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| | ::<math>F(x)=\int_{\cos (x)}^5 \frac{1}{1+u^{10}}~du.</math> | | ::<math>F(x)=\int_{\cos (x)}^5 \frac{1}{1+u^{10}}~du.</math> |
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| | + | [[009B Sample Midterm 3, Problem 2 Solution|'''<u>Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Foundations:
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| − | |What does Part 1 of the Fundamental Theorem of Calculus say is the derivative of <math style="vertical-align: -16px">G(x)=\int_x^5 \frac{1}{1+u^{10}}~du?</math>
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| − | ::First, we need to switch the bounds of integration.
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| − | ::So, we have <math style="vertical-align: -16px">G(x)=-\int_5^x \frac{1}{1+u^{10}}~du.</math>
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| − | ::By Part 1 of the Fundamental Theorem of Calculus, <math style="vertical-align: -16px">G'(x)=-\frac{1}{1+x^{10}}.</math>
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| − | |}
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| − | '''Solution:'''
| + | [[009B Sample Midterm 3, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| − | {| 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>[a,b]</math> and let <math>F(x)=\int_a^x f(t)~dt.</math>
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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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| − | |'''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>F</math> be any antiderivative of <math>f.</math>
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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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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |First, we have <math>F(x)=-\int_5^{\cos (x)} \frac{1}{1+u^{10}}~du.</math>
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| − | |Now, let <math>g(x)=\cos(x)</math> and <math>G(x)=\int_5^x \frac{1}{1+u^{10}}~du.</math>
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| − | |So, <math>F(x)=-G(g(x)).</math>
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| − | |Hence, <math>F'(x)=-G'(g(x))g'(x)</math> 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>g'(x)=-\sin(x).</math>
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| − | | By the Fundamental Theorem of Calculus, <math>G'(x)=\frac{1}{1+x^{10}}.</math>
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| − | |Hence, <math>F'(x)=-\frac{1}{1+\cos^{10}x}(-\sin(x))=\frac{\sin(x)}{1+\cos^{10}x}.</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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| − | |'''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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| − | |<math>F'(x)=\frac{\sin(x)}{1+\cos^{10}x}</math>
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| − | |}
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| | [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] | | [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] |
