TBIL-Cal Improper Integrals (TI8)

$\int_{a}^{1} \frac{1}{x^{2}} d x$ approaches $0 .$
$\int_{a}^{1} \frac{1}{x^{2}} d x$ approaches a finite constant greater than $0 .$
$\int_{a}^{1} \frac{1}{x^{2}} d x$ approaches $\infty .$
There is not enough information.

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Activity 5.8.3.

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Compute the following definite integrals, again using

\int \frac{1}{x^{2}} d x = - \frac{1}{x} + C .

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(a)

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\int_{1}^{100} \frac{1}{x^{2}} d x = {[- \frac{1}{x}]}_{1}^{100}

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(b)

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\int_{1}^{10000} \frac{1}{x^{2}} d x

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(c)

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\int_{1}^{1000000} \frac{1}{x^{2}} d x

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Activity 5.8.4.

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What do you notice about

\int_{1}^{b} \frac{1}{x^{2}} d x

b

approached

\infty

in Activity 5.8.3?

$\int_{1}^{b} \frac{1}{x^{2}} d x$ approaches $0 .$
$\int_{1}^{b} \frac{1}{x^{2}} d x$ approaches a finite constant greater than $0 .$
$\int_{1}^{b} \frac{1}{x^{2}} d x$ approaches $\infty .$
There is not enough information.

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Activity 5.8.5.

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Recall

\int \frac{1}{\sqrt{x}} d x = 2 \sqrt{x} + C .

Compute the following definite integrals.

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(a)

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\int_{1 / 100}^{1} \frac{1}{\sqrt{x}} d x = {[2 \sqrt{x}]}_{1 / 100}^{1}

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(b)

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\int_{1 / 10000}^{1} \frac{1}{\sqrt{x}} d x

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(c)

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\int_{1 / 1000000}^{1} \frac{1}{\sqrt{x}} d x

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Activity 5.8.6.

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(a)

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What do you notice about the integral

\int_{a}^{1} \frac{1}{\sqrt{x}} d x

a

approached 0 in Activity 5.8.5?

$\int_{a}^{1} \frac{1}{\sqrt{x}} d x$ approaches $0 .$
$\int_{a}^{1} \frac{1}{\sqrt{x}} d x$ approaches a finite constant greater than $0 .$
$\int_{a}^{1} \frac{1}{\sqrt{x}} d x$ approaches $\infty .$
There is not enough information.

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(b)

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

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Activity 5.8.7.

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Compute the following definite integrals using

\int \frac{1}{\sqrt{x}} d x = 2 \sqrt{x} + C .

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(a)

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\int_{1}^{100} \frac{1}{\sqrt{x}} d x = {[2 \sqrt{x}]}_{1}^{100}

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(b)

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\int_{1}^{10000} \frac{1}{\sqrt{x}} d x

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(c)

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\int_{1}^{1000000} \frac{1}{\sqrt{x}} d x

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Activity 5.8.8.

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(a)

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What do you notice about the integral

\int_{1}^{b} \frac{1}{\sqrt{x}} d x

b

approached

\infty

in Activity 5.8.7?

$\int_{1}^{b} \frac{1}{\sqrt{x}} d x$ approaches $0 .$
$\int_{1}^{b} \frac{1}{\sqrt{x}} d x$ approaches a finite constant greater than $0 .$
$\int_{1}^{b} \frac{1}{\sqrt{x}} d x$ approaches $\infty .$
There is not enough information.

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(b)

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

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Definition 5.8.9.

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For a function

f (x)

and a constant

a,

we let

\int_{a}^{\infty} f (x) d x

denote

\int_{a}^{\infty} f (x) d x = lim_{b \to \infty} (\int_{a}^{b} f (x) d x) .

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If this limit is a defined real number, then we say

\int_{a}^{\infty} f (x) d x

is convergent. Otherwise, it is divergent.

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Similarly,

\int_{- \infty}^{b} f (x) d x = lim_{a \to - \infty} (\int_{a}^{b} f (x) d x) .

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Activity 5.8.10.

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Which of these limits is equal to

\int_{1}^{\infty} \frac{1}{x^{2}} d x ?

$lim_{b \to \infty} \int_{1}^{b} \frac{1}{x^{2}} d x$
$lim_{b \to \infty} {[- \frac{1}{x}]}_{1}^{b}$
$lim_{b \to \infty} [- \frac{1}{b} + 1]$
All of these.

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Activity 5.8.11.

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Given the result of Activity 5.8.10, what is

\int_{1}^{\infty} \frac{1}{x^{2}} d x ?

$0$
$1$
$\infty$
$- \infty$

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Activity 5.8.12.

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Does

\int_{1}^{\infty} \frac{1}{\sqrt{x}} d x

converge or diverge?

Converges because $lim_{b \to 0^{+}} [2 \sqrt{b} - 2]$ converges.
Diverges because $lim_{b \to 0^{+}} [2 \sqrt{b} - 2]$ diverges.
Converges because $lim_{b \to \infty} [2 \sqrt{b} - 2]$ converges.
Diverges because $lim_{b \to \infty} [2 \sqrt{b} - 2]$ diverges.

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Definition 5.8.13.

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For a function

f (x)

with a vertical asymptote at

x = c > a,

we let

\int_{a}^{c} f (x) d x

denote

\int_{a}^{c} f (x) d x = lim_{b \to c^{-}} (\int_{a}^{b} f (x) d x) .

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For a function

f (x)

with a vertical asymptote at

x = c < b,

we let

\int_{c}^{b} f (x) d x

denote

\int_{c}^{b} f (x) d x = lim_{a \to c^{+}} (\int_{a}^{b} f (x) d x) .

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Activity 5.8.14.

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Which of these limits is equal to

\int_{0}^{1} \frac{1}{\sqrt{x}} d x ?

$lim_{a \to 0^{+}} \int_{a}^{1} \frac{1}{\sqrt{x}} d x$
$lim_{a \to 0^{+}} {[2 \sqrt{x}]}_{a}^{1}$
$lim_{a \to 0^{+}} [2 - 2 \sqrt{a}]$
All of these.

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Activity 5.8.15.

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Given the this result, what is

\int_{0}^{1} \frac{1}{\sqrt{x}} d x ?

$0$
$1$
$2$
$\infty$

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Activity 5.8.16.

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Does

\int_{0}^{1} \frac{1}{x^{2}} d x

converge or diverge?

Converges because $lim_{a \to 0^{+}} [- 1 + \frac{1}{a}]$ converges.
Diverges because $lim_{a \to 0^{+}} [- 1 + \frac{1}{a}]$ diverges.
Converges because $lim_{a \to 1^{-}} [- 1 + \frac{1}{a}]$ converges.
Diverges because $lim_{a \to 1^{-}} [- 1 + \frac{1}{a}]$ diverges.

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Activity 5.8.17.

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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)

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\int_{- 2}^{+ \infty} \frac{1}{\sqrt{x + 6}} d x .

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(b)

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\int_{- 4}^{- 2} \frac{1}{{(x + 4)}^{\frac{4}{3}}} d x .

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(c)

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\int_{- 5}^{0} \frac{1}{{(x + 5)}^{\frac{5}{9}}} d x .

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(d)

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\int_{10}^{+ \infty} \frac{1}{{(x - 8)}^{\frac{4}{3}}} d x .

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Fact 5.8.18.

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Suppose that

0 < p

and

p \neq 1 .

Applying the integration power rule gives us the indefinite integral

\int \frac{1}{x^{p}} d x = \frac{1}{(1 - p)} x^{1 - p} + C .

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Activity 5.8.19.

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(a)

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0 < p < 1,

which of the following statements must be true? Select all that apply.

$1 - p < 0$
$1 - p > 0$
$1 - p < 1$
$\int_{1}^{\infty} \frac{1}{x^{p}} d x$ converges.
$\int_{1}^{\infty} \frac{1}{x^{p}} d x$ diverges.

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(b)

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p > 1,

which of the following statements must be true? Select all that apply.

$1 - p < 0$
$1 - p > 0$
$1 - p < 1$
$\int_{1}^{\infty} \frac{1}{x^{p}} d x$ converges.
$\int_{1}^{\infty} \frac{1}{x^{p}} d x$ diverges.

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Activity 5.8.20.

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(a)

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0 < p < 1,

which of the following statements must be true?

$\int_{0}^{1} \frac{1}{x^{p}} d x$ converges.
$\int_{0}^{1} \frac{1}{x^{p}} d x$ diverges.

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(b)

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p > 1,

which of the following statements must be true?

$\int_{0}^{1} \frac{1}{x^{p}} d x$ converges.
$\int_{0}^{1} \frac{1}{x^{p}} d x$ diverges.

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Activity 5.8.21.

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Consider when

p = 1 .

Then

\frac{1}{x^{p}} = \frac{1}{x}

and

\int \frac{1}{x^{p}} d x = \int \frac{1}{x} d x = \ln | x | + C .

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(a)

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What can we conclude about

\int_{1}^{\infty} \frac{1}{x} d x ?

$\int_{1}^{\infty} \frac{1}{x} d x$ converges.
$\int_{1}^{\infty} \frac{1}{x} d x$ diverges.
There is not enough information to determine whether this integral converges or diverges.

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(b)

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What can we conclude about

\int_{0}^{1} \frac{1}{x} d x ?

$\int_{0}^{1} \frac{1}{x} d x$ converges.
$\int_{0}^{1} \frac{1}{x} d x$ diverges.
There is not enough information to determine whether this integral converges or diverges.

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Fact 5.8.22.

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Let

c, p > 0 .

$\int_{0}^{c} \frac{1}{x^{p}} d x$ converges if and only if $p < 1 .$

$\int_{c}^{\infty} \frac{1}{x^{p}} d x$ converges if and only if $p > 1 .$

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Activity 5.8.23.

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Consider the plots of

f (x), g (x), h (x)

where

0 < g (x) < f (x) < h (x) .

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\int_{1}^{\infty} f (x) d x

is convergent, what can we say about

g (x), h (x) ?

$\int_{1}^{\infty} g (x) d x$ and $\int_{1}^{\infty} h (x) d x$ are both convergent.
$\int_{1}^{\infty} g (x) d x$ and $\int_{1}^{\infty} h (x) d x$ are both divergent.
Whether or not $\int_{1}^{\infty} g (x) d x$ and $\int_{1}^{\infty} h (x) d x$ are convergent or divergent cannot be determined.
$\int_{1}^{\infty} g (x) d x$ is convergent and $\int_{1}^{\infty} h (x) d x$ is divergent.
$\int_{1}^{\infty} g (x) d x$ is convergent and $\int_{1}^{\infty} h (x) d x$ could be either convergent or divergent.

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Activity 5.8.24.

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Consider the plots of

f (x), g (x), h (x)

where

0 < g (x) < f (x) < h (x) .

\int_{1}^{\infty} f (x) d x

is divergent, what can we say about

g (x), h (x) ?

$\int_{1}^{\infty} g (x) d x$ and $\int_{1}^{\infty} h (x) d x$ are both convergent.
$\int_{1}^{\infty} g (x) d x$ and $\int_{1}^{\infty} h (x) d x$ are both divergent.
Whether or not $\int_{1}^{\infty} g (x) d x$ and $\int_{1}^{\infty} h (x) d x$ are convergent or divergent cannot be determined.
$\int_{1}^{\infty} g (x) d x$ could be either convergent or divergent and $\int_{1}^{\infty} h (x) d x$ is divergent.
$\int_{1}^{\infty} g (x) d x$ is convergent and $\int_{1}^{\infty} h (x) d x$ is divergent.

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Activity 5.8.25.

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Consider the plots of

f (x), g (x), h (x)

where

0 < g (x) < f (x) < h (x) .

Plots of positive functions f(x), g(x) and h(x). — Figure 115. Plots of $f (x), g (x), h (x)$

\int_{0}^{1} f (x) d x

is convergent, what can we say about

g (x)

and

h (x) ?

$\int_{0}^{1} g (x) d x$ and $\int_{0}^{1} h (x) d x$ are both convergent.
$\int_{0}^{1} g (x) d x$ and $\int_{0}^{1} h (x) d x$ are both divergent.
Whether or not $\int_{0}^{1} g (x) d x$ and $\int_{0}^{1} h (x) d x$ are convergent or divergent cannot be determined.
$\int_{0}^{1} g (x) d x$ is convergent and $\int_{0}^{1} h (x) d x$ is divergent.
$\int_{0}^{1} g (x) d x$ is convergent and $\int_{0}^{1} h (x) d x$ can either be convergent or divergent.

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Activity 5.8.26.

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Consider the plots of

f (x), g (x), h (x)

where

0 < g (x) < f (x) < h (x) .

\int_{0}^{1} f (x) d x

is divergent, what can we say about

g (x)

and

h (x) ?

$\int_{0}^{1} g (x) d x$ and $\int_{0}^{1} h (x) d x$ are both convergent.
$\int_{0}^{1} g (x) d x$ and $\int_{0}^{1} h (x) d x$ are both divergent.
Whether or not $\int_{0}^{1} g (x) d x$ and $\int_{0}^{1} h (x) d x$ are convergent or divergent cannot be determined.
$\int_{0}^{1} g (x) d x$ can be either convergent or divergent and $\int_{0}^{1} h (x) d x$ is divergent.
$\int_{0}^{1} g (x) d x$ is convergent and $\int_{0}^{1} h (x) d x$ is divergent.

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Fact 5.8.27.

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Let

f (x), g (x)

be functions such that for

a < x < b,

0 \leq f (x) \leq g (x) .

Then

0 \leq \int_{a}^{b} f (x) d x \leq \int_{a}^{b} g (x) d x .

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In particular:

If $\int_{a}^{b} g (x) d x$ converges, so does the smaller $\int_{a}^{b} f (x) d x .$
If $\int_{a}^{b} f (x) d x$ diverges, so does the bigger $\int_{a}^{b} g (x) d x .$