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

Revision as of 14:07, 2 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) $\int {\frac {2x^{2}+1}{2x^{2}+x}}~dx$

c) $\int \sin ^{3}x~dx$

Foundations:
Review $u$ -substitution
Integration by parts
Partial fraction decomposition
Trig identities

Solution:

(a)

Step 1:
We first distribute to get $\int e^{x}(x+\sin(e^{x}))~dx=\int e^{x}x~dx+\int e^{x}\sin(e^{x})~dx$ .
Now, for the first integral on the right hand side of the last equation, we use integration by parts.
Let $u=x$ and $dv=e^{x}dx$ . Then, $du=dx$ and $v=e^{x}$ . So, we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int e^{x}(x+\sin(e^{x}))~dx={\bigg (}xe^{x}-\int e^{x}~dx{\bigg )}+\int e^{x}\sin(e^{x})~dx=xe^{x}-e^{x}+\int e^{x}\sin(e^{x})~dx}

Step 2:
Now, for the one remaining integral, we use $u$ -substitution.
Let $u=e^{x}$ . Then, $du=e^{x}dx$ . So, we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int e^{x}(x+\sin(e^{x}))~dx=xe^{x}-e^{x}+\int \sin(u)~du=xe^{x}-e^{x}-\cos(u)+C=xe^{x}-e^{x}-\cos(e^{x})+C} .

(b)

Step 1:
First, we add and subtract $x$ from the numerator. So, we have
$\int {\frac {2x^{2}+1}{2x^{2}+x}}~dx=\int {\frac {2x^{2}+x-x+1}{2x^{2}+x}}~dx=\int {\frac {2x^{2}+x}{2x^{2}+x}}~dx+\int {\frac {1-x}{2x^{2}+x}}~dx=\int ~dx+\int {\frac {1-x}{2x^{2}+x}}~dx$ .

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 \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} .

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 \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} .
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 \int\sin^3x~dx=\int -(1-u^2)~du=-u+\frac{u^3}{3}+C=-\cos x+\frac{\cos^3x}{3}+C} .

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}

Return to Sample Exam

@@ Line 51: / Line 51: @@
 !Step 1: &nbsp;
 |-
-|
+|First, we add and subtract <math>x</math> from the numerator. So, we have
-|-
-|
-|-
-|
 |-
-|
+|<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>.
 |}
@@ Line 63: / Line 59: @@
 !Step 2: &nbsp;
 |-
-|
+|Now, we need to use partial fraction decomposition for the second integral.
+|-
+|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>.
+|-
+|If we let <math>x=0</math>, the last equation becomes <math>1=A</math>.
+|-
+|If we let <math>x=-\frac{1}{2}</math>, then we get <math>\frac{3}{2}=-\frac{1}{2}B</math>. Thus, <math>B=-3</math>.
+|-
+|So, in summation, we have <math>\frac{1-x}{2x^2+x}=\frac{1}{x}+\frac{-3}{2x+1}</math>.
 |}
@@ Line 69: / Line 75: @@
 !Step 3: &nbsp;
 |-
-|
+|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>.
 |}
+{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+!Step 4: &nbsp;
+|-
+|For the final remaining integral, we use <math>u</math>-substitution.
+|-
+|Let <math>u=2x+1</math>. Then, <math>du=2dx</math> and <math>\frac{du}{2}=dx</math>.
+|-
+|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>.
+|-
+|Therefore, the final answer is
+|-
+|<math>\int \frac{2x^2+1}{2x^2+x}~dx=x+\ln x-\frac{3}{2}\ln (2x+1) +C</math>
+|}
 '''(c)'''
@@ Line 103: / Line 124: @@
 |'''(a)''' <math>xe^x-e^x-\cos(e^x)+C</math>
 |-
-|'''(b)'''
+|'''(b)''' <math>x+\ln x-\frac{3}{2}\ln (2x+1) +C</math>
 |-
 |'''(c)''' <math>-\cos x+\frac{\cos^3x}{3}+C</math>
 |}
 [[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

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

Revision as of 14:07, 2 February 2016

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