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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Plots of positive functions f(x), g(x) where f(x) is an upper bound of g(x). — Figure 111. Plots of \(f(x), g(x), h(x)\)
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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 \(0 < g(x) < f(x) < h(x)\text{.}\)

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 \(0 < g(x) < f(x) < h(x)\text{.}\)

Figure 113. Plots of \(f(x), g(x), h(x)\)
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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 \(0 < g(x) < f(x) < h(x)\text{.}\)

Plots of positive functions f(x), g(x) and h(x). — Figure 114. Plots of \(f(x), g(x), h(x)\)
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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?