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

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!Step 1:    
 
!Step 1:    
 
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|We need to use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -2px">u=1+x^3.</math> Then, <math style="vertical-align: 0px">du=3x^2dx</math> and&thinsp; <math style="vertical-align: -13px">\frac{du}{3}=x^2dx.</math>
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|We need to use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -2px">u=1+x^3.</math>  
 +
|-
 +
|Then, <math style="vertical-align: 0px">du=3x^2dx</math> and&thinsp; <math style="vertical-align: -13px">\frac{du}{3}=x^2dx.</math>
 
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|-
 
|Therefore, the integral becomes&thinsp; <math style="vertical-align: -13px">\frac{1}{3}\int \sqrt{u}~du.</math>
 
|Therefore, the integral becomes&thinsp; <math style="vertical-align: -13px">\frac{1}{3}\int \sqrt{u}~du.</math>
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
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|Again, we need to use <math>u</math>-substitution. Let <math style="vertical-align: -5px">u=\sin(x).</math> Then, <math style="vertical-align: -5px">du=\cos(x)dx.</math> Also, we need to change the bounds of integration.
+
|Again, we need to use <math>u</math>-substitution. Let <math style="vertical-align: -5px">u=\sin(x).</math> Then, <math style="vertical-align: -5px">du=\cos(x)dx.</math>  
 +
|-
 +
|Also, we need to change the bounds of integration.
 +
|-
 +
|Plugging in our values into the equation <math style="vertical-align: -5px">u=\sin(x),</math>
 
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|-
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|Plugging in our values into the equation <math style="vertical-align: -5px">u=\sin(x),</math> we get <math style="vertical-align: -15px">u_1=\sin\bigg(\frac{\pi}{4}\bigg)=\frac{\sqrt{2}}{2}</math> and <math style="vertical-align: -16px">u_2=\sin\bigg(\frac{\pi}{2}\bigg)=1.</math>
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|we get <math style="vertical-align: -15px">u_1=\sin\bigg(\frac{\pi}{4}\bigg)=\frac{\sqrt{2}}{2}</math> and <math style="vertical-align: -16px">u_2=\sin\bigg(\frac{\pi}{2}\bigg)=1.</math>
 
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|-
 
|Therefore, the integral becomes <math style="vertical-align: -19px">\int_{\frac{\sqrt{2}}{2}}^1 \frac{1}{u^2}~du.</math>
 
|Therefore, the integral becomes <math style="vertical-align: -19px">\int_{\frac{\sqrt{2}}{2}}^1 \frac{1}{u^2}~du.</math>

Revision as of 07:46, 7 February 2017

Evaluate the indefinite and definite 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}{2}} \frac{\cos(x)}{\sin^2(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 \int \frac{\ln x}{x}~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=\ln(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=\frac{1}{x}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 \frac{\ln x}{x}~dx=\int u~du=\frac{u^2}{2}+C=\frac{(\ln x)^2}{2}+C.}


Solution:

(a)

Step 1:  
We need to 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=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:  
Again, we need to 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=\sin(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=\cos(x)dx.}
Also, we need to change the bounds of integration.
Plugging in our values into the equation 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=\sin(x),}
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 u_1=\sin\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=\sin\bigg(\frac{\pi}{2}\bigg)=1.}
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 \int_{\frac{\sqrt{2}}{2}}^1 \frac{1}{u^2}~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 _{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\cos(x)}{\sin^2(x)}~dx=\int_{\frac{\sqrt{2}}{2}}^1 \frac{1}{u^2}~du=\left.\frac{-1}{u}\right|_{\frac{\sqrt{2}}{2}}^1=-\frac{1}{1}-\frac{-1}{\frac{\sqrt{2}}{2}}=-1+\sqrt{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)   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 -1+\sqrt{2}}

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