Difference between revisions of "009B Sample Final 1, Problem 4"

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Line 69: Line 69:
 
!Step 1:    
 
!Step 1:    
 
|-
 
|-
−
|First, we add and subtract <math>x</math> from the numerator. So, we have
+
|First, we add and subtract <math>x</math> from the numerator.  
 
|-
 
|-
−
|<math>\int \frac{2x^2+1}{2x^2+x}~dx=\int\frac{2x^2+x-x+1}{2x^2+x}~dx=\int \frac{2x^2+x}{2x^2+x}~dx+\int\frac{1-x}{2x^2+x}~dx=\int ~dx+\int\frac{1-x}{2x^2+x}~dx </math>.
+
|So, we have
 +
|-
 +
|
 +
::<math>\begin{array}{rcl}
 +
\displaystyle{\int \frac{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{\int\frac{2x^2+x-x+1}{2x^2+x}~dx}\\
 +
&&\\
 +
& = & \displaystyle{\int \frac{2x^2+x}{2x^2+x}~dx+\int\frac{1-x}{2x^2+x}~dx}\\
 +
&&\\
 +
& = & \displaystyle{\int ~dx+\int\frac{1-x}{2x^2+x}~dx}\\
 +
\end{array}</math>
 
|}
 
|}
  
Line 81: Line 90:
 
|Since <math>2x^2+x=x(2x+1)</math>, we let <math>\frac{1-x}{2x^2+x}=\frac{A}{x}+\frac{B}{2x+1}</math>.
 
|Since <math>2x^2+x=x(2x+1)</math>, we let <math>\frac{1-x}{2x^2+x}=\frac{A}{x}+\frac{B}{2x+1}</math>.
 
|-
 
|-
−
|Multiplying both sides of the last equation by <math>x(2x+1)</math>, we get <math>1-x=A(2x+1)+Bx</math>.
+
|Multiplying both sides of the last equation by <math>x(2x+1)</math>,  
 +
|-
 +
|we get <math>1-x=A(2x+1)+Bx</math>.
 
|-
 
|-
 
|If we let <math>x=0</math>, the last equation becomes <math>1=A</math>.
 
|If we let <math>x=0</math>, the last equation becomes <math>1=A</math>.
Line 95: Line 106:
 
|If we plug in the last equation from Step 2 into our final integral in Step 1, we have  
 
|If we plug in the last equation from Step 2 into our final integral in Step 1, we have  
 
|-
 
|-
−
|<math>\int \frac{2x^2+1}{2x^2+x}~dx=\int ~dx+\int\frac{1}{x}~dx+\int\frac{-3}{2x+1}~dx=x+\ln x+ \int\frac{-3}{2x+1}~dx</math>.
+
|
 +
::<math>\begin{array}{rcl}
 +
\displaystyle{\int \frac{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{\int ~dx+\int\frac{1}{x}~dx+\int\frac{-3}{2x+1}~dx}\\
 +
&&\\
 +
& = & \displaystyle{x+\ln x+ \int\frac{-3}{2x+1}~dx}\\
 +
\end{array}</math>
 
|}
 
|}
  
Line 107: Line 123:
 
|Thus, our final integral becomes
 
|Thus, our final integral becomes
 
|-
 
|-
−
|<math>\int \frac{2x^2+1}{2x^2+x}~dx=x+\ln x+ \int\frac{-3}{2x+1}~dx=x+\ln x+\int\frac{-3}{2u}~du=x+\ln x-\frac{3}{2}\ln u +C</math>.
+
|
 +
::<math>\begin{array}{rcl}
 +
\displaystyle{\int \frac{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{x+\ln x+ \int\frac{-3}{2x+1}~dx}\\
 +
&&\\
 +
& = & \displaystyle{x+\ln x+\int\frac{-3}{2u}~du}\\
 +
&&\\
 +
& = & \displaystyle{x+\ln x-\frac{3}{2}\ln u +C}\\
 +
\end{array}</math>
 
|-
 
|-
 
|Therefore, the final answer is
 
|Therefore, the final answer is
Line 120: Line 143:
 
|First, we write <math style="vertical-align: -13px">\int\sin^3x~dx=\int \sin^2 x \sin x~dx</math>.
 
|First, we write <math style="vertical-align: -13px">\int\sin^3x~dx=\int \sin^2 x \sin x~dx</math>.
 
|-
 
|-
−
|Using the identity <math style="vertical-align: -2px">\sin^2x+\cos^2x=1</math>, we get <math style="vertical-align: -1px">\sin^2x=1-\cos^2x</math>. If we use this identity, we have
+
|Using the identity <math style="vertical-align: -2px">\sin^2x+\cos^2x=1</math>, we get <math style="vertical-align: -1px">\sin^2x=1-\cos^2x</math>.  
 +
|-
 +
|If we use this identity, we have
 
|-
 
|-
 
| &nbsp; &nbsp; <math style="vertical-align: -13px">\int\sin^3x~dx=\int (1-\cos^2x)\sin x~dx</math>.
 
| &nbsp; &nbsp; <math style="vertical-align: -13px">\int\sin^3x~dx=\int (1-\cos^2x)\sin x~dx</math>.
Line 130: Line 155:
 
!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
−
|Now, we proceed by <math>u</math>-substitution. Let <math>u=\cos x</math>. Then, <math>du=-\sin x dx</math>. So we have
+
|Now, we proceed by <math>u</math>-substitution. Let <math>u=\cos x</math>. Then, <math>du=-\sin x dx</math>.  
 +
|-
 +
|So we have
 
|-
 
|-
−
|<math style="vertical-align: -13px">\int\sin^3x~dx=\int -(1-u^2)~du=-u+\frac{u^3}{3}+C=-\cos x+\frac{\cos^3x}{3}+C</math>.
+
|
 +
::<math>\begin{array}{rcl}
 +
\displaystyle{\int\sin^3x~dx} & = & \displaystyle{\int -(1-u^2)~du}\\
 +
&&\\
 +
& = & \displaystyle{-u+\frac{u^3}{3}+C}\\
 +
&&\\
 +
& = & \displaystyle{-\cos x+\frac{\cos^3x}{3}+C}\\
 +
\end{array}</math>
 
|-
 
|-
 
|
 
|

Revision as of 11:51, 10 February 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 e^x(x+\sin(e^x))~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{2x^2+1}{2x^2+x}~dx}

c) 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 \sin^3x~dx}


Foundations:  
Review 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
Integration by parts
Partial fraction decomposition
Trig identities

Solution:

(a)

Step 1:  
We first distribute 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 e^x(x+\sin(e^x))~dx=\int e^xx~dx+\int e^x\sin(e^x)~dx} .
Now, for the first integral on the right hand side of the last equation, we use integration by parts.
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} 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 dv=e^xdx} . 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=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 v=e^x} .
So, 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 \begin{array}{rcl} \displaystyle{\int e^x(x+\sin(e^x))~dx=} & = & \displaystyle{\bigg(xe^x-\int e^x~dx \bigg)+\int e^x\sin(e^x)~dx}\\ &&\\ & = & \displaystyle{xe^x-e^x+\int e^x\sin(e^x)~dx}\\ \end{array}}
Step 2:  
Now, for the one remaining integral, 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=e^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=e^xdx} .
So, 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 \begin{array}{rcl} \displaystyle{\int e^x(x+\sin(e^x))~dx} & = & \displaystyle{xe^x-e^x+\int \sin(u)~du}\\ &&\\ & = & \displaystyle{xe^x-e^x-\cos(u)+C}\\ &&\\ & = & \displaystyle{xe^x-e^x-\cos(e^x)+C}\\ \end{array}}

(b)

Step 1:  
First, we add and subtract 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 x} from the numerator.
So, 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 \begin{array}{rcl} \displaystyle{\int \frac{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{\int\frac{2x^2+x-x+1}{2x^2+x}~dx}\\ &&\\ & = & \displaystyle{\int \frac{2x^2+x}{2x^2+x}~dx+\int\frac{1-x}{2x^2+x}~dx}\\ &&\\ & = & \displaystyle{\int ~dx+\int\frac{1-x}{2x^2+x}~dx}\\ \end{array}}
Step 2:  
Now, we need to use partial fraction decomposition for the second integral.
Since 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^2+x=x(2x+1)} , we 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 \frac{1-x}{2x^2+x}=\frac{A}{x}+\frac{B}{2x+1}} .
Multiplying both sides of the last equation by 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 x(2x+1)} ,
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 1-x=A(2x+1)+Bx} .
If we 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 x=0} , the last equation 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 1=A} .
If we 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 x=-\frac{1}{2}} , then 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 \frac{3}{2}=-\frac{1}{2}B} . 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 B=-3} .
So, in summation, 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 \frac{1-x}{2x^2+x}=\frac{1}{x}+\frac{-3}{2x+1}} .
Step 3:  
If we plug in the last equation from Step 2 into our final integral in Step 1, 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 \begin{array}{rcl} \displaystyle{\int \frac{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{\int ~dx+\int\frac{1}{x}~dx+\int\frac{-3}{2x+1}~dx}\\ &&\\ & = & \displaystyle{x+\ln x+ \int\frac{-3}{2x+1}~dx}\\ \end{array}}
Step 4:  
For the final remaining integral, 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=2x+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=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}{2}=dx} .
Thus, our final 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 \begin{array}{rcl} \displaystyle{\int \frac{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{x+\ln x+ \int\frac{-3}{2x+1}~dx}\\ &&\\ & = & \displaystyle{x+\ln x+\int\frac{-3}{2u}~du}\\ &&\\ & = & \displaystyle{x+\ln x-\frac{3}{2}\ln u +C}\\ \end{array}}
Therefore, the final answer is
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{2x^2+1}{2x^2+x}~dx=x+\ln x-\frac{3}{2}\ln (2x+1) +C}

(c)

Step 1:  
First, we write 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\sin^3x~dx=\int \sin^2 x \sin x~dx} .
Using the identity 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 \sin^2x+\cos^2x=1} , 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 \sin^2x=1-\cos^2x} .
If we use this identity, 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\sin^3x~dx=\int (1-\cos^2x)\sin x~dx} .
Step 2:  
Now, we proceed by 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=\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} .
So 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 \begin{array}{rcl} \displaystyle{\int\sin^3x~dx} & = & \displaystyle{\int -(1-u^2)~du}\\ &&\\ & = & \displaystyle{-u+\frac{u^3}{3}+C}\\ &&\\ & = & \displaystyle{-\cos x+\frac{\cos^3x}{3}+C}\\ \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 xe^x-e^x-\cos(e^x)+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 x+\ln x-\frac{3}{2}\ln (2x+1) +C}
(c) 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 -\cos x+\frac{\cos^3x}{3}+C}

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