MathAdmin: Created page with "(a) Express the indefinite integral $\int \sin(x^2)~dx$ as a power series. (b) Expr..."

2017-03-12T16:53:59Z

Created page with "(a) Express the indefinite integral <math style="vertical-align: -13px">\int \sin(x^2)~dx</math> as a power series. (b) Expr..."

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(a) Express the indefinite integral  <math style="vertical-align: -13px">\int \sin(x^2)~dx</math>  as a power series.

(b) Express the definite integral  <math style="vertical-align: -14px">\int_0^1 \sin(x^2)~dx</math>  as a number series.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|What is the power series of  <math style="vertical-align: -1px">\sin x?</math>
|-
|        The power series of  <math style="vertical-align: -1px"> \sin x</math>  is   <math>\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}.</math>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|The power series of  <math style="vertical-align: -1px"> \sin x</math>  is   <math>\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}.</math>
|-
|So, the power series of  <math style="vertical-align: -5px">\sin(x^2)</math>   is
|-
| 
|-
|        <math>\begin{array}{rcl}
\displaystyle{\sin(x^2)} & = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^n(x^2)^{2n+1}}{(2n+1)!}}\\
&&\\
& = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^nx^{4n+2}}{(2n+1)!}.}
\end{array}</math>
|-
| 
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, to express the indefinite integral as a power series, we have
|-
| 
|-
|        <math>\begin{array}{rcl}
\displaystyle{\int \sin(x^2)~dx} & = & \displaystyle{\int \sum_{n=0}^\infty \frac{(-1)^nx^{4n+2}}{(2n+1)!}~dx}\\
&&\\
& = & \displaystyle{\sum_{n=0}^\infty \int \frac{(-1)^nx^{4n+2}}{(2n+1)!}~dx}\\
&&\\
& = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^n x^{4n+3}}{(4n+3)(2n+1)!}.}
\end{array}</math>
|-
| 
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|From part (a), we have
|-
|        <math>\int \sin(x^2)~dx=\sum_{n=0}^\infty \frac{(-1)^n x^{4n+3}}{(4n+3)(2n+1)!}.</math>
|-
|Now, we have
|-
|        <math>\int_0^1 \sin(x^2)~dx=\sum_{n=0}^\infty \frac{(-1)^n x^{4n+3}}{(4n+3)(2n+1)!}\bigg|_0^1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Hence, we have
|-
| 
|-
|        <math>\begin{array}{rcl}
\displaystyle{\int_0^1 \sin(x^2)~dx} & = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^n (1)^{4n+3}}{(4n+3)(2n+1)!}-\sum_{n=0}^\infty \frac{(-1)^n (0)^{4n+3}}{(4n+3)(2n+1)!}}\\
&&\\
& = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^n}{(4n+3)(2n+1)!}-0}\\
&&\\
& = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^n}{(4n+3)(2n+1)!}}
\end{array}</math>
|-
| 
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     <math>\sum_{n=0}^\infty \frac{(-1)^n x^{4n+3}}{(4n+3)(2n+1)!}</math>
|-
|    '''(b)'''     <math>\sum_{n=0}^\infty \frac{(-1)^n}{(4n+3)(2n+1)!}</math>
|}
[[009C_Sample_Final_2|'''Return to Sample Exam''']]

009C Sample Final 2, Problem 6 - Revision history

MathAdmin: Created page with "(a) Express the indefinite integral \int \sin(x^2)~dx as a power series. (b) Expr..."

MathAdmin: Created page with "(a) Express the indefinite integral $\int \sin(x^2)~dx$ as a power series. (b) Expr..."