# 3.7 Improper integrals  (Page 3/6)

 Page 3 / 6

Evaluate ${\int }_{-3}^{+\infty }{e}^{\text{−}x}dx.$ State whether the improper integral converges or diverges.

${e}^{3},$ converges

## Integrating a discontinuous integrand

Now let’s examine integrals of functions containing an infinite discontinuity in the interval over which the integration occurs. Consider an integral of the form ${\int }_{a}^{b}f\left(x\right)dx,$ where $f\left(x\right)$ is continuous over $\left[a,b\right)$ and discontinuous at $b.$ Since the function $f\left(x\right)$ is continuous over $\left[a,t\right]$ for all values of $t$ satisfying $a the integral ${\int }_{a}^{t}f\left(x\right)dx$ is defined for all such values of $t.$ Thus, it makes sense to consider the values of ${\int }_{a}^{t}f\left(x\right)dx$ as $t$ approaches $b$ for $a That is, we define ${\int }_{a}^{b}f\left(x\right)dx=\underset{t\to {b}^{-}}{\text{lim}}{\int }_{a}^{t}f\left(x\right)dx,$ provided this limit exists. [link] illustrates ${\int }_{a}^{t}f\left(x\right)dx$ as areas of regions for values of $t$ approaching $b.$

We use a similar approach to define ${\int }_{a}^{b}f\left(x\right)dx,$ where $f\left(x\right)$ is continuous over $\left(a,b\right]$ and discontinuous at $a.$ We now proceed with a formal definition.

## Definition

1. Let $f\left(x\right)$ be continuous over $\left[a,b\right).$ Then,
${\int }_{a}^{b}f\left(x\right)dx=\underset{t\to {b}^{-}}{\text{lim}}{\int }_{a}^{t}f\left(x\right)dx.$
2. Let $f\left(x\right)$ be continuous over $\left(a,b\right].$ Then,
${\int }_{a}^{b}f\left(x\right)dx=\underset{t\to {a}^{+}}{\text{lim}}{\int }_{t}^{b}f\left(x\right)dx.$

In each case, if the limit exists, then the improper integral is said to converge. If the limit does not exist, then the improper integral is said to diverge.
3. If $f\left(x\right)$ is continuous over $\left[a,b\right]$ except at a point $c$ in $\left(a,b\right),$ then
${\int }_{a}^{b}f\left(x\right)dx={\int }_{a}^{c}f\left(x\right)dx+{\int }_{c}^{b}f\left(x\right)dx,$

provided both ${\int }_{a}^{c}f\left(x\right)dx$ and ${\int }_{c}^{b}f\left(x\right)dx$ converge. If either of these integrals diverges, then ${\int }_{a}^{b}f\left(x\right)dx$ diverges.

The following examples demonstrate the application of this definition.

## Integrating a discontinuous integrand

Evaluate ${\int }_{0}^{4}\frac{1}{\sqrt{4-x}}dx,$ if possible. State whether the integral converges or diverges.

The function $f\left(x\right)=\frac{1}{\sqrt{4-x}}$ is continuous over $\left[0,4\right)$ and discontinuous at 4. Using [link] from the definition, rewrite ${\int }_{0}^{4}\frac{1}{\sqrt{4-x}}dx$ as a limit:

$\begin{array}{ccccc}\hfill {\int }_{0}^{4}\frac{1}{\sqrt{4-x}}dx& =\underset{t\to {4}^{-}}{\text{lim}}{\int }_{0}^{t}\frac{1}{\sqrt{4-x}}dx\hfill & & & \text{Rewrite as a limit.}\hfill \\ & =\underset{t\to {4}^{-}}{\text{lim}}\left(-2\sqrt{4-x}\right)|{}_{\begin{array}{c}\\ 0\end{array}}^{\begin{array}{c}t\\ \end{array}}\hfill & & & \text{Find the antiderivative.}\hfill \\ & =\underset{t\to {4}^{-}}{\text{lim}}\left(-2\sqrt{4-t}+4\right)\hfill & & & \text{Evaluate the antiderivative.}\hfill \\ & =4.\hfill & & & \text{Evaluate the limit.}\hfill \end{array}$

The improper integral converges.

## Integrating a discontinuous integrand

Evaluate ${\int }_{0}^{2}x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}dx.$ State whether the integral converges or diverges.

Since $f\left(x\right)=x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}x$ is continuous over $\left(0,2\right]$ and is discontinuous at zero, we can rewrite the integral in limit form using [link] :

$\begin{array}{ccccc}\hfill {\int }_{0}^{2}x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}dx& =\underset{t\to {0}^{+}}{\text{lim}}{\int }_{t}^{2}x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}dx\hfill & & & \text{Rewrite as a limit.}\hfill \\ & =\underset{t\to {0}^{+}}{\text{lim}}\left(\frac{1}{2}{x}^{2}\text{ln}\phantom{\rule{0.1em}{0ex}}x-\frac{1}{4}{x}^{2}\right)|{}_{\begin{array}{c}\\ t\end{array}}^{\begin{array}{c}2\\ \end{array}}\hfill & & & \begin{array}{c}\text{Evaluate}\phantom{\rule{0.2em}{0ex}}{\int }^{\text{​}}x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}\text{using integration by parts}\hfill \\ \text{with}\phantom{\rule{0.2em}{0ex}}u=\text{ln}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}dv=x.\hfill \end{array}\hfill \\ & =\underset{t\to {0}^{+}}{\text{lim}}\left(2\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}2-1-\frac{1}{2}{t}^{2}\text{ln}\phantom{\rule{0.1em}{0ex}}t+\frac{1}{4}{t}^{2}\right).\hfill & & & \text{Evaluate the antiderivative.}\hfill \\ & =2\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}2-1.\hfill & & & \begin{array}{c}\text{Evaluate the limit.}\phantom{\rule{0.2em}{0ex}}\underset{t\to {0}^{+}}{\text{lim}}{t}^{2}\text{ln}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.2em}{0ex}}\text{is indeterminate.}\hfill \\ \text{To evaluate it, rewrite as a quotient and apply}\hfill \\ \text{L’Hôpital’s rule.}\hfill \end{array}\hfill \end{array}$

The improper integral converges.

## Integrating a discontinuous integrand

Evaluate ${\int }_{-1}^{1}\frac{1}{{x}^{3}}dx.$ State whether the improper integral converges or diverges.

Since $f\left(x\right)=1\text{/}{x}^{3}$ is discontinuous at zero, using [link] , we can write

${\int }_{-1}^{1}\frac{1}{{x}^{3}}dx={\int }_{-1}^{0}\frac{1}{{x}^{3}}dx+{\int }_{0}^{1}\frac{1}{{x}^{3}}dx.$

If either of the two integrals diverges, then the original integral diverges. Begin with ${\int }_{-1}^{0}\frac{1}{{x}^{3}}dx:$

$\begin{array}{ccccc}\hfill {\int }_{-1}^{0}\frac{1}{{x}^{3}}dx& =\underset{t\to {0}^{-}}{\text{lim}}{\int }_{-1}^{t}\frac{1}{{x}^{3}}dx\hfill & & & \text{Rewrite as a limit.}\hfill \\ & =\underset{t\to {0}^{-}}{\text{lim}}\left(-\frac{1}{2{x}^{2}}\right)|{}_{\begin{array}{c}\\ -1\end{array}}^{\begin{array}{c}t\\ \end{array}}\hfill & & & \text{Find the antiderivative.}\hfill \\ & =\underset{t\to {0}^{-}}{\text{lim}}\left(-\frac{1}{2{t}^{2}}+\frac{1}{2}\right)\hfill & & & \text{Evaluate the antiderivative.}\hfill \\ & =\text{+}\infty .\hfill & & & \text{Evaluate the limit.}\hfill \end{array}$

Therefore, ${\int }_{-1}^{0}\frac{1}{{x}^{3}}dx$ diverges. Since ${\int }_{-1}^{0}\frac{1}{{x}^{3}}dx$ diverges, ${\int }_{-1}^{1}\frac{1}{{x}^{3}}dx$ diverges.

how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
is it 3×y ?
J, combine like terms 7x-4y
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!