Difference between revisions of "009B Sample Final 1, Problem 6"

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<span class="exam"> Evaluate the improper integrals:
 
<span class="exam"> Evaluate the improper integrals:
  
−
<span class="exam">(a) <math>\int_0^{\infty} xe^{-x}~dx</math>
+
<span class="exam">(a) &nbsp; <math>\int_0^{\infty} xe^{-x}~dx</math>
  
−
<span class="exam">(b) <math>\int_1^4 \frac{dx}{\sqrt{4-x}}</math>
+
<span class="exam">(b) &nbsp; <math>\int_1^4 \frac{dx}{\sqrt{4-x}}</math>
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Foundations: &nbsp;  
 
!Foundations: &nbsp;  
 
|-
 
|-
−
|'''1.''' How could you write <math style="vertical-align: -14px">\int_0^{\infty} f(x)~dx</math> so that you can integrate?
+
|'''1.''' How could you write &nbsp; <math style="vertical-align: -14px">\int_0^{\infty} f(x)~dx</math> so that you can integrate?
 
|-
 
|-
 
|  
 
|  
−
::You can write <math>\int_0^{\infty} f(x)~dx=\lim_{a\rightarrow\infty} \int_0^a f(x)~dx.</math>
+
&nbsp; &nbsp; &nbsp; &nbsp; You can write &nbsp; <math>\int_0^{\infty} f(x)~dx=\lim_{a\rightarrow\infty} \int_0^a f(x)~dx.</math>
 
|-
 
|-
−
|'''2.''' How could you write <math>\int_{-1}^1 \frac{1}{x}~dx</math> ?
+
|'''2.''' How could you write &nbsp; <math>\int_{-1}^1 \frac{1}{x}~dx?</math>
 
|-
 
|-
 
|
 
|
−
::The problem is that&thinsp; <math>\frac{1}{x}</math> &thinsp;is not continuous at <math style="vertical-align: 0px">x=0</math>.
+
&nbsp; &nbsp; &nbsp; &nbsp; The problem is that &nbsp;<math>\frac{1}{x}</math>&nbsp; is not continuous at &nbsp;<math style="vertical-align: 0px">x=0.</math>
 
|-
 
|-
 
|
 
|
−
::So, you can write <math style="vertical-align: -15px">\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</math>.
+
&nbsp; &nbsp; &nbsp; &nbsp; So, you can write &nbsp;<math style="vertical-align: -15px">\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.</math>
 
|-
 
|-
−
|'''3.''' How would you integrate <math style="vertical-align: -13px">\int xe^x\,dx</math> ?
+
|'''3.''' How would you integrate &nbsp;<math style="vertical-align: -13px">\int xe^x\,dx?</math>
 
|-
 
|-
 
|
 
|
−
::You can use integration by parts.  
+
&nbsp; &nbsp; &nbsp; &nbsp;You can use integration by parts.  
 
|-
 
|-
 
|
 
|
−
::Let <math style="vertical-align: 0px">u=x</math> and <math style="vertical-align: 0px">dv=e^xdx</math>.
+
&nbsp; &nbsp; &nbsp; &nbsp;Let &nbsp;<math style="vertical-align: 0px">u=x</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">dv=e^xdx.</math>
 
|}
 
|}
  
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
−
|First, we write <math style="vertical-align: -14px">\int_0^{\infty} xe^{-x}~dx=\lim_{a\rightarrow \infty} \int_0^a xe^{-x}~dx</math>.
+
|First, we write &nbsp;<math style="vertical-align: -14px">\int_0^{\infty} xe^{-x}~dx=\lim_{a\rightarrow \infty} \int_0^a xe^{-x}~dx.</math>
 
|-
 
|-
−
|Now, we proceed using integration by parts. Let <math style="vertical-align: 0px">u=x</math> and <math style="vertical-align: 0px">dv=e^{-x}dx</math>. Then, <math style="vertical-align: 0px">du=dx</math> and <math style="vertical-align: 0px">v=-e^{-x}</math>.
+
|Now, we proceed using integration by parts.  
 +
|-
 +
|Let &nbsp;<math style="vertical-align: 0px">u=x</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">dv=e^{-x}dx.</math>
 +
|-
 +
|Then, &nbsp;<math style="vertical-align: 0px">du=dx</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">v=-e^{-x}.</math>
 
|-
 
|-
 
|Thus, the integral becomes
 
|Thus, the integral becomes
 
|-
 
|-
 
|
 
|
−
::<math>\int_0^{\infty} xe^{-x}~dx=\lim_{a\rightarrow \infty} \left.-xe^{-x}\right|_0^a-\int_0^a-e^{-x}\,dx.</math>
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>\int_0^{\infty} xe^{-x}~dx=\lim_{a\rightarrow \infty} \left.-xe^{-x}\right|_0^a-\int_0^a-e^{-x}\,dx.</math>
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
−
|For the remaining integral, we need to use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: 0px">u=-x</math>. Then, <math style="vertical-align: 0px">du=-dx</math>.
+
|For the remaining integral, we need to use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.  
 +
|-
 +
|Let &nbsp;<math style="vertical-align: 0px">u=-x.</math>&nbsp; Then, &nbsp;<math style="vertical-align: 0px">du=-dx.</math>
 
|-
 
|-
 
|Since the integral is a definite integral, we need to change the bounds of integration.
 
|Since the integral is a definite integral, we need to change the bounds of integration.
 
|-
 
|-
−
|Plugging in our values into the equation <math style="vertical-align: 0px">u=-x</math>, we get <math style="vertical-align: -5px">u_1=0</math> and <math style="vertical-align: -3px">u_2=-a</math>.
+
|Plugging in our values into the equation &nbsp;<math style="vertical-align: -4px">u=-x,</math>&nbsp; we get
 +
|-
 +
|<math style="vertical-align: -5px">u_1=0</math>&nbsp; and &nbsp;<math style="vertical-align: -3px">u_2=-a.</math>
 
|-
 
|-
 
|Thus, the integral becomes
 
|Thus, the integral becomes
 
|-
 
|-
 
|
 
|
−
::<math>\begin{array}{rcl}
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>\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{\int_0^{\infty} xe^{-x}~dx} & = & \displaystyle{\lim_{a\rightarrow \infty} -xe^{-x}\bigg|_0^a-\int_0^{-a}e^{u}~du}\\
 
&&\\
 
&&\\
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|-
 
|-
 
|
 
|
−
::<math>\begin{array}{rcl}
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 
\displaystyle{\int_0^{\infty} xe^{-x}~dx} & = & \displaystyle{\lim_{a\rightarrow \infty} -ae^{-a}-(e^{-a}-1)}\\
 
\displaystyle{\int_0^{\infty} xe^{-x}~dx} & = & \displaystyle{\lim_{a\rightarrow \infty} -ae^{-a}-(e^{-a}-1)}\\
 
&&\\
 
&&\\
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|-
 
|-
 
|
 
|
−
::<math>\begin{array}{rcl}
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 
\displaystyle{\int_0^{\infty} xe^{-x}~dx} & = & \displaystyle{\lim_{a\rightarrow \infty} \frac{-1}{e^a}+1}\\
 
\displaystyle{\int_0^{\infty} xe^{-x}~dx} & = & \displaystyle{\lim_{a\rightarrow \infty} \frac{-1}{e^a}+1}\\
 
&&\\
 
&&\\
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
−
|First, we write <math>\int_1^4 \frac{dx}{\sqrt{4-x}}=\lim_{a\rightarrow 4} \int_1^a\frac{dx}{\sqrt{4-x}}</math>.
+
|First, we write &nbsp;<math>\int_1^4 \frac{dx}{\sqrt{4-x}}=\lim_{a\rightarrow 4} \int_1^a\frac{dx}{\sqrt{4-x}}.</math>
 
|-
 
|-
−
|Now, we proceed by <math style="vertical-align: 0px">u</math>-substitution. We let <math style="vertical-align: -1px">u=4-x</math>. Then, <math style="vertical-align: 0px">du=-dx</math>.
+
|Now, we proceed by &nbsp;<math style="vertical-align: 0px">u</math>-substitution.  
 +
|-
 +
|We let &nbsp;<math style="vertical-align: -1px">u=4-x.</math>&nbsp; Then, &nbsp;<math style="vertical-align: 0px">du=-dx.</math>
 
|-
 
|-
 
|Since the integral is a definite integral, we need to change the bounds of integration.  
 
|Since the integral is a definite integral, we need to change the bounds of integration.  
 
|-
 
|-
−
|Plugging in our values into the equation <math style="vertical-align: -1px">u=4-x</math>, we get <math style="vertical-align: -5px">u_1=4-1=3</math>&thinsp; and <math style="vertical-align: -3px">u_2=4-a</math>.
+
|Plugging in our values into the equation &nbsp;<math style="vertical-align: -4px">u=4-x,</math>&nbsp; we get  
 +
|-
 +
|<math style="vertical-align: -5px">u_1=4-1=3</math>&nbsp; and &nbsp;<math style="vertical-align: -3px">u_2=4-a.</math>
 
|-
 
|-
 
|Thus, the integral becomes
 
|Thus, the integral becomes
 
|-
 
|-
 
|
 
|
−
::<math>\int_1^4 \frac{dx}{\sqrt{4-x}}\,=\,\lim_{a\rightarrow 4} \int_3^{4-a}\frac{-1}{\sqrt{u}}~du.</math>
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>\int_1^4 \frac{dx}{\sqrt{4-x}}\,=\,\lim_{a\rightarrow 4} \int_3^{4-a}\frac{-1}{\sqrt{u}}~du.</math>
 
|}
 
|}
  
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|-
 
|-
 
|
 
|
−
::<math>\begin{array}{rcl}
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>\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{\int_1^4 \frac{dx}{\sqrt{4-x}}} & = & \displaystyle{\lim_{a\rightarrow 4} -2u^{\frac{1}{2}}\bigg|_{3}^{4-a}}\\
 
&&\\
 
&&\\
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
−
|'''(a)''' &nbsp;<math style="vertical-align: -3px">1</math>
+
|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp;<math style="vertical-align: -3px">1</math>
 
|-
 
|-
−
|'''(b)''' &nbsp;<math style="vertical-align: -4px">2\sqrt{3}</math>
+
|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp;<math style="vertical-align: -4px">2\sqrt{3}</math>
 
|}
 
|}
 
[[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Revision as of 09:33, 27 February 2017

Evaluate the improper integrals:

(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 \int_0^{\infty} xe^{-x}~dx}

(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 \int_1^4 \frac{dx}{\sqrt{4-x}}}

Foundations:  
1. How could you 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} f(x)~dx} so that you can integrate?

        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_0^{\infty} f(x)~dx=\lim_{a\rightarrow\infty} \int_0^a f(x)~dx.}

2. How could you 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?}

        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  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=4-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=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}}


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}}

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