Difference between revisions of "009B Sample Final 1, Problem 2"
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<span class="exam"> We would like to evaluate | <span class="exam"> We would like to evaluate | ||
| − | + | ::<math>\frac{d}{dx}\bigg(\int_{-1}^{x} \sin(t^2)2t\,dt\bigg).</math> | |
| − | + | <span class="exam">(a) Compute <math style="vertical-align: -15px">f(x)=\int_{-1}^{x} \sin(t^2)2t\,dt.</math> | |
| − | + | <span class="exam">(b) Find <math style="vertical-align: -5px">f'(x).</math> | |
| − | + | <span class="exam">(c) State the Fundamental Theorem of Calculus. | |
| − | + | <span class="exam">(d) Use the Fundamental Theorem of Calculus to compute <math style="vertical-align: -15px">\frac{d}{dx}\bigg(\int_{-1}^{x} \sin(t^2)2t\,dt\bigg)</math> without first computing the integral. | |
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
!Foundations: | !Foundations: | ||
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| − | |How would you integrate <math>\int e^{x^2}2x~dx?</math> | + | |How would you integrate <math>\int e^{x^2}2x~dx?</math> |
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| − | + | You could use <math style="vertical-align: -1px">u</math>-substitution. | |
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| + | | Let <math style="vertical-align: 0px">u=x^2.</math> Then, <math style="vertical-align: 0px">du=2xdx.</math> | ||
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| − | + | So, we get <math style="vertical-align: -14px">\int e^u~du=e^u+C=e^{x^2}+C.</math> | |
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'''Solution:''' | '''Solution:''' | ||
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!Step 1: | !Step 1: | ||
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| − | |We proceed using <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: 0px">u=t^2.</math> Then, <math style="vertical-align: 0px">du=2t\,dt.</math> | + | |We proceed using <math style="vertical-align: 0px">u</math>-substitution. |
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| + | |Let <math style="vertical-align: 0px">u=t^2.</math> Then, <math style="vertical-align: 0px">du=2t\,dt.</math> | ||
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|Since this is a definite integral, we need to change the bounds of integration. | |Since this is a definite integral, we need to change the bounds of integration. | ||
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| − | |Plugging our values into the equation <math style="vertical-align: -4px">u=t^2,</math> we get <math style="vertical-align: -5px">u_1=(-1)^2=1</math> and <math style="vertical-align: -3px">u_2=x^2.</math> | + | |Plugging our values into the equation <math style="vertical-align: -4px">u=t^2,</math> we get |
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| + | | <math style="vertical-align: -5px">u_1=(-1)^2=1</math> and <math style="vertical-align: -3px">u_2=x^2.</math> | ||
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| − | + | <math>\begin{array}{rcl} | |
f(x) & = & \displaystyle{\int_{-1}^{x} \sin(t^2)2t~dt}\\ | f(x) & = & \displaystyle{\int_{-1}^{x} \sin(t^2)2t~dt}\\ | ||
&&\\ | &&\\ | ||
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!Step 1: | !Step 1: | ||
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| − | |From part (a), we have <math style="vertical-align: -5px">f(x)=-\cos(x^2)+\cos(1).</math> | + | |From part (a), we have <math style="vertical-align: -5px">f(x)=-\cos(x^2)+\cos(1).</math> |
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!Step 2: | !Step 2: | ||
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| − | |If we take the derivative, we get <math style="vertical-align: -5px">f'(x)=\sin(x^2)2x,</math> since <math style="vertical-align: -5px">\cos(1)</math> is | + | |If we take the derivative, we get <math style="vertical-align: -5px">f'(x)=\sin(x^2)2x,</math> since <math style="vertical-align: -5px">\cos(1)</math> is a constant. |
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|The Fundamental Theorem of Calculus has two parts. | |The Fundamental Theorem of Calculus has two parts. | ||
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| − | |''' | + | |'''The Fundamental Theorem of Calculus, Part 1''' |
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| − | | | + | | Let <math>f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: -14px">F(x)=\int_a^x f(t)~dt.</math> |
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| − | | | + | | Then, <math style="vertical-align: 0px">F</math> is a differentiable function on <math style="vertical-align: -5px">(a,b)</math> and <math style="vertical-align: -5px">F'(x)=f(x).</math> |
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!Step 2: | !Step 2: | ||
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| − | |''' | + | |'''The Fundamental Theorem of Calculus, Part 2''' |
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| − | | | + | | Let <math>f</math> be continuous on <math>[a,b]</math> and let <math style="vertical-align: 0px">F</math> be any antiderivative of <math>f.</math> |
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| − | | | + | | Then, <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a).</math> |
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| − | + | <math>\frac{d}{dx}\bigg(\int_{-1}^{x} \sin(t^2)2t~dt\bigg)\,=\,\sin(x^2)2x.</math> | |
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
!Final Answer: | !Final Answer: | ||
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| − | | '''(a)''' <math>f(x)=-\cos(x^2)+\cos(1)</math> | + | | '''(a)''' <math>f(x)=-\cos(x^2)+\cos(1)</math> |
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| − | | | + | | '''(b)''' <math>f'(x)=\sin(x^2)2x</math> |
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| − | | | + | | '''(c)''' See above |
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| − | | '''(d)''' <math style="vertical-align: -5px">\sin(x^2)2x</math> | + | | '''(d)''' <math style="vertical-align: -5px">\sin(x^2)2x</math> |
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[[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | [[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | ||
Latest revision as of 08:41, 10 April 2017
We would like to evaluate
(a) Compute Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f(x)=\int _{-1}^{x}\sin(t^{2})2t\,dt.}
(b) Find Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f'(x).}
(c) State the Fundamental Theorem of Calculus.
(d) Use the Fundamental Theorem of Calculus to compute without first computing the integral.
| Foundations: |
|---|
| How would you integrate Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int e^{x^{2}}2x~dx?} |
|
You could use -substitution. |
| Let Then, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle du=2xdx.} |
|
So, we get Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int e^{u}~du=e^{u}+C=e^{x^{2}}+C.} |
Solution:
(a)
| Step 1: |
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| We proceed using -substitution. |
| Let Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=t^{2}.} Then, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle du=2t\,dt.} |
| Since this is a definite integral, we need to change the bounds of integration. |
| Plugging our values into the equation Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=t^{2},} we get |
| and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u_{2}=x^{2}.} |
| Step 2: |
|---|
| So, we have |
|
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}f(x)&=&\displaystyle {\int _{-1}^{x}\sin(t^{2})2t~dt}\\&&\\&=&\displaystyle {\int _{1}^{x^{2}}\sin(u)~du}\\&&\\&=&\displaystyle {-\cos(u){\bigg |}_{1}^{x^{2}}}\\&&\\&=&\displaystyle {-\cos(x^{2})+\cos(1)}.\\\end{array}}} |
(b)
| Step 1: |
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| From part (a), we have |
| Step 2: |
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| If we take the derivative, we get Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f'(x)=\sin(x^{2})2x,} since Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \cos(1)} is a constant. |
(c)
| Step 1: |
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| The Fundamental Theorem of Calculus has two parts. |
| The Fundamental Theorem of Calculus, Part 1 |
| Let be continuous on and let |
| Then, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle F} is a differentiable function on and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle F'(x)=f(x).} |
| Step 2: |
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| The Fundamental Theorem of Calculus, Part 2 |
| Let be continuous on and let Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle F} be any antiderivative of Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f.} |
| Then, |
| (d) |
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| By the Fundamental Theorem of Calculus, Part 1, |
|
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| Final Answer: |
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| (a) |
| (b) Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f'(x)=\sin(x^{2})2x} |
| (c) See above |
| (d) 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(x^2)2x} |