# 009B Sample Final 1, Problem 6

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Evaluate the improper integrals:

a) ${\displaystyle \int _{0}^{\infty }xe^{-x}~dx}$
b) ${\displaystyle \int _{1}^{4}{\frac {dx}{\sqrt {4-x}}}}$
Foundations:
1. How could you write ${\displaystyle \int _{0}^{\infty }f(x)~dx}$ so that you can integrate?
You can write ${\displaystyle \int _{0}^{\infty }f(x)~dx=\lim _{a\rightarrow \infty }\int _{0}^{a}f(x)~dx.}$
2. How could you write ${\displaystyle \int _{-1}^{1}{\frac {1}{x}}~dx}$ ?
The problem is that  ${\displaystyle {\frac {1}{x}}}$  is not continuous at ${\displaystyle x=0}$.
So, you can write ${\displaystyle \int _{-1}^{1}{\frac {1}{x}}~dx=\lim _{a\rightarrow 0^{-}}\int _{-1}^{a}{\frac {1}{x}}~dx+\lim _{a\rightarrow 0^{+}}\int _{a}^{1}{\frac {1}{x}}~dx}$.
3. How would you integrate ${\displaystyle \int xe^{x}\,dx}$ ?
You can use integration by parts.
Let ${\displaystyle u=x}$ and ${\displaystyle dv=e^{x}dx}$.

Solution:

(a)

Step 1:
First, we write ${\displaystyle \int _{0}^{\infty }xe^{-x}~dx=\lim _{a\rightarrow \infty }\int _{0}^{a}xe^{-x}~dx}$.
Now, we proceed using integration by parts. Let ${\displaystyle u=x}$ and ${\displaystyle dv=e^{-x}dx}$. Then, ${\displaystyle du=dx}$ and ${\displaystyle v=-e^{-x}}$.
Thus, the integral becomes
${\displaystyle \int _{0}^{\infty }xe^{-x}~dx=\lim _{a\rightarrow \infty }\left.-xe^{-x}\right|_{0}^{a}-\int _{0}^{a}-e^{-x}\,dx.}$
Step 2:
For the remaining integral, we need to use ${\displaystyle u}$-substitution. Let ${\displaystyle u=-x}$. Then, ${\displaystyle du=-dx}$.
Since the integral is a definite integral, we need to change the bounds of integration.
Plugging in our values into the equation ${\displaystyle u=-x}$, we get ${\displaystyle u_{1}=0}$ and ${\displaystyle u_{2}=-a}$.
Thus, the integral becomes
${\displaystyle {\begin{array}{rcl}\displaystyle {\int _{0}^{\infty }xe^{-x}~dx}&=&\displaystyle {\lim _{a\rightarrow \infty }-xe^{-x}{\bigg |}_{0}^{a}-\int _{0}^{-a}e^{u}~du}\\&&\\&=&\displaystyle {\lim _{a\rightarrow \infty }-xe^{-x}{\bigg |}_{0}^{a}-e^{u}{\bigg |}_{0}^{-a}}\\&&\\&=&\displaystyle {\lim _{a\rightarrow \infty }-ae^{-a}-(e^{-a}-1)}.\\\end{array}}}$
Step 3:
Now, we evaluate to get
${\displaystyle {\begin{array}{rcl}\displaystyle {\int _{0}^{\infty }xe^{-x}~dx}&=&\displaystyle {\lim _{a\rightarrow \infty }-ae^{-a}-(e^{-a}-1)}\\&&\\&=&\displaystyle {\lim _{a\rightarrow \infty }{\frac {-a}{e^{a}}}-{\frac {1}{e^{a}}}+1}\\&&\\&=&\displaystyle {\lim _{a\rightarrow \infty }{\frac {-a-1}{e^{a}}}+1}.\\\end{array}}}$
Using L'Hôpital's Rule, we get
${\displaystyle {\begin{array}{rcl}\displaystyle {\int _{0}^{\infty }xe^{-x}~dx}&=&\displaystyle {\lim _{a\rightarrow \infty }{\frac {-1}{e^{a}}}+1}\\&&\\&=&\displaystyle {0+1}\\&&\\&=&\displaystyle {1}.\\\end{array}}}$

(b)

Step 1:
First, we write ${\displaystyle \int _{1}^{4}{\frac {dx}{\sqrt {4-x}}}=\lim _{a\rightarrow 4}\int _{1}^{a}{\frac {dx}{\sqrt {4-x}}}}$.
Now, we proceed by ${\displaystyle u}$-substitution. We let ${\displaystyle u=4-x}$. Then, ${\displaystyle du=-dx}$.
Since the integral is a definite integral, we need to change the bounds of integration.
Plugging in our values into the equation ${\displaystyle u=4-x}$, we get ${\displaystyle u_{1}=4-1=3}$  and ${\displaystyle u_{2}=4-a}$.
Thus, the integral becomes
${\displaystyle \int _{1}^{4}{\frac {dx}{\sqrt {4-x}}}\,=\,\lim _{a\rightarrow 4}\int _{3}^{4-a}{\frac {-1}{\sqrt {u}}}~du.}$
Step 2:
We integrate to get
${\displaystyle {\begin{array}{rcl}\displaystyle {\int _{1}^{4}{\frac {dx}{\sqrt {4-x}}}}&=&\displaystyle {\lim _{a\rightarrow 4}-2u^{\frac {1}{2}}{\bigg |}_{3}^{4-a}}\\&&\\&=&\displaystyle {\lim _{a\rightarrow 4}-2{\sqrt {4-a}}+2{\sqrt {3}}}\\&&\\&=&\displaystyle {2{\sqrt {3}}}.\\\end{array}}}$
(a)  ${\displaystyle 1}$
(b)  ${\displaystyle 2{\sqrt {3}}}$