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

Revision as of 16:36, 2 February 2016

Evaluate the improper integrals:

a)

\int _{0}^{\infty }xe^{-x}~dx

b)

\int _{1}^{4}{\frac {dx}{\sqrt {4-x}}}

Foundations:
Review integration by parts

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
$\int _{0}^{\infty }xe^{-x}~dx=\lim _{a\rightarrow \infty }\left.-xe^{-x}\right\|_{0}^{a}-\int _{0}^{-a}e^{u}du=\lim _{a\rightarrow \infty }\left.-xe^{-x}\right\|_{0}^{a}-\left.e^{u}\right\|_{0}^{-a}=\lim _{a\rightarrow \infty }-ae^{-a}-(e^{-a}-1)$

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

(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
$\int _{1}^{4}{\frac {dx}{\sqrt {4-x}}}=\lim _{a\rightarrow 4}\int _{3}^{4-a}{\frac {-du}{\sqrt {u}}}$ .

Step 2:
We integrate to get
$\int _{1}^{4}{\frac {dx}{\sqrt {4-x}}}=\lim _{a\rightarrow 4}\left.-2u^{\frac {1}{2}}\right\|_{3}^{4-a}=\lim _{a\rightarrow 4}-2{\sqrt {4-a}}+2{\sqrt {3}}=2{\sqrt {3}}$

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

@@ Line 59: / Line 59: @@
 !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>.
 |-
-|
+|Now, we proceed by <math>u</math>-substitution. We let <math>u=4-x</math>. Then, <math>du=-dx</math>.
 |-
-|
+|Since the integral is a definite integral, we need to change the bounds of integration.
+|-
+|Plugging in our values into the equation <math>u=4-x</math>, we get <math>u_1=4-1=3</math> and <math>u_2=4-a</math>.
+|-
+|Thus, the integral becomes
+|-
+|<math>\int_1^4 \frac{dx}{\sqrt{4-x}}=\lim_{a\rightarrow 4} \int_3^{4-a}\frac{-du}{\sqrt{u}}</math>.
 |}
@@ Line 69: / Line 75: @@
 !Step 2: &nbsp;
 |-
-|
+|We integrate to get
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 3: &nbsp;
-|-
-|
 |-
-|
+|<math>\int_1^4 \frac{dx}{\sqrt{4-x}}=\lim_{a\rightarrow 4} \left.-2u^{\frac{1}{2}}\right|_{3}^{4-a}=\lim_{a\rightarrow 4}-2\sqrt{4-a}+2\sqrt{3}=2\sqrt{3}</math>
 |}
@@ Line 85: / Line 85: @@
 |'''(a)''' <math>1</math>
 |-
-|'''(b)'''
+|'''(b)''' <math>2\sqrt{3}</math>
 |}
 [[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]