Difference between revisions of "022 Exam 2 Sample A, Problem 6"

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(Created page with "<span class="exam">Find the area under the curve of  <math style="vertical-align: -60%">y\,=\,\frac{8}{\sqrt{x}}</math>  between <math style="vertical-align: -5%...")
 
 
(6 intermediate revisions by the same user not shown)
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!Foundations: &nbsp;  
 
!Foundations: &nbsp;  
 
|-
 
|-
|This problem requires two rules of integration.  In particular, you need
+
|For solving the problem, we only require the use of the power rule for integration:
|-
 
|'''Integration by substitution (U - sub):''' If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -15%;">g</math> are differentiable functions, then
 
 
|-
 
|-
 +
|
 +
::<math style="vertical-align: -70%;">\int x^n\,dx\,=\,\frac{x^{n+1}}{n+1} +C,</math>&thinsp; for <math style="vertical-align: -25%;">n\neq -1.</math>
  
|<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>(f\circ g)'(x) = f'(g(x))\cdot g'(x).</math>
 
|-
 
|<br>'''The Product Rule:'''  If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -15%;">g</math> are differentiable functions, then
 
|-
 
|<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>(fg)'(x) = f'(x)\cdot g(x)+f(x)\cdot g'(x).</math>
 
|-
 
|<br>'''The Quotient Rule:'''  If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -15%;">g</math> are differentiable functions and <math style="vertical-align: -21%;">g(x) \neq 0</math>&thinsp;, then
 
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|<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>\left(\frac{f}{g}\right)'(x) = \frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{\left(g(x)\right)^2}. </math>
 
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|Additionally, we will need our power rule for differentiation:
 
|-
 
|
 
::<math style="vertical-align: -21%;">\left(x^n\right)'\,=\,nx^{n-1},</math> for <math style="vertical-align: -25%;">n\neq 0</math>,
 
|-
 
|as well as the derivative of natural log:
 
 
|-
 
|-
|
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|Geometrically, we need to integrate the region between the <math style="vertical-align: 0%">x</math>-axis, the curve, and the vertical lines <math style="vertical-align: -4%">x = 1</math> and <math style="vertical-align: -2%">x = 4</math>.
::<math>\left(\ln x\right)'\,=\,\frac{1}{x}.</math>
 
|<br>
 
 
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|Set up the integral:
 
|Set up the integral:
 
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|-
|<math>\int_1^4 \frac{8}{\sqrt{x}} dx</math>
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|
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::<math>\int_1^{\,4} \frac{8}{\sqrt{x}} \,dx.</math>
 
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|-
 
|-
 
|
 
|
::<math>
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::<math>\begin{array}{rcl}
\begin{array}{rcl}
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\displaystyle{\int_1^{\,4} \frac{8}{\sqrt{x}}\,dx} & = & \displaystyle {\int_1^{\,4} 8x^{-1/2}\,dx}\\
\int_1^4 \frac{8}{\sqrt{x}}dx & = & \frac{x^{1/2}}
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\\
 +
& = & \displaystyle{\frac{8 x^{1/2}}{1/2} \Bigr|_{x=1}^4}\\
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\\
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& = & 16x^{1/2}  \Bigr|_{x=1}^4.
 
\end{array}</math>
 
\end{array}</math>
 
|}
 
|}
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!Step 3: &nbsp;
 
!Step 3: &nbsp;
 
|-
 
|-
| Now we need to substitute back into our original variables using our original substitution <math>u = 3x + 2</math>
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| Now we need to evaluate to get:
 
|-
 
|-
| to get <math>\frac{\log(u)}{3} = \frac{\log(3x + 2}{3}</math>
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|
|}
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::<math>16x^{1/2}  \Bigr|_{x=1}^4\,=\,16\cdot 4^{1/2} - 16\cdot 1^{1/2} \,=\, 32 - 16 \,=\, 16.</math>
 
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 4: &nbsp;
 
|-
 
| Since this integral is an indefinite integral we have to remember to add ''' "+ C" ''' at the end.
 
 
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!Final Answer: &nbsp;
 
!Final Answer: &nbsp;
 
|-
 
|-
|<math>\int \frac{1}{3x + 2} dx = \frac{\ln(3x + 2)}{3} + C</math>
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|
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::<math>\int_1^{\,4} \frac{8}{\sqrt{x}} \,dx\,=\,16.</math>
 
|}
 
|}
  
 
[[022_Exam_2_Sample_A|'''<u>Return to Sample Exam</u>''']]
 
[[022_Exam_2_Sample_A|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 07:54, 16 May 2015

Find the area under the curve of    between and .

Foundations:  
For solving the problem, we only require the use of the power rule for integration:
  for
Geometrically, we need to integrate the region between the -axis, the curve, and the vertical lines and .

 Solution:

Step 1:  
Set up the integral:
Step 2:  
Using the power rule we have:
Step 3:  
Now we need to evaluate to get:
Final Answer:  

Return to Sample Exam