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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)2tdt\bigg)</math>.
| + | ::<math>\frac{d}{dx}\bigg(\int_{-1}^{x} \sin(t^2)2t\,dt\bigg).</math> |
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| − | <span class="exam">a) Compute <math>f(x)=\int_{-1}^{x} \sin(t^2)2t~dt</math>. | + | <span class="exam">(a) Compute <math style="vertical-align: -15px">f(x)=\int_{-1}^{x} \sin(t^2)2t\,dt.</math> |
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| − | <span class="exam">b) Find <math>f'(x)</math>. | + | <span class="exam">(b) Find <math style="vertical-align: -5px">f'(x).</math> |
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| − | <span class="exam">c) State the fundamental theorem of calculus. | + | <span class="exam">(c) State the Fundamental Theorem of Calculus. |
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| − | <span class="exam">d) Use the fundamental theorem of calculus to compute <math>\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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| − | |<math>u</math>-substitution | |
| − | |}
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| − | '''Solution:'''
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| − | '''(a)''' | + | [[009B Sample Final 1, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |We proceed using <math>u</math>-substitution. Let <math>u=t^2</math>. Then, <math>du=2tdt</math>.
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| − | |-
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| − | |Since this is a definite integral, we need to change the bounds of integration.
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| − | |-
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| − | |Plugging in our values into the equation <math>u=t^2</math>, we get <math>u_1=(-1)^2=1</math> and <math>u_2=x^2</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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| − | |-
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| − | ||So, we have
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| − | |-
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| − | |<math>f(x)=\int_{-1}^{x} \sin(t^2)2t~dt=\int_{1}^{x^2} \sin(u)du=\left.-\cos(u)\right|_{1}^{x^2}=-\cos(x^2)+\cos(1)</math>.
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| − | |}
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| − |
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| − | '''(b)'''
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |From part (a), we have <math>f(x)=-\cos(x^2)+\cos(1)</math>.
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| − | |}
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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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| − | |-
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| − | |If we take the derivative, we get <math>f'(x)=\sin(x^2)2x</math>.
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| − | |}
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| − |
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| − | '''(c)'''
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |The Fundamental Theorem of Calculus has two parts.
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| − | |-
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| − | |'''The Fundamental Theorem of Calculus, Part 1'''
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| − | |-
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| − | |Let <math>f</math> be continuous on <math>[a,b]</math> and let <math>F(x)=\int_a^x f(t)~dt</math>.
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| − | |-
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| − | |Then, <math>F</math> is a differentiable function on <math>(a,b)</math> and <math>F'(x)=f(x)</math>.
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| − | |}
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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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| − | |-
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| − | |'''The Fundamental Theorem of Calculus, Part 2'''
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| − | |-
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| − | |Let <math>f</math> be continuous on <math>[a,b]</math> and let <math>F</math> be any antiderivative of <math>f</math>.
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| − | |-
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| − | |Then, <math>\int_a^b f(x)~dx=F(b)-F(a)</math>
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| − | |}
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| − |
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| − | '''(d)'''
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |By the '''Fundamental Theorem of Calculus, Part 1''',
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| − | |-
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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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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | |-
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| − | |'''(a)''' <math>f(x)=-\cos(x^2)+\cos(1)</math>
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| − | |-
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| − | |'''(b)''' <math>f'(x)=\sin(x^2)2x</math>
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| − | |-
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| − | |'''(c)''' '''The Fundamental Theorem of Calculus, Part 1'''
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| − | |-
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| − | |Let <math>f</math> be continuous on <math>[a,b]</math> and let <math>F(x)=\int_a^x f(t)~dt</math>.
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| − | |-
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| − | |Then, <math>F</math> is a differentiable function on <math>(a,b)</math> and <math>F'(x)=f(x)</math>.
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| − | |-
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| − | |'''The Fundamental Theorem of Calculus, Part 2'''
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| − | |-
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| − | |Let <math>f</math> be continuous on <math>[a,b]</math> and let <math>F</math> be any antiderivative of <math>f</math>.
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| − | |-
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| − | |Then, <math>\int_a^b f(x)~dx=F(b)-F(a)</math>.
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| − | |-
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| − | |'''(d)''' <math>\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>''']] |
