Difference between revisions of "009B Sample Midterm 2, Problem 2"

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|-
 
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::You could use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -2px">u=x^2+x.</math> Then, <math style="vertical-align: -4px">du=(2x+1)~dx.</math>
+
&nbsp; &nbsp; You could use <math style="vertical-align: 0px">u</math>-substitution.  
 +
|-
 +
|&nbsp; &nbsp; Let <math style="vertical-align: -2px">u=x^2+x.</math> Then, <math style="vertical-align: -4px">du=(2x+1)~dx.</math>
 +
|-
 +
|
 +
&nbsp; &nbsp; Thus,
 
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|-
 
|
 
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::Thus, <math style="vertical-align: -12px">\int (2x+1)\sqrt{x^2+x}~dx=\int \sqrt{u}=\frac{2}{3}u^{3/2}+C=\frac{2}{3}(x^2+x)^{3/2}+C.</math>
+
&nbsp; &nbsp; <math>\begin{array}{rcl}
 +
\displaystyle{\int (2x+1)\sqrt{x^2+x}~dx} & = & \displaystyle{\int \sqrt{u}~du}\\
 +
&&\\
 +
& = & \displaystyle{\frac{2}{3}u^{3/2}+C}\\
 +
&&\\
 +
& = & \displaystyle{\frac{2}{3}(x^2+x)^{3/2}+C.}
 +
\end{array}</math>
 
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|}
  
Line 27: Line 38:
 
|We multiply the product inside the integral to get  
 
|We multiply the product inside the integral to get  
 
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|-
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| &nbsp;&nbsp; <math style="vertical-align: -16px">\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt=\int_1^2 \bigg(8t^3-10+12-\frac{15}{t^3}\bigg)~dt=\int_1^2 (8t^3+2-15t^{-3})~dt.</math>
+
|  
 +
&nbsp; &nbsp; <math>\begin{array}{rcl}
 +
\displaystyle{\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt} & = & \displaystyle{\int_1^2 \bigg(8t^3-10+12-\frac{15}{t^3}\bigg)~dt}\\
 +
&&\\
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& = & \displaystyle{\int_1^2 (8t^3+2-15t^{-3})~dt.}
 +
\end{array}</math>
 
|}
 
|}
  
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|We now evaluate to get
 
|We now evaluate to get
 
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| &nbsp;&nbsp; <math style="vertical-align: -16px">\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt=2(2)^4+2(2)+\frac{15}{2(2)^2}-\bigg(2+2+\frac{15}{2}\bigg)=36+\frac{15}{8}-4-\frac{15}{2}=\frac{211}{8}.</math>
+
|  
 +
&nbsp; &nbsp; <math>\begin{array}{rcl}
 +
\displaystyle{\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt} & = & \displaystyle{2(2)^4+2(2)+\frac{15}{2(2)^2}-\bigg(2+2+\frac{15}{2}\bigg)}\\
 +
&&\\
 +
& = & \displaystyle{36+\frac{15}{8}-4-\frac{15}{2}}\\
 +
&&\\
 +
& = & \displaystyle{\frac{211}{8}.}
 +
\end{array}</math>
 
|}
 
|}
  
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
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|We use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -2px">u=x^4+2x^2+4.</math> Then, <math style="vertical-align: -5px">du=(4x^3+4x)dx</math> and <math style="vertical-align: -14px">\frac{du}{4}=(x^3+x)dx.</math> Also, we need to change the bounds of integration.  
+
|We use <math style="vertical-align: 0px">u</math>-substitution.  
 
|-
 
|-
−
|Plugging in our values into the equation <math style="vertical-align: -4px">u=x^4+2x^2+4,</math> we get <math style="vertical-align: -5px">u_1=0^4+2(0)^2+4=4</math> and <math style="vertical-align: -5px">u_2=2^4+2(2)^2+4=28.</math>
+
|Let <math style="vertical-align: -2px">u=x^4+2x^2+4.</math> Then, <math style="vertical-align: -5px">du=(4x^3+4x)dx</math> and <math style="vertical-align: -14px">\frac{du}{4}=(x^3+x)dx.</math>  
 
|-
 
|-
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|Therefore, the integral becomes&nbsp; <math style="vertical-align: -14px">\frac{1}{4}\int_4^{28}\sqrt{u}~du.</math>
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|Also, we need to change the bounds of integration.  
 
|-
 
|-
−
|
+
|Plugging in our values into the equation <math style="vertical-align: -4px">u=x^4+2x^2+4,</math>
 +
|-
 +
|we get <math style="vertical-align: -5px">u_1=0^4+2(0)^2+4=4</math> and <math style="vertical-align: -5px">u_2=2^4+2(2)^2+4=28.</math>
 +
|-
 +
|Therefore, the integral becomes
 +
|-
 +
|&nbsp; &nbsp; <math style="vertical-align: -14px">\frac{1}{4}\int_4^{28}\sqrt{u}~du.</math>
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
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|We now have:
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|We now have
 
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| &nbsp;&nbsp; <math style="vertical-align: -16px">\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx=\frac{1}{4}\int_4^{28}\sqrt{u}~du=\left.\frac{1}{6}u^{\frac{3}{2}}\right|_4^{28}=\frac{1}{6}(28^{\frac{3}{2}}-4^{\frac{3}{2}})=\frac{1}{6}((\sqrt{28})^3-(\sqrt{4})^3)=\frac{1}{6}((2\sqrt{7})^3-2^3).</math>
+
|  
 +
&nbsp; &nbsp; <math>\begin{array}{rcl}
 +
\displaystyle{\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx} & = & \displaystyle{\frac{1}{4}\int_4^{28}\sqrt{u}~du}\\
 +
&&\\
 +
& = & \displaystyle{\left.\frac{1}{6}u^{\frac{3}{2}}\right|_4^{28}}\\
 +
&&\\
 +
& = & \displaystyle{\frac{1}{6}(28^{\frac{3}{2}}-4^{\frac{3}{2}})}\\
 +
&&\\
 +
& = & \displaystyle{\frac{1}{6}((\sqrt{28})^3-(\sqrt{4})^3)}\\
 +
&&\\
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& = & \displaystyle{\frac{1}{6}((2\sqrt{7})^3-2^3).}
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\end{array}</math>
 
|-
 
|-
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|So, we have
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|Therefore,  
 
|-
 
|-
 
| &nbsp;&nbsp; <math style="vertical-align: -16px">\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx=\frac{28\sqrt{7}-4}{3}.</math>
 
| &nbsp;&nbsp; <math style="vertical-align: -16px">\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx=\frac{28\sqrt{7}-4}{3}.</math>

Revision as of 08:43, 7 February 2017

Evaluate

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_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt}
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_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~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 (2x+1)\sqrt{x^2+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=x^2+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=(2x+1)~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+1)\sqrt{x^2+x}~dx} & = & \displaystyle{\int \sqrt{u}~du}\\ &&\\ & = & \displaystyle{\frac{2}{3}u^{3/2}+C}\\ &&\\ & = & \displaystyle{\frac{2}{3}(x^2+x)^{3/2}+C.} \end{array}}


Solution:

(a)

Step 1:  
We multiply the product inside the integral 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_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt} & = & \displaystyle{\int_1^2 \bigg(8t^3-10+12-\frac{15}{t^3}\bigg)~dt}\\ &&\\ & = & \displaystyle{\int_1^2 (8t^3+2-15t^{-3})~dt.} \end{array}}

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_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt=\left. 2t^4+2t+\frac{15}{2}t^{-2}\right|_1^2.}
We now evaluate 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_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt} & = & \displaystyle{2(2)^4+2(2)+\frac{15}{2(2)^2}-\bigg(2+2+\frac{15}{2}\bigg)}\\ &&\\ & = & \displaystyle{36+\frac{15}{8}-4-\frac{15}{2}}\\ &&\\ & = & \displaystyle{\frac{211}{8}.} \end{array}}

(b)

Step 1:  
We 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^4+2x^2+4.} 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=(4x^3+4x)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}{4}=(x^3+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=x^4+2x^2+4,}
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=0^4+2(0)^2+4=4} 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=2^4+2(2)^2+4=28.}
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}{4}\int_4^{28}\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 \begin{array}{rcl} \displaystyle{\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx} & = & \displaystyle{\frac{1}{4}\int_4^{28}\sqrt{u}~du}\\ &&\\ & = & \displaystyle{\left.\frac{1}{6}u^{\frac{3}{2}}\right|_4^{28}}\\ &&\\ & = & \displaystyle{\frac{1}{6}(28^{\frac{3}{2}}-4^{\frac{3}{2}})}\\ &&\\ & = & \displaystyle{\frac{1}{6}((\sqrt{28})^3-(\sqrt{4})^3)}\\ &&\\ & = & \displaystyle{\frac{1}{6}((2\sqrt{7})^3-2^3).} \end{array}}

Therefore,
   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_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx=\frac{28\sqrt{7}-4}{3}.}


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{211}{8}}
(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 \frac{28\sqrt{7}-4}{3}}

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