Difference between revisions of "009B Sample Midterm 1, Problem 1"

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!Step 1:  
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|We need to use substitution. Let <math>u=1+x^3</math>. Then, <math>du=3x^2dx</math> and <math>\frac{du}{3}=x^2dx</math>.
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|Therefore, the integral becomes <math>\frac{1}{3}\int \sqrt{u}du</math>.
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|<math>\int x^2\sqrt{1+x^3}dx=\frac{1}{3}\int \sqrt{u}du=\frac{2}{9}u^{\frac{3}{2}}+C=\frac{2}{9}(1+x^3)^{\frac{3}{2}}+C</math>.
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[[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

Revision as of 13:56, 26 January 2016

Evaluate the indefinite and definite integrals.

a) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int x^2\sqrt{1+x^3}dx}
b) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int _{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\cos(x)}{\sin^2(x)}dx}


Foundations:  

Solution:

(a)

Step 1:  
We need to use substitution. Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=1+x^3} . Then, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle du=3x^2dx} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{du}{3}=x^2dx} .
Therefore, the integral becomes Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{3}\int \sqrt{u}du} .
Step 2:  
We now have:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int x^2\sqrt{1+x^3}dx=\frac{1}{3}\int \sqrt{u}du=\frac{2}{9}u^{\frac{3}{2}}+C=\frac{2}{9}(1+x^3)^{\frac{3}{2}}+C} .

(b)

Step 1:  
Step 2:  
Final Answer:  
(a) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{2}{9}(1+x^3)^{\frac{3}{2}}+C}
(b)

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