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

From Math Wiki
Jump to navigation Jump to search
 
(5 intermediate revisions by the same user not shown)
Line 4: Line 4:
 
!Foundations:    
 
!Foundations:    
 
|-
 
|-
−
|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
 
−
|-
 
−
|<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>
 
−
|-
 
−
|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:
 
 
|-
 
|-
−
|
+
|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>
 
 
|}
 
|}
  
Line 38: Line 20:
 
|Set up the integral:
 
|Set up the integral:
 
|-
 
|-
−
|<math>\int_1^4 \frac{8}{\sqrt{x}} dx</math>
+
|
 +
::<math>\int_1^{\,4} \frac{8}{\sqrt{x}} \,dx.</math>
 
|}
 
|}
  
Line 48: Line 31:
 
|
 
|
 
::<math>\begin{array}{rcl}
 
::<math>\begin{array}{rcl}
−
\int_1^4 \frac{8}{\sqrt{x}}dx & = & \frac{8 x^{1/2}}{2} \vert_1^4\\
+
\displaystyle{\int_1^{\,4} \frac{8}{\sqrt{x}}\,dx} & = & \displaystyle {\int_1^{\,4} 8x^{-1/2}\,dx}\\
−
& = & 4x^{1/2} \vert_1^4
+
\\
 +
& = & \displaystyle{\frac{8 x^{1/2}}{1/2} \Bigr|_{x=1}^4}\\
 +
\\
 +
& = & 16x^{1/2} \Bigr|_{x=1}^4.
 
\end{array}</math>
 
\end{array}</math>
 
|}
 
|}
Line 58: Line 44:
 
| Now we need to evaluate to get:
 
| Now we need to evaluate to get:
 
|-
 
|-
−
| <math>4\cdot 4^{1/2} - 4\cdot 1^{1/2} = 8 - 4 = 4</math>  
+
|
 +
::<math>16x^{1/2}  \Bigr|_{x=1}^4\,=\,16\cdot 4^{1/2} - 16\cdot 1^{1/2} \,=\, 32 - 16 \,=\, 16.</math>  
 
|}
 
|}
  
Line 64: Line 51:
 
!Final Answer: &nbsp;
 
!Final Answer: &nbsp;
 
|-
 
|-
−
|4
+
|
 +
::<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 06:54, 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

Retrieved from "https://wiki.math.ucr.edu/index.php?title=022_Exam_2_Sample_A,_Problem_6&oldid=571"