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| | <span class="exam"> This problem has three parts: | | <span class="exam"> This problem has three parts: |
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| − | <span class="exam">(a) State the Fundamental Theorem of Calculus. | + | <span class="exam">(a) State the both parts of the fundamental theorem of calculus. |
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| − | <span class="exam">(b) Compute <math style="vertical-align: -15px">\frac{d}{dx}\int_0^{\cos (x)}\sin (t)~dt.</math> | + | <span class="exam">(b) Compute <math style="vertical-align: -15px">\frac{d}{dx}\int_0^{\cos (x)}\sin (t)~dt</math>. |
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| − | <span class="exam">(c) Evaluate <math style="vertical-align: -14px">\int_{0}^{\pi/4}\sec^2 x~dx.</math> | + | <span class="exam">(c) Evaluate <math style="vertical-align: -14px">\int_{0}^{\pi/4}\sec^2 x~dx</math>. |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009B Sample Midterm 2, Problem 1 Solution|'''<u>Solution</u>''']] |
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| − | |'''1.''' What does Part 1 of the Fundamental Theorem of Calculus say about <math style="vertical-align: -15px">\frac{d}{dx}\int_0^x\sin(t)~dt?</math> | |
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| − | Part 1 of the Fundamental Theorem of Calculus says that
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| − | | <math style="vertical-align: -15px">\frac{d}{dx}\int_0^x\sin(t)~dt=\sin(x).</math>
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| − | |'''2.''' What does Part 2 of the Fundamental Theorem of Calculus say about <math style="vertical-align: -15px">\int_a^b\sec^2x~dx</math> where <math style="vertical-align: -5px">a,b</math> are constants?
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| − | Part 2 of the Fundamental Theorem of Calculus says that
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| − | | <math style="vertical-align: -15px">\int_a^b\sec^2x~dx=F(b)-F(a)</math> where <math style="vertical-align: 0px">F</math> is any antiderivative of <math style="vertical-align: 0px">\sec^2x.</math>
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| − | '''Solution:''' | + | [[009B Sample Midterm 2, Problem 1 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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| − | |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 style="vertical-align: -4px">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 style="vertical-align: -4px">f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: 0px">F</math> be any antiderivative of <math style="vertical-align: -4px">f.</math>
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| − | |Then,
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| − | | <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a).</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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| − | |Let <math style="vertical-align: -15px">F(x)=\int_0^{\cos (x)}\sin (t)~dt.</math>
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| − | |The problem is asking us to find <math style="vertical-align: -5px">F'(x).</math>
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| − | |Let <math style="vertical-align: -5px">g(x)=\cos(x)</math> and <math style="vertical-align: -14px">G(x)=\int_0^x \sin(t)~dt.</math>
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| − | |Then,
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| − | | <math style="vertical-align: -5px">F(x)=G(g(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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| − | |If we take the derivative of both sides of the last equation,
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| − | |we get
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| − | | <math style="vertical-align: -5px">F'(x)=G'(g(x))g'(x)</math>
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| − | |by the Chain Rule.
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 3:
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| − | |Now, <math style="vertical-align: -5px">g'(x)=-\sin(x)</math> and <math style="vertical-align: -5px">G'(x)=\sin(x)</math>
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| − | |by the '''Fundamental Theorem of Calculus, Part 1'''.
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| − | |Since
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| − | | <math style="vertical-align: -4px">G'(g(x))=\sin(g(x))=\sin(\cos(x)),</math>
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| − | |we have
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| − | | <math style="vertical-align: -5px">F'(x)=G'(g(x))\cdot g'(x)=\sin(\cos(x))\cdot(-\sin(x)).</math>
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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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| − | | Using the '''Fundamental Theorem of Calculus, Part 2''', we have
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| − | | <math>\int_{0}^{\frac{\pi}{4}}\sec^2 x~dx=\tan(x)\biggr|_{0}^{\pi/4}.</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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| − | |So, we get
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| − | | <math style="vertical-align: -16px">\int_{0}^{\frac{\pi}{4}}\sec^2 x~dx=\tan \bigg(\frac{\pi}{4}\bigg)-\tan (0)=1.</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)''' See solution above.
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| − | | '''(b)''' <math style="vertical-align: -5px">\sin(\cos(x))\cdot(-\sin(x))</math>
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| − | | '''(c)''' <math style="vertical-align: -3px">1</math>
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| − | |}
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| | [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] | | [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] |
