|
|
| Line 5: |
Line 5: |
| | <span class="exam">(b) <math> \int_0^1 \frac{3\ln x}{\sqrt{x}}~dx</math> | | <span class="exam">(b) <math> \int_0^1 \frac{3\ln x}{\sqrt{x}}~dx</math> |
| | | | |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009B Sample Final 2, Problem 7 Solution|'''<u>Solution</u>''']] |
| − | |-
| |
| − | |'''1.''' How could you write <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>
| |
| − | |-
| |
| − | |'''2.''' How could you write <math>\int_{0}^1 \frac{1}{x}~dx?</math>
| |
| − | |-
| |
| − | |
| |
| − | The problem is that <math>\frac{1}{x}</math> is not continuous at <math style="vertical-align: 0px">x=0.</math>
| |
| − | |-
| |
| − | |
| |
| − | So, you can write <math style="vertical-align: -15px">\int_{0}^1 \frac{1}{x}~dx=\lim_{a\rightarrow 0} \int_{a}^1 \frac{1}{x}~dx.</math>
| |
| − | |}
| |
| | | | |
| | | | |
| − | '''Solution:''' | + | [[009B Sample Final 2, Problem 7 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| | | | |
| − | '''(a)'''
| |
| | | | |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 1:
| |
| − | |-
| |
| − | |First, we write
| |
| − | |-
| |
| − | | <math>\int_1^\infty \frac{\ln x}{x^4}~dx=\lim_{a\rightarrow \infty} \int_1^a \frac{\ln x}{x^4}~dx.</math>
| |
| − | |-
| |
| − | |Now, we use integration by parts.
| |
| − | |-
| |
| − | |Let <math style="vertical-align: -2px">u=\ln x</math> and <math style="vertical-align: -13px">dv=\frac{1}{x^4}dx.</math>
| |
| − | |-
| |
| − | |Then, <math style="vertical-align: -13px">du=\frac{1}{x}dx</math> and <math style="vertical-align: -13px">v=\frac{1}{-3x^3}.</math>
| |
| − | |-
| |
| − | |Using integration by parts, we get
| |
| − | |-
| |
| − | | <math>\begin{array}{rcl}
| |
| − | \displaystyle{\int_1^\infty \frac{\ln x}{x^4}~dx} & = & \displaystyle{\lim_{a\rightarrow \infty} \frac{\ln x}{-3x^3}\bigg|_1^a+\int_1^a \frac{1}{3x^4}~dx}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{\lim_{a\rightarrow \infty} \frac{\ln x}{-3x^3}-\frac{1}{9x^3}\bigg|_1^a.}
| |
| − | \end{array}</math>
| |
| − | |}
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 2:
| |
| − | |-
| |
| − | |Now, using L'Hopital's Rule, we get
| |
| − | |-
| |
| − | | <math>\begin{array}{rcl}
| |
| − | \displaystyle{\int_1^\infty \frac{\ln x}{x^4}~dx} & = & \displaystyle{\lim_{a\rightarrow \infty} \frac{\ln a}{-3a^3}-\frac{1}{9a^3}-\bigg(\frac{\ln 1}{-3}-\frac{1}{9}\bigg)}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{\lim_{a\rightarrow \infty} \frac{\ln(a)}{-3a^3}+0+0+\frac{1}{9}}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\ln(x)}{-3x^3}+\frac{1}{9}}\\
| |
| − | &&\\
| |
| − | & \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{1}{x}}{-9x^2}+\frac{1}{9}}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{\frac{1}{9}.}
| |
| − | \end{array}</math>
| |
| − | |}
| |
| − |
| |
| − | '''(b)'''
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 1:
| |
| − | |-
| |
| − | |First, we write
| |
| − | |-
| |
| − | | <math>\int_0^1 \frac{3\ln x}{\sqrt{x}}~dx=\lim_{a\rightarrow 0^+} \int_a^1 \frac{3\ln x}{\sqrt{x}}~dx.</math>
| |
| − | |-
| |
| − | |Now, we use integration by parts.
| |
| − | |-
| |
| − | |Let <math style="vertical-align: -2px">u=3\ln x</math> and <math style="vertical-align: -19px">dv=\frac{1}{\sqrt{x}}dx.</math>
| |
| − | |-
| |
| − | |Then, <math style="vertical-align: -13px">du=\frac{3}{x}dx</math> and <math style="vertical-align: -5px">v=2\sqrt{x}.</math>
| |
| − | |-
| |
| − | |Using integration by parts, we get
| |
| − | |-
| |
| − | | <math>\begin{array}{rcl}
| |
| − | \displaystyle{\int_0^1 \frac{3\ln x}{\sqrt{x}}~dx} & = & \displaystyle{\lim_{a\rightarrow 0^+} (3\ln x)(2\sqrt{x})\bigg|_a^1-\int_a^1 \frac{6}{\sqrt{x}}~dx}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{\lim_{a\rightarrow 0^+} 6\sqrt{x}\ln(x)-12\sqrt{x}\bigg|_a^1.}
| |
| − | \end{array}</math>
| |
| − | |}
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 2:
| |
| − | |-
| |
| − | |Now, using L'Hopital's Rule, we get
| |
| − | |-
| |
| − | | <math>\begin{array}{rcl}
| |
| − | \displaystyle{\int_0^1 \frac{3\ln x}{\sqrt{x}}~dx} & = & \displaystyle{\lim_{a\rightarrow 0^+} (6\sqrt{1}\ln(1)-12\sqrt{1})-(6\sqrt{a}\ln(a)-12\sqrt{a})}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{\lim_{a\rightarrow 0^+} -12 -6\sqrt{a}\ln(a) +12\sqrt{a}}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{\lim_{a\rightarrow 0^+} -12 -6\sqrt{a}\ln(a)+0}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{\lim_{x\rightarrow 0^+} -12-6\sqrt{x}\ln(x)}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{-12-\lim_{x\rightarrow 0^+} \frac{6\ln(x)}{\frac{1}{\sqrt{x}}}}\\
| |
| − | &&\\
| |
| − | & \overset{L'H}{=} & \displaystyle{-12-\lim_{x\rightarrow 0^+} \frac{\frac{6}{x}}{-\frac{1}{2x^{3/2}}}}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{-12+\lim_{x\rightarrow 0^+} 12\sqrt{x}}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{-12.}
| |
| − | \end{array}</math>
| |
| − | |-
| |
| − | |
| |
| − | |}
| |
| − |
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Final Answer:
| |
| − | |-
| |
| − | | '''(a)''' <math>\frac{1}{9}</math>
| |
| − | |-
| |
| − | | '''(b)''' <math>-12</math>
| |
| − | |}
| |
| | [[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] | | [[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] |
