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
${\int}_{a}^{b}f(x)dx,$ where
$f(x)$ is continuous over
$[a,b)$ and discontinuous at
$b.$ Since the function
$f(x)$ is continuous over
$[a,t]$ for all values of
$t$ satisfying
$a<t<b,$ the integral
${\int}_{a}^{t}f(x)dx$ is defined for all such values of
$t.$ Thus, it makes sense to consider the values of
${\int}_{a}^{t}f(x)dx$ as
$t$ approaches
$b$ for
$a<t<b.$ That is, we define
${\int}_{a}^{b}f(x)dx=\underset{t\to {b}^{-}}{\text{lim}}{\displaystyle {\int}_{a}^{t}f(x)dx}},$ provided this limit exists.
[link] illustrates
${\int}_{a}^{t}f(x)dx$ as areas of regions for values of
$t$ approaching
$b.$
We use a similar approach to define
${\int}_{a}^{b}f(x)dx,$ where
$f(x)$ is continuous over
$(a,b]$ and discontinuous at
$a.$ We now proceed with a formal definition.
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.
If
$f(x)$ is continuous over
$[a,b]$ except at a point
$c$ in
$(a,b),$ then
provided both
${\int}_{a}^{c}f(x)dx$ and
${\int}_{c}^{b}f(x)dx$ converge. If either of these integrals diverges, then
${\int}_{a}^{b}f(x)dx$ 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
$[0,4)$ and discontinuous at 4. Using
[link] from the definition, rewrite
${\int}_{0}^{4}\frac{1}{\sqrt{4-x}}dx$ as a limit:
$\begin{array}{ccccc}\hfill {\displaystyle {\int}_{0}^{4}\frac{1}{\sqrt{4-x}}dx}& =\underset{t\to {4}^{-}}{\text{lim}}{\displaystyle {\int}_{0}^{t}\frac{1}{\sqrt{4-x}}dx}\hfill & & & \text{Rewrite as a limit.}\hfill \\ & =\underset{t\to {4}^{-}}{\text{lim}}\left(\mathrm{-2}\sqrt{4-x}\right)|{}_{\begin{array}{c}\\ 0\end{array}}^{\begin{array}{c}t\\ \end{array}}\hfill & & & \text{Find the antiderivative.}\hfill \\ & =\underset{t\to {4}^{-}}{\text{lim}}\left(\mathrm{-2}\sqrt{4-t}+4\right)\hfill & & & \text{Evaluate the antiderivative.}\hfill \\ & =4.\hfill & & & \text{Evaluate the limit.}\hfill \end{array}$
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
$(0,2]$ and is discontinuous at zero, we can rewrite the integral in limit form using
[link] :
$\begin{array}{ccccc}\hfill {\displaystyle {\int}_{0}^{2}x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}dx}& =\underset{t\to {0}^{+}}{\text{lim}}{\displaystyle {\int}_{t}^{2}x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}dx}\hfill & & & \text{Rewrite as a limit.}\hfill \\ & =\underset{t\to {0}^{+}}{\text{lim}}\left(\frac{1}{2}{x}^{2}\text{ln}\phantom{\rule{0.1em}{0ex}}x-\frac{1}{4}{x}^{2}\right)|{}_{\begin{array}{c}\\ t\end{array}}^{\begin{array}{c}2\\ \end{array}}\hfill & & & \begin{array}{c}\text{Evaluate}\phantom{\rule{0.2em}{0ex}}{{\displaystyle \int}}^{\text{}}x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}\text{using integration by parts}\hfill \\ \text{with}\phantom{\rule{0.2em}{0ex}}u=\text{ln}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}dv=x.\hfill \end{array}\hfill \\ & =\underset{t\to {0}^{+}}{\text{lim}}\left(2\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}2-1-\frac{1}{2}{t}^{2}\text{ln}\phantom{\rule{0.1em}{0ex}}t+\frac{1}{4}{t}^{2}\right).\hfill & & & \text{Evaluate the antiderivative.}\hfill \\ & =2\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}2-1.\hfill & & & \begin{array}{c}\text{Evaluate the limit.}\phantom{\rule{0.2em}{0ex}}\underset{t\to {0}^{+}}{\text{lim}}{t}^{2}\text{ln}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.2em}{0ex}}\text{is indeterminate.}\hfill \\ \text{To evaluate it, rewrite as a quotient and apply}\hfill \\ \text{L\u2019H\xf4pital\u2019s rule.}\hfill \end{array}\hfill \end{array}$
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.
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Leaves accumulate on the forest floor at a rate of 2 g/cm2/yr and also decompose at a rate of 90% per year. Write a differential equation governing the number of grams of leaf litter per square centimeter of forest floor, assuming at time 0 there is no leaf litter on the ground. Does this amount approach a steady value? What is that value?
You have a cup of coffee at temperature 70°C, which you let cool 10 minutes before you pour in the same amount of milk at 1°C as in the preceding problem. How does the temperature compare to the previous cup after 10 minutes?