystyle \int\frac{1}{x^2}\, dx=-\frac{1}{x}+C\text{.}\)

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(a)

\(\displaystyle \int_{1}^{100} \frac{1}{x^2}\, dx=\left[-\frac{1}{x}\right]_{1}^{100}\)

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(b)

\(\displaystyle \int_{1}^{10000} \frac{1}{x^2}\, dx\)

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(c)

\(\displaystyle \int_{1}^{1000000} \frac{1}{x^2}\, dx\)

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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?

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\(\displaystyle \int_{1}^b \frac{1}{x^2}\, dx\) approaches \(0\text{.}\)
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\(\displaystyle \int_{1}^b \frac{1}{x^2}\, dx\) approaches a finite constant greater than \(0\text{.}\)
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\(\displaystyle \int_{1}^b \frac{1}{x^2}\, dx\) approaches \(\infty\text{.}\)
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There is not enough information.
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Activity 5.8.5.

Recall \(\displaystyle \int\frac{1}{\sqrt x}\, dx=2\sqrt{x}+C\text{.}\) Compute the following definite integrals.

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(a)

\(\displaystyle \int_{1/100}^1 \frac{1}{\sqrt{x}}\, dx=\left[2\sqrt{x}\right]_{1/100}^1\)

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(b)

\(\displaystyle \int_{1/10000}^1 \frac{1}{\sqrt{x}}\, dx\)

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(c)

\(\displaystyle \int_{1/1000000}^1 \frac{1}{\sqrt{x}}\, dx\)

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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?

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\(\displaystyle \int_{a}^1 \frac{1}{\sqrt{x}}\, dx\) approaches \(0\text{.}\)
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\(\displaystyle \int_{a}^1 \frac{1}{\sqrt{x}}\, dx\) approaches a finite constant greater than \(0\text{.}\)
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\(\displaystyle \int_{a}^1 \frac{1}{\sqrt{x}}\, dx\) approaches \(\infty\text{.}\)
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There is not enough information.
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(b)

How does this compare to what you found in Activity 5.8.1?

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Activity 5.8.7.

Compute the following definite integrals using \(\displaystyle \int\frac{1}{\sqrt x}\, dx=2\sqrt{x}+C\text{.}\)

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(a)

\(\displaystyle \int_{1}^{100} \frac{1}{\sqrt{x}}\, dx = \left[2\sqrt{x}\right]_{1}^{100}\)

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(b)

\(\displaystyle \int_{1}^{10000} \frac{1}{\sqrt{x}}\, dx\)

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(c)

\(\displaystyle \int_{1}^{1000000} \frac{1}{\sqrt{x}}\, dx\)

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Activity 5.8.8.

(a)

What do you notice about the integral \(\displaystyle \int_{1}^b \frac{1}{\sqrt{x}}\, dx\) as \(b\) approached \(\infty\) in Activity 5.8.7?

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\(\displaystyle \int_{1}^b \frac{1}{\sqrt{x}}\,dx\) approaches \(0\text{.}\)
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\(\displaystyle \int_{1}^b \frac{1}{\sqrt{x}}\, dx\) approaches a finite constant greater than \(0\text{.}\)
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\(\displaystyle \int_{1}^b \frac{1}{\sqrt{x}}\, dx\) approaches \(\infty\text{.}\)
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There is not enough information.
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(b)

How does this compare to what you found in Activity 5.8.3?

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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.

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Similarly,

\begin{equation*} \int_{-\infty}^b f(x)\, dx=\lim_{a\to-\infty}\left( \int_a^b f(x) \, dx\right). \end{equation*}

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Activity 5.8.10.

Which of these limits is equal to \(\displaystyle\int_1^\infty \frac{1}{x^2} \, dx\text{?}\)

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\(\displaystyle \displaystyle\lim_{b\to\infty}\int_1^b\frac{1}{x^2}\, dx\)

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\(\displaystyle \displaystyle\lim_{b\to\infty}\left[-\frac{1}{x}\right]_1^b\)

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\(\displaystyle \displaystyle\lim_{b\to\infty}\left[-\frac{1}{b}+1\right]\)

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All of these.

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Activity 5.8.11.

Given the result of Activity 5.8.10, what is \(\displaystyle\int_1^\infty \frac{1}{x^2}\, dx\text{?}\)

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\(\displaystyle 0\)

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\(\displaystyle 1\)

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\(\displaystyle \infty\)

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\(\displaystyle -\infty\)

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Activity 5.8.12.

Does \(\displaystyle\int_1^\infty \frac{1}{\sqrt x}\, dx\) converge or diverge?

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Converges because \(\displaystyle\lim_{b\to 0^+}\left[2\sqrt b-2\right]\) converges.

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Diverges because \(\displaystyle\lim_{b\to 0^+}\left[2\sqrt b-2\right]\) diverges.

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Converges because \(\displaystyle\lim_{b\to \infty}\left[2\sqrt b-2\right]\) converges.

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Diverges because \(\displaystyle\lim_{b\to \infty}\left[2\sqrt b-2\right]\) diverges.

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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*}

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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*}

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Activity 5.8.14.

Which of these limits is equal to \(\displaystyle\int_0^1 \frac{1}{\sqrt x} \, dx\text{?}\)

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\(\displaystyle \displaystyle\lim_{a\to 0^+}\int_a^1\frac{1}{\sqrt x}\, dx\)

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\(\displaystyle \displaystyle\lim_{a\to 0^+}\left[2\sqrt x\right]_a^1\)

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\(\displaystyle \displaystyle\lim_{a\to 0^+}\left[2-2\sqrt a\right]\)

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All of these.

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Activity 5.8.15.

Given the this result, what is \(\displaystyle\int_0^1 \frac{1}{\sqrt x}\, dx\text{?}\)

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\(\displaystyle 0\)

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\(\displaystyle 1\)

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\(\displaystyle 2\)

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\(\displaystyle \infty\)

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Activity 5.8.16.

Does \(\displaystyle\int_0^1 \frac{1}{x^2}\, dx\) converge or diverge?

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Converges because \(\displaystyle\lim_{a\to 0^+}\left[-1+\frac{1}{a}\right]\) converges.

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Diverges because \(\displaystyle\lim_{a\to 0^+}\left[-1+\frac{1}{a}\right]\) diverges.

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Converges because \(\displaystyle\lim_{a\to 1^-}\left[-1+\frac{1}{a}\right]\) converges.

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Diverges because \(\displaystyle\lim_{a\to 1^-}\left[-1+\frac{1}{a}\right]\) diverges.

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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.

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(a)

\(\displaystyle\int_{ -2 }^{ +\infty } \frac{1}{\sqrt{x + 6}}\, dx.\)

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(b)

\(\displaystyle\int_{ -4 }^{ -2 } \frac{1}{{\left(x + 4\right)}^{\frac{4}{3}}}\, dx.\)

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(c)

\(\displaystyle\int_{ -5 }^{ 0 } \frac{1}{{\left(x + 5\right)}^{\frac{5}{9}}}\, dx.\)

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(d)

\(\displaystyle\int_{ 10 }^{ +\infty } \frac{1}{{\left(x - 8\right)}^{\frac{4}{3}}}\, dx.\)

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Fact 5.8.18.

Suppose that \(0< p\) and \(p\neq 1\text{.}\) Applying the integration power rule gives us the indefinite integral \(\displaystyle \int \frac{1}{x^p}\, dx=\frac{1}{(1-p)}x^{1-p}+C\text{.}\)

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Activity 5.8.19.

(a)

If \(0< p< 1\text{,}\) which of the following statements must be true? Select all that apply.

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\(\displaystyle 1-p< 0\)
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\(\displaystyle 1-p> 0\)
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\(\displaystyle 1-p< 1\)
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\(\displaystyle \int_1^\infty \frac{1}{x^p}\, dx\) converges.
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\(\displaystyle \int_1^\infty \frac{1}{x^p}\, dx\) diverges.
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(b)

If \(p>1\text{,}\) which of the following statements must be true? Select all that apply.

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\(\displaystyle 1-p< 0\)
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\(\displaystyle 1-p> 0\)
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\(\displaystyle 1-p< 1\)
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\(\displaystyle \int_1^\infty \frac{1}{x^p}\, dx\) converges.
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\(\displaystyle \int_1^\infty \frac{1}{x^p}\, dx\) diverges.
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Activity 5.8.20.

(a)

If \(0< p< 1\text{,}\) which of the following statements must be true?

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\(\displaystyle \int_0^1 \frac{1}{x^p}\, dx\) converges.
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\(\displaystyle \int_0^1 \frac{1}{x^p}\, dx\) diverges.
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(b)

If \(p>1\text{,}\) which of the following statements must be true?

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\(\displaystyle \int_0^1 \frac{1}{x^p}\, dx\) converges.
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\(\displaystyle \int_0^1 \frac{1}{x^p}\, dx\) diverges.
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Activity 5.8.21.

Consider when \(p=1\text{.}\) Then \(\dfrac{1}{x^p}=\dfrac{1}{x}\) and \(\displaystyle \int \frac{1}{x^p}\, dx=\displaystyle \int \frac{1}{x}\, dx=\ln|x|+C\text{.}\)

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(a)

What can we conclude about \(\displaystyle \int_1^\infty \frac{1}{x}\, dx\text{?}\)

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\(\displaystyle \int_1^\infty \frac{1}{x}\, dx\) converges.
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\(\displaystyle \int_1^\infty \frac{1}{x}\, dx\) diverges.
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There is not enough information to determine whether this integral converges or diverges.
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(b)

What can we conclude about \(\displaystyle \int_0^1 \frac{1}{x}\, dx\text{?}\)

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\(\displaystyle \int_0^1 \frac{1}{x}\, dx\) converges.
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\(\displaystyle \int_0^1 \frac{1}{x}\, dx\) diverges.
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There is not enough information to determine whether this integral converges or diverges.
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Fact 5.8.22.

Let \(c, p>0\text{.}\)

\(\displaystyle \int_0^c \frac{1}{x^p}\, dx\) converges if and only if \(p< 1\text{.}\)

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\(\displaystyle \int_c^\infty \frac{1}{x^p}\, dx\) converges if and only if \(p > 1\text{.}\)

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Activity 5.8.23.

Consider the plots of \(f(x), g(x), h(x) \) where \(0 < g(x) < f(x) < h(x)\text{.}\)

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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.

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\(\displaystyle \int_1^\infty g(x)\, dx\) and \(\displaystyle \int_1^\infty h(x)\, dx\) are both divergent.

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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.

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\(\displaystyle \int_1^\infty g(x)\, dx\) is convergent and \(\displaystyle \int_1^\infty h(x)\, dx\) is divergent.

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\(\displaystyle \int_1^\infty g(x)\, dx\) is convergent and \(\displaystyle \int_1^\infty h(x)\, dx\) could be either convergent or divergent.

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Activity 5.8.24.

Consider the plots of \(f(x), g(x), h(x) \) where \(0 < g(x) < f(x) < h(x)\text{.}\)

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If \(\displaystyle \int_1^\infty f(x)\, dx\) is divergent, 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.

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\(\displaystyle \int_1^\infty g(x)\, dx\) and \(\displaystyle \int_1^\infty h(x)\, dx\) are both divergent.

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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.

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\(\displaystyle \int_1^\infty g(x)\, dx\) could be either convergent or divergent and \(\displaystyle \int_1^\infty h(x)\, dx\) is divergent.

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\(\displaystyle \int_1^\infty g(x)\, dx\) is convergent and \(\displaystyle \int_1^\infty h(x)\, dx\) is divergent.

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Activity 5.8.25.

Consider the plots of \(f(x), g(x), h(x) \) where \(0 < g(x) < f(x) < h(x)\text{.}\)

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Cursor down	Explore next lower level
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If \(\displaystyle \int_0^1 f(x)\, dx\) is convergent, what can we say about \(g(x)\) and \(h(x)\text{?}\)

\(\displaystyle \int_0^1 g(x)\, dx\) and \(\displaystyle \int_0^1 h(x)\, dx\) are both convergent.

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\(\displaystyle \int_0^1 g(x)\, dx\) and \(\displaystyle \int_0^1 h(x)\, dx\) are both divergent.

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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.

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\(\displaystyle \int_0^1 g(x)\, dx\) is convergent and \(\displaystyle \int_0^1 h(x)\, dx\) is divergent.

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\(\displaystyle \int_0^1 g(x)\, dx\) is convergent and \(\displaystyle \int_0^1 h(x)\, dx\) can either be convergent or divergent.

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Activity 5.8.26.

Consider the plots of \(f(x), g(x), h(x) \) where \(0 < g(x) < f(x) < h(x)\text{.}\)

Diagram Exploration Keyboard Controls

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B	Activate menu driven exploration
Escape	Leave exploration mode
Cursor down	Explore next lower level
Cursor up	Explore next upper level
Cursor right	Explore next element on level
Cursor left	Explore previous element on level
X	Toggle expert mode
W	Extra details if available
Space	Repeat speech
M	Activate step magnification
Comma	Activate direct magnification
N	Deactivate magnification
Z	Toggle subtitles
C	Cycle contrast settings
T	Monochrome colours
L	Toggle language (if available)
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If \(\displaystyle \int_0^1 f(x)\, dx\) is divergent, what can we say about \(g(x)\) and \(h(x)\text{?}\)

\(\displaystyle \int_0^1 g(x)\, dx\) and \(\displaystyle \int_0^1 h(x)\, dx\) are both convergent.

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\(\displaystyle \int_0^1 g(x)\, dx\) and \(\displaystyle \int_0^1 h(x)\, dx\) are both divergent.

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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.

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\(\displaystyle \int_0^1 g(x)\, dx\) can be either convergent or divergent and \(\displaystyle \int_0^1 h(x)\, dx\) is divergent.

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\(\displaystyle \int_0^1 g(x)\, dx\) is convergent and \(\displaystyle \int_0^1 h(x)\, dx\) is divergent.

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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 \(\displaystyle\int_a^b g(x)\, dx\) converges, so does the smaller \(\displaystyle\int_a^b f(x)\, dx\text{.}\)
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If \(\displaystyle\int_a^b f(x)\, dx\) diverges, so does the bigger \(\displaystyle\int_a^b g(x)\, dx\text{.}\)
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Activity 5.8.28.

Compare \(\dfrac{1}{x^3+1}\) 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.

\(\displaystyle \dfrac{1}{x}\)

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\(\displaystyle \dfrac{1}{\sqrt{x}}\)

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\(\displaystyle \dfrac{1}{x^2}\)

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\(\displaystyle \dfrac{1}{x^3}\)

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Activity 5.8.29.

Comparing \(\dfrac{1}{x^3-4}\) 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?