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

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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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::<math>\begin{array}{rcl}
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\displaystyle{\int 2x(x^2+1)^3~dx} & = & \displaystyle{\int u^3~du}\\
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&&\\
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& = & \displaystyle{\frac{u^4}{4}+C}\\
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&& \\
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& = & \displaystyle{\frac{(x^2+1)^4}{4}+C.}\\
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\end{array}</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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::::<math>\begin{array}{rcl}
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\displaystyle{\int x^2\sin (x^3) ~dx} & = & \displaystyle{\frac{-1}{3}\cos(u)+C}\\
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&&\\
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& = & \displaystyle{\frac{-1}{3}\cos(x^3)+C.}\\
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\end{array}</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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::<math>\begin{array}{rcl}
 +
\displaystyle{\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^2(x)\sin (x)~dx} & = & \displaystyle{\int_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}} -u^2~du}\\
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&&\\
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& = & \displaystyle{\left.\frac{-u^3}{3}\right|_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}}}\\
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&&\\
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& = & \displaystyle{0.} \\
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\end{array}</math>
 
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Revision as of 16:02, 29 March 2016

Compute the following 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\sin (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}{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 \begin{array}{rcl} \displaystyle{\int 2x(x^2+1)^3~dx} & = & \displaystyle{\int u^3~du}\\ &&\\ & = & \displaystyle{\frac{u^4}{4}+C}\\ && \\ & = & \displaystyle{\frac{(x^2+1)^4}{4}+C.}\\ \end{array}}

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 \begin{array}{rcl} \displaystyle{\int x^2\sin (x^3) ~dx} & = & \displaystyle{\frac{-1}{3}\cos(u)+C}\\ &&\\ & = & \displaystyle{\frac{-1}{3}\cos(x^3)+C.}\\ \end{array}}

(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 \begin{array}{rcl} \displaystyle{\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^2(x)\sin (x)~dx} & = & \displaystyle{\int_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}} -u^2~du}\\ &&\\ & = & \displaystyle{\left.\frac{-u^3}{3}\right|_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}}}\\ &&\\ & = & \displaystyle{0.} \\ \end{array}}
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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