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

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
${\begin{array}{rcl}\displaystyle {\int _{0}^{\infty }xe^{-x}~dx}&=&\displaystyle {\lim _{a\rightarrow \infty }-xe^{-x}{\bigg |}_{0}^{a}-\int _{0}^{-a}e^{u}~du}\\&&\\&=&\displaystyle {\lim _{a\rightarrow \infty }-xe^{-x}{\bigg |}_{0}^{a}-e^{u}{\bigg |}_{0}^{-a}}\\&&\\&=&\displaystyle {\lim _{a\rightarrow \infty }-ae^{-a}-(e^{-a}-1)}.\\\end{array}}$

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

3

(b)

Step 1:
First, we write $\int _{1}^{4}{\frac {dx}{\sqrt {4-x}}}=\lim _{a\rightarrow 4}\int _{1}^{a}{\frac {dx}{\sqrt {4-x}}}$.
Now, we proceed by $u$-substitution. We let $u=4-x$. 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 $u=4-x$, we get $u_{1}=4-1=3$  and $u_{2}=4-a$.
Thus, the integral becomes
$\int _{1}^{4}{\frac {dx}{\sqrt {4-x}}}\,=\,\lim _{a\rightarrow 4}\int _{3}^{4-a}{\frac {-1}{\sqrt {u}}}~du.$

Step 2:
We integrate to get
${\begin{array}{rcl}\displaystyle {\int _{1}^{4}{\frac {dx}{\sqrt {4-x}}}}&=&\displaystyle {\lim _{a\rightarrow 4}-2u^{\frac {1}{2}}{\bigg |}_{3}^{4-a}}\\&&\\&=&\displaystyle {\lim _{a\rightarrow 4}-2{\sqrt {4-a}}+2{\sqrt {3}}}\\&&\\&=&\displaystyle {2{\sqrt {3}}}.\\\end{array}}$

4

Final Answer:
(a)  $1$
(b)  $2{\sqrt {3}}$

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