009B Sample Final 1, Problem 6 Detailed Solution

Evaluate the improper integrals:

(a)   $\int _{0}^{\infty }xe^{-x}~dx$

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

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Background Information:
1. How could you write   Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int _{0}^{\infty }f(x)~dx} so that you can integrate?
You can write   Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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  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 \frac{1}{x}}   is not continuous at  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 x=0.}
So, you can write  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}^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  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 xe^x\,dx?}
You can use integration by parts.
Let  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=x}   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 dv=e^xdx.}

Solution:

(a)

Step 1:
First, we write  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_0^{\infty} xe^{-x}~dx=\lim_{a\rightarrow \infty} \int_0^a xe^{-x}~dx.}
Now, we proceed using integration by parts.
Let  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=x}   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 dv=e^{-x}dx.}
Then,  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 du=dx}   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 v=-e^{-x}.}
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_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  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} -substitution.
Let  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=-x.}   Then,  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 du=-dx.}
Since the integral is a definite integral, we need to change the bounds of integration.
Plugging in our values into the equation  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=-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=0}   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=-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 \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
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_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
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_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  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_1^a\frac{dx}{\sqrt{4-x}}.}
Now, we proceed by  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} -substitution.
We let  $u=4-x.$  Then,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=4-x,}   we get
$u_{1}=4-1=3$  and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u_{2}=4-a.}
Thus, the integral becomes
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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}}}

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

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