Difference between revisions of "009B Sample Midterm 3, Problem 2"

Revision as of 18:18, 29 March 2016

State the fundamental theorem of calculus, and use this theorem to find the derivative of

F(x)=\int _{\cos(x)}^{5}{\frac {1}{1+u^{10}}}~du.

Foundations:
What does Part 1 of the Fundamental Theorem of Calculus say is the derivative of $G(x)=\int _{x}^{5}{\frac {1}{1+u^{10}}}~du?$
First, we need to switch the bounds of integration.
So, we have $G(x)=-\int _{5}^{x}{\frac {1}{1+u^{10}}}~du.$
By Part 1 of the Fundamental Theorem of Calculus, $G'(x)=-{\frac {1}{1+x^{10}}}.$

Solution:

Step 1:
The Fundamental Theorem of Calculus, Part 1
Let $f$ be continuous on $[a,b]$ and let $F(x)=\int _{a}^{x}f(t)~dt.$
Then, $F$ is a differentiable function on $(a,b)$ and $F'(x)=f(x).$
The Fundamental Theorem of Calculus, Part 2
Let $f$ be continuous on $[a,b]$ and let $F$ be any antiderivative of $f.$
Then, $\int _{a}^{b}f(x)~dx=F(b)-F(a).$

Step 2:
First, we have $F(x)=-\int _{5}^{\cos(x)}{\frac {1}{1+u^{10}}}~du.$
Now, let $g(x)=\cos(x)$ and $G(x)=\int _{5}^{x}{\frac {1}{1+u^{10}}}~du.$
So, $F(x)=-G(g(x)).$
Hence, $F'(x)=-G'(g(x))g'(x)$ by the Chain Rule.

Step 3:
Now, $g'(x)=-\sin(x).$
By the Fundamental Theorem of Calculus, $G'(x)={\frac {1}{1+x^{10}}}.$
Hence, $F'(x)=-{\frac {1}{1+\cos ^{10}x}}(-\sin(x))={\frac {\sin(x)}{1+\cos ^{10}x}}.$

Final Answer:
The Fundamental Theorem of Calculus, Part 1
Let $f$ be continuous on $[a,b]$ and let $F(x)=\int _{a}^{x}f(t)~dt.$
Then, $F$ is a differentiable function on $(a,b)$ and $F'(x)=f(x).$
The Fundamental Theorem of Calculus, Part 2
Let $f$ be continuous on $[a,b]$ and let $F$ be any antiderivative of $f.$
Then, $\int _{a}^{b}f(x)~dx=F(b)-F(a).$
$F'(x)={\frac {\sin(x)}{1+\cos ^{10}x}}$

Return to Sample Exam

@@ Line 23: / Line 23: @@
 !Step 1: &nbsp;
 |-
-|The Fundamental Theorem of Calculus has two parts.
+|'''The Fundamental Theorem of Calculus, Part 1'''
-|-
-|'''The Fundamental Theorem of Calculus, Part 1'''
 |-
-|Let <math>f</math> be continuous on <math>[a,b]</math> and let <math>F(x)=\int_a^x f(t)~dt.</math>
+|Let <math style="vertical-align: -5px">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>
 |-
-|Then, <math>F</math> is a differentiable function on <math>(a,b)</math> and <math>F'(x)=f(x).</math>
+|Then, <math style="vertical-align: -1px">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>
 |-
 |'''The Fundamental Theorem of Calculus, Part 2'''
 |-
-|Let <math>f</math> be continuous on <math>[a,b]</math> and let <math>F</math> be any antiderivative of <math>f.</math>
+|Let <math style="vertical-align: -5px">f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: -1px">F</math> be any antiderivative of <math style="vertical-align: -5px">f.</math>
 |-
-|Then, <math>\int_a^b f(x)~dx=F(b)-F(a).</math>
+|Then, <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a).</math>
 |}
@@ Line 41: / Line 39: @@
 !Step 2: &nbsp;
 |-
-|First, we have <math>F(x)=-\int_5^{\cos (x)} \frac{1}{1+u^{10}}~du.</math>
+|First, we have <math style="vertical-align: -15px">F(x)=-\int_5^{\cos (x)} \frac{1}{1+u^{10}}~du.</math>
 |-
-|Now, let <math>g(x)=\cos(x)</math> and <math>G(x)=\int_5^x \frac{1}{1+u^{10}}~du.</math>
+|Now, let <math style="vertical-align: -5px">g(x)=\cos(x)</math> and <math style="vertical-align: -15px">G(x)=\int_5^x \frac{1}{1+u^{10}}~du.</math>
 |-
-|So, <math>F(x)=-G(g(x)).</math>
+|So, <math style="vertical-align: -5px">F(x)=-G(g(x)).</math>
 |-
-|Hence, <math>F'(x)=-G'(g(x))g'(x)</math> by the Chain Rule.
+|Hence, <math style="vertical-align: -5px">F'(x)=-G'(g(x))g'(x)</math> by the Chain Rule.
 |}
@@ Line 53: / Line 51: @@
 !Step 3: &nbsp;
 |-
-|Now, <math>g'(x)=-\sin(x).</math>
+|Now, <math style="vertical-align: -5px">g'(x)=-\sin(x).</math>
 |-
-| By the Fundamental Theorem of Calculus, <math>G'(x)=\frac{1}{1+x^{10}}.</math>
+| By the Fundamental Theorem of Calculus, <math style="vertical-align: -15px">G'(x)=\frac{1}{1+x^{10}}.</math>
 |-
-|Hence, <math>F'(x)=-\frac{1}{1+\cos^{10}x}(-\sin(x))=\frac{\sin(x)}{1+\cos^{10}x}.</math>
+|Hence, <math style="vertical-align: -15px">F'(x)=-\frac{1}{1+\cos^{10}x}(-\sin(x))=\frac{\sin(x)}{1+\cos^{10}x}.</math>
 |-
 |
@@ Line 67: / Line 65: @@
 |'''The Fundamental Theorem of Calculus, Part 1'''
 |-
-|Let <math>f</math> be continuous on <math>[a,b]</math> and let <math>F(x)=\int_a^x f(t)~dt.</math>
+|Let <math style="vertical-align: -5px">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>
 |-
-|Then, <math>F</math> is a differentiable function on <math>(a,b)</math> and <math>F'(x)=f(x).</math>
+|Then, <math style="vertical-align: -1px">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>
 |-
 |'''The Fundamental Theorem of Calculus, Part 2'''
 |-
-|Let <math>f</math> be continuous on <math>[a,b]</math> and let <math>F</math> be any antiderivative of <math>f.</math>
+|Let <math style="vertical-align: -5px">f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: -1px">F</math> be any antiderivative of <math style="vertical-align: -5px">f.</math>
 |-
-|Then, <math>\int_a^b f(x)~dx=F(b)-F(a).</math>
+|Then, <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a).</math>
 |-
 |<math>F'(x)=\frac{\sin(x)}{1+\cos^{10}x}</math>
 |}
 [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

Difference between revisions of "009B Sample Midterm 3, Problem 2"

Revision as of 18:18, 29 March 2016

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