Difference between revisions of "009B Sample Midterm 3, Problem 3"

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::You could use <math>u</math>-substitution. Let <math>u=x^2+1</math>. Then, <math>du=2xdx</math>.
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::You could use <math>u</math>-substitution. Let <math>u=x^2+1.</math> Then, <math>du=2x~dx.</math> Thus,
 
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::Thus, <math>\int 2x(x^2+1)^3~dx=\int u^3~du=\frac{u^4}{4}+C=\frac{(x^2+1)^4}{4}+C</math>.
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::<math>\int 2x(x^2+1)^3~dx=\int u^3~du=\frac{u^4}{4}+C=\frac{(x^2+1)^4}{4}+C.</math>
 
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
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|We proceed using <math>u</math>-substitution. Let <math>u=x^3</math>. Then, <math>du=3x^2dx</math> and <math>\frac{du}{3}=x^2dx</math>.
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|We proceed using <math>u</math>-substitution. Let <math>u=x^3.</math> Then, <math>du=3x^2~dx</math> and <math>\frac{du}{3}=x^2~dx.</math>
 
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|Therefore, we have
 
|Therefore, we have
 
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|<math>\int x^2\sin (x^3) ~dx=\int \frac{\sin(u)}{3}~du</math>
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::<math>\int x^2\sin (x^3) ~dx=\int \frac{\sin(u)}{3}~du.</math>
 
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|We integrate to get
 
|We integrate to get
 
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|<math>\int x^2\sin (x^3) ~dx=\frac{-1}{3}\cos(u)+C=\frac{-1}{3}\cos(x^3)+C</math>
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::<math>\int x^2\sin (x^3) ~dx=\frac{-1}{3}\cos(u)+C=\frac{-1}{3}\cos(x^3)+C.</math>
 
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
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|Again, we proceed using u substitution. Let <math>u=\cos(x)</math>. Then, <math>du=-\sin(x)dx</math>.
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|Again, we proceed using u substitution. Let <math>u=\cos(x).</math> Then, <math>du=-\sin(x)~dx.</math>  
 
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|Since this is a definite integral, we need to change the bounds of integration.  
 
|Since this is a definite integral, we need to change the bounds of integration.  
 
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|We have <math>u_1=\cos\bigg(-\frac{\pi}{4}\bigg)=\frac{\sqrt{2}}{2}</math> and <math>u_2=\cos\bigg(\frac{\pi}{4}\bigg)=\frac{\sqrt{2}}{2}</math>.
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|We have <math>u_1=\cos\bigg(-\frac{\pi}{4}\bigg)=\frac{\sqrt{2}}{2}</math> and <math>u_2=\cos\bigg(\frac{\pi}{4}\bigg)=\frac{\sqrt{2}}{2}.</math>
 
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|So, we get  
 
|So, we get  
 
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|<math>\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^2(x)\sin (x)~dx=\int_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}} -u^2~du=\left.\frac{-u^3}{3}\right|_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}}=0</math>
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::<math>\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^2(x)\sin (x)~dx=\int_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}} -u^2~du=\left.\frac{-u^3}{3}\right|_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}}=0.</math>
 
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Revision as of 15:36, 29 March 2016

Compute the following integrals:

a)
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}{4}} \cos^2(x)\sin (x)~dx}


Foundations:  
How would you integrate 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 2x(x^2+1)^3~dx?}
You could use 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} -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=x^2+1.} 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=2x~dx.} Thus,
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 2x(x^2+1)^3~dx=\int u^3~du=\frac{u^4}{4}+C=\frac{(x^2+1)^4}{4}+C.}

Solution:

(a)

Step 1:  
We proceed using 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} -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=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^2~dx} 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^2~dx.}
Therefore, we 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\sin (x^3) ~dx=\int \frac{\sin(u)}{3}~du.}
Step 2:  
We integrate to get
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\sin (x^3) ~dx=\frac{-1}{3}\cos(u)+C=\frac{-1}{3}\cos(x^3)+C.}

(b)

Step 1:  
Again, we proceed using u 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=\cos(x).} 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=-\sin(x)~dx.}
Since this is a definite integral, we need to change the bounds of integration.
We 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 u_1=\cos\bigg(-\frac{\pi}{4}\bigg)=\frac{\sqrt{2}}{2}} 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 u_2=\cos\bigg(\frac{\pi}{4}\bigg)=\frac{\sqrt{2}}{2}.}
Step 2:  
So, we get
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}{4}} \cos^2(x)\sin (x)~dx=\int_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}} -u^2~du=\left.\frac{-u^3}{3}\right|_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}}=0.}
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{-1}{3}\cos(x^3)+C}
(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 0}

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