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| | <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. | | <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;"
| + | <hr> |
| − | !Foundations:
| + | [[009B Sample Final 1, Problem 2 Solution|'''<u>Solution</u>''']] |
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| − | |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:''' | + | [[009B Sample Final 1, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | '''(a)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |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.
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| − | |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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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | ||So, we have
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| − | <math>\begin{array}{rcl}
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| − | f(x) & = & \displaystyle{\int_{-1}^{x} \sin(t^2)2t~dt}\\
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| − | &&\\
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| − | & = & \displaystyle{\int_{1}^{x^2} \sin(u)~du}\\
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| − | &&\\
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| − | & = & \displaystyle{-\cos(u)\bigg|_{1}^{x^2}}\\
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| − | &&\\
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| − | & = & \displaystyle{-\cos(x^2)+\cos(1)}.\\
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| − | \end{array}</math>
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| − | |}
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| − | '''(b)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |From part (a), we have <math style="vertical-align: -5px">f(x)=-\cos(x^2)+\cos(1).</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !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 a constant.
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| − | |}
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| − | '''(c)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |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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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !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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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !(d)
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| − | |By the '''Fundamental Theorem of Calculus, Part 1''',
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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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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | | '''(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>
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| − | |}
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| | [[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | | [[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] |
