009B Sample Final 1, Problem 6

Evaluate the improper integrals:

a) $\int _{0}^{\infty }xe^{-x}~dx$

b) $\int _{1}^{4}{\frac {dx}{\sqrt {4-x}}}$

Foundations:
Review integration by parts

Solution:

(a)

Step 1:
First, we write $\int _{0}^{\infty }xe^{-x}~dx=\lim _{a\rightarrow \infty }\int _{0}^{a}xe^{-x}~dx$.
Now, we proceed using integration by parts. Let $u=x$ and $dv=e^{-x}dx$. Then, $du=dx$ and $v=-e^{-x}$.
Thus, the integral becomes
$\int _{0}^{\infty }xe^{-x}~dx=\lim _{a\rightarrow \infty }\left.-xe^{-x}\right|_{0}^{a}-\int _{0}^{a}-e^{-x}dx$

Step 2:
For the remaining integral, we need to use $u$-substitution. Let $u=-x$. Then, $du=-dx$.
Since the integral is a definite integral, we need to change the bounds of integration.
Plugging in our values into the equation $u=-x$, we get $u_{1}=0$ and $u_{2}=-a$.
Thus, the integral becomes
$\int _{0}^{\infty }xe^{-x}~dx=\lim _{a\rightarrow \infty }\left.-xe^{-x}\right|_{0}^{a}-\int _{0}^{-a}e^{u}du=\lim _{a\rightarrow \infty }\left.-xe^{-x}\right|_{0}^{a}-\left.e^{u}\right|_{0}^{-a}=\lim _{a\rightarrow \infty }-ae^{-a}-(e^{-a}-1)$

Step 3:
Now, we evaluate to get
$\int _{0}^{\infty }xe^{-x}~dx=\lim _{a\rightarrow \infty }-ae^{-a}-(e^{-a}-1)=\lim _{a\rightarrow \infty }{\frac {-a}{e^{a}}}-{\frac {1}{e^{a}}}+1=\lim _{a\rightarrow \infty }{\frac {-a-1}{e^{a}}}+1$.
Using L'Hopital's Rule, we get
$\int _{0}^{\infty }xe^{-x}~dx=\lim _{a\rightarrow \infty }{\frac {-1}{e^{a}}}+1=0+1=1$.

(b)

Step 1:

Step 2:

Step 3:

Final Answer:
(a) $1$
(b)

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