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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_1=4-1=3} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_2=4-a} .
Thus, the integral becomes
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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}}

4

Final Answer:
(a) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1}
(b) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2\sqrt{3}}

Return to Sample Exam