Difference between revisions of "022 Exam 2 Sample B, Problem 7"

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(Created page with "<span class="exam">Find the antiderivatives: ::<span class="exam">(a) <math> \int xe^{3x^2+1}\,dx.</math> <br> ::<span class="exam">(b) <math>\int_2^54x - 5\,dx.</math> {| c...")
 
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!Foundations: &nbsp;  
 
!Foundations: &nbsp;  
 
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|This problem requires two rules of integration.  In particular, you need
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|This problem requires '''Integration by substitution (''u'' - sub):''' If <math style="vertical-align: -20%">u = g(x)</math>&thinsp; is a differentiable functions whose range is in the domain of <math style="vertical-align: -20%">f</math>, then
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|'''Integration by substitution (''u'' - sub):''' If <math style="vertical-align: -20%">u = g(x)</math>&thinsp; is a differentiable functions whose range is in the domain of <math style="vertical-align: -20%">f</math>, then
 
 
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::<math>\int g'(x)f(g(x)) dx \,=\, \int f(u) du.</math>
 
::<math>\int g'(x)f(g(x)) dx \,=\, \int f(u) du.</math>
 
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|We also need the derivative of the natural log since we will recover natural log from integration:
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|We also need our power rule for integration:
 
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::<math>\left(ln(x)\right)' \,=\, \frac{1}{x}.</math>
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::<math style="vertical-align: -70%;">\int x^n dx \,=\, \frac{x^{n + 1}}{n + 1}+C,</math>&thinsp; for <math style="vertical-align: -23%;">n\neq 0</math>.
 
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!Step 1: &nbsp;
 
!Step 1: &nbsp;
 
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|Use a ''u''-substitution with <math style="vertical-align: -8%">u = 3x + 2.</math> This means <math style="vertical-align: 0%">du = 3\,dx</math>, or <math style="vertical-align: -20%">dx=du/3</math>. After substitution we have
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|(a) Use a ''u''-substitution with <math style="vertical-align: -8%">u = 3x^2 + 1.</math> This means <math style="vertical-align: 0%">du = 6x\,dx</math>, or <math style="vertical-align: -20%">dx=du/6</math>. After substitution we have
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::<math>\int \frac{1}{3x + 2} \,=\, \int \frac{1}{u}\,\frac{du}{3}\,=\,\frac{1}{3}\int\frac{1}{u}\,du.</math>
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::<math>\int x e^{3x^2+1}\, dx &= \frac{1}{6} \int e^{u}\, du.</math>
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|(b) We need to use the power rule to find that <math>\int_2^5 4x - 5 \, dx = 2x^2 - 5x \Bigr|_2^5</math>
 
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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|We can now take the integral remembering the special rule:
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|(a)
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::<math>\frac{1}{6} \int e^{u}\, du = \frac{1}{6}e^u.</math>
 
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|(b) We just need to evaluate at the endpoints to finish the problem:
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::<math>\frac{1}{3}\int\frac{1}{u}\,du\,=\, \frac{\log(u)}{3}.</math>
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<math>\begin{array}{rcl}2x^2 - 5x \Bigr|_2^5 & = & 2(5^2) - 5(5) -(2(2)^2 - 5(2)\\
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& = & 50 - 25 -(8 - 10)\\
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& = & 25 +2\\
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& = & 27
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\end{array}</math>
 
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!Step 3: &nbsp;
 
!Step 3: &nbsp;
 
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| Now we need to substitute back into our original variables using our original substitution <math style="vertical-align: -5%">u = 3x + 2</math>
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|(a) Now we need to substitute back into our original variables using our original substitution <math style="vertical-align: -5%">u = 3x^2 + 1</math>
 
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| to find&nbsp; <math>\frac{\log(u)}{3} = \frac{\log(3x + 2)}{3}.</math>
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| to find&nbsp; <math>\frac{1}{6}e^u = \frac{e^{3x^2 + 1}}{6}.</math>
 
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!Final Answer: &nbsp;
 
!Final Answer: &nbsp;
 
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|(a)
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::<math>\int \frac{1}{3x + 2}\,dx \,=\, \frac{\ln(3x + 2)}{3} + C.</math>
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::<math>\frac{e^{3x^2 + 1}}{6} + C.</math>
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|(b) 27
 
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[[022_Exam_2_Sample_A|'''<u>Return to Sample Exam</u>''']]
 
[[022_Exam_2_Sample_A|'''<u>Return to Sample Exam</u>''']]

Revision as of 17:28, 15 May 2015

Find the antiderivatives:

(a)


(b)
Foundations:  
This problem requires Integration by substitution (u - sub): If   is a differentiable functions whose range is in the domain of , then
We also need our power rule for integration:
  for .

 Solution:

Step 1:  
(a) Use a u-substitution with This means , or . After substitution we have
Failed to parse (syntax error): {\displaystyle \int x e^{3x^2+1}\, dx &= \frac{1}{6} \int e^{u}\, du.}
(b) We need to use the power rule to find that
Step 2:  
(a)
(b) We just need to evaluate at the endpoints to finish the problem:

Step 3:  
(a) Now we need to substitute back into our original variables using our original substitution
to find 
Step 4:  
Since this integral is an indefinite integral we have to remember to add a constant  at the end.
Final Answer:  
(a)
(b) 27

Return to Sample Exam

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