# Math Analysis issue

Maryanne
Visitor
2016-08-25 07:27:11
#1
let be the function f:R->R, f(x) = ex (x-1).

a) Show that
$\int_{0}^{2}f(x)e^{-x} \mathrm{d}x= 0$.

b) Prove that area delimited by the graph of f, the Ox, axis and the lines of equations x = 1 and x = 2 has area equal with e.

## RE: Math Analysis issue

Steve McLoud
Visitor
2016-08-25 07:58:52
#2
$\int_{0}^{2}f(x)e^{-x}dx = \int_{0}^{2} e^{x}(x-1)e^{-x}dx =$

$= \int_{0}^{2}(x-1)dx = \left.\begin{matrix} \left ( \frac{x^{2}}{2}-x \right ) \end{matrix}\right|_{0}^{2} = (\frac{4}{2}-2)-0 = 2-2=0$

## RE: Math Analysis issue

SteveC
Member
2016-09-08 14:55:59
#3
For the second issue we have:
$\textit{S} = \int_{1}^{2}|f(x)|dx = \int_{1}^{2}(x-1)e^{x}dx=$

and we use part integration:
$= (x-1)e^{x}|_{0}^{2}-\int_{1}^{2}e^{x}dx = (x-1)e^{x}|_{1}^{2}-e^{x}|_{1}^{2} = (x-2)e^{x}|_{1}^{2} =$

## RE: Math Analysis issue

SteveC
Member
2016-09-08 14:59:46
#4
$=0-(-1)e = e$