Definite Integral Problem

[Inverted Composer]

\(\displaystyle \displaystyle{ \int_1^{2}\, }\) \(\displaystyle \frac{x^{2}-x-1 }{x^{2}}e^{-x}\, dx\,\)

I tried integration by parts, but the result doesn't match up. Any tips?

 

[Ishuda]

\(\displaystyle \displaystyle{ \int_1^{2}\, }\) \(\displaystyle \frac{x^{2}-x-1 }{x^{2}}e^{-x}\, dx\,\)

I tried integration by parts, but the result doesn't match up. Any tips?

Break it up
\(\displaystyle \displaystyle{ \int_1^{2}\, }\) \(\displaystyle \frac{x^{2}-x-1 }{x^{2}}e^{-x}\, dx\,\)
=\(\displaystyle \displaystyle{ \int_1^{2}\, }\, e^{-x}\, dx - \displaystyle{ \int_1^{2}\, }x^{-1}\, e^{-x}\, dx - \displaystyle{ \int_1^{2}\, }x^{-2}\, e^{-x}\, dx\)
The last two involve the Exponential Integral
Ei(x) = \(\displaystyle - \displaystyle{ \int_{-x}^{\infty}\, }x^{-1}\, e^{-x}\, dx\)
and you would need a table look up or MatLab or ... See
http://en.wikipedia.org/wiki/Exponential_integral
for example

However, one could use u=x^-1 and dv=e^-x on the middle integral and see what happens.

 

[Inverted Composer]

Solved it as follows:

\(\displaystyle \large{
{ \displaystyle{ \int_1^{2}\, }\frac{x^{2}-x-1 }{x^{2}}e^{-x}\, dx\,}

}\)

\(\displaystyle \large\displaystyle{
\mbox{a)} {-\frac{1}{2e^2}}\qquad {b)} {-\frac{1}{e}}\qquad {c)} {-\frac{1}{e^2} +\frac{1}{2e}}\qquad {d)} {\frac{1}{2e^2}}\qquad {e)} {-\frac{1}{e^2}-\frac{1}{2e}}
}\)


\(\displaystyle \large\displaystyle{

I= { \int_1^{2}\, }\frac{x^{2}-x-1 }{x^{2}}e^{-x}\, dx\, = \underbrace{{ \int_1^{2}\, }e^{-x}\, dx\,}_{I_1}-{ \int_1^{2}\, }\frac{1}{x}e^{-x}\, dx\,-\underbrace{{ \int_1^{2}\, }\frac{1}{x^2}e^{-x}\, dx\,}_{I_2}

}\)

\(\displaystyle \large\displaystyle{
I_1= { \int_1^{2}\, }e^{-x}\, dx\, = \left. -e^{-x} \right|_{1}^{2}= -(\frac{1}{e^2}-\frac{1}{e}) = \frac{1}{e}- \frac{1}{e^2}
}\)


Using the formula:

\(\displaystyle \large \displaystyle{
{ \int\, }\frac{e^{ax} }{x^{2}}\, dx\,= -\frac{e^{ax}}{x} + a{\int}\,\frac{e^{ax}}{x}\,dx\,
}\)

\(\displaystyle \large\displaystyle{

I_2= { \int_1^{2}\, }\frac{1}{x^2}e^{-x}\, dx\, = \left. -\frac{e^{-x}}{x} \right|_{1}^{2} - { \int_1^{2}\, }\frac{1}{x}e^{-x}\, dx\, = -(\frac{1}{2e^2}-\frac{1}{e}) -{ \int_1^{2}\, }\frac{1}{x}e^{-x}\, dx\,


}\)


\(\displaystyle \large\displaystyle{

I= I_1-{ \int_1^{2}\, }\frac{1}{x}e^{-x}\, dx\,-I_2 = \frac{1}{e}- \frac{1}{e^2} -{ \int_1^{2}\, }\frac{1}{x}e^{-x}\, dx\,-\Big(-(\frac{1}{2e^2}-\frac{1}{e}) -{ \int_1^{2}\, }\frac{1}{x}e^{-x}\, dx\,\Big)
}\)


\(\displaystyle \large\displaystyle{
I= \frac{1}{e}- \frac{1}{e^2}+\frac{1}{2e^2}-\frac{1}{e}-{ \int_1^{2}\, }\frac{1}{x}e^{-x}\, dx\,+{ \int_1^{2}\, }\frac{1}{x}e^{-x}\, dx\,
}\)

\(\displaystyle \large\displaystyle{
I= \frac{1-2}{2e^2}= -\frac{1}{2e^2}
}\)