009B Sample Final 2, Problem 1 - Revision history

Kayla Murray at 00:24, 3 December 2017

2017-12-03T00:24:09Z

2017-03-04T22:41:10Z

2017-03-04T22:37:14Z

2017-03-04T22:32:27Z

2017-03-03T20:53:40Z

2017-02-19T02:22:50Z

2017-02-18T03:07:07Z

@@ Line 9: / Line 9: @@
 ::<math>\frac{d}{dx}\int_1^{\frac{1}{x}} \sin t~dt</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009B Sample Final 2, Problem 1 Solution|'''<u>Solution</u>''']]
-'''(a)'''
+[[009B Sample Final 2, Problem 1 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 60: / Line 60: @@
 |The Fundamental Theorem of Calculus Part 2 says that
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\int_0^1 \frac{d}{dx}(e^{\arctan(x)})~dx=F(1)-F(0)</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\int_0^1 \frac{d}{dx}\bigg(e^{\arctan(x)}\bigg)~dx=F(1)-F(0)</math>
 |-
-|where &nbsp;<math>F(x)</math>&nbsp; is any antiderivative of &nbsp;<math>\frac{d}{dx}(e^{\arctan(x)}).</math>
+|where &nbsp;<math style="vertical-align: -5px">F(x)</math>&nbsp; is any antiderivative of &nbsp;<math style="vertical-align: -15px">\frac{d}{dx}\bigg(e^{\arctan(x)}\bigg).</math>
 |-
 |Thus, we can take
@@ Line 68: / Line 68: @@
 |&nbsp; &nbsp; &nbsp; &nbsp;<math>F(x)=e^{\arctan(x)}</math>
 |-
-|since then <math>F'(x)=\frac{d}{dx}(e^{\arctan(x)}).</math>
+|since then <math style="vertical-align: -15px">F'(x)=\frac{d}{dx}\bigg(e^{\arctan(x)}\bigg).</math>
 |}
@@ Line 77: / Line 77: @@
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\int_0^1 \frac{d}{dx}(e^{\arctan(x)})~dx} & = & \displaystyle{F(1)-F(0)}\\
+\displaystyle{\int_0^1 \frac{d}{dx}\bigg(e^{\arctan(x)}\bigg)~dx} & = & \displaystyle{F(1)-F(0)}\\
 &&\\
-& = & \displaystyle{e^{\arctan(1}-e^{\arctan(0)}}\\
+& = & \displaystyle{e^{\arctan(1)}-e^{\arctan(0)}}\\
 &&\\
 & = & \displaystyle{e^{\frac{\pi}{4}}-e^0}\\

@@ Line 52: / Line 52: @@
 !Step 1: &nbsp;
 |-
+|The Fundamental Theorem of Calculus Part 2 says that
 |-
+|since then <math>F'(x)=\frac{d}{dx}(e^{\arctan(x)}).</math>
 |}
@@ Line 60: / Line 68: @@
 !Step 2: &nbsp;
 |-
+|Now, we have
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |}
@@ Line 70: / Line 86: @@
 !Step 1: &nbsp;
 |-
+|Using the Fundamental Theorem of Calculus Part 1 and the Chain Rule, we have
 |-
+|
@@ Line 78: / Line 96: @@
 !Step 2: &nbsp;
 |-
+|Hence, we have
 |-
+|
@@ Line 88: / Line 108: @@
 |&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp; See above
 |-
-|'''(b)'''
+|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp; <math>e^{\frac{\pi}{4}}-1</math>
 |-
-|'''(c)'''
+|&nbsp; &nbsp;'''(c)'''&nbsp; &nbsp; <math>\sin\bigg(\frac{1}{x}\bigg)\bigg(-\frac{1}{x^2}\bigg)</math>
 |}
 [[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
-::<span class="exam">a) State '''both parts''' of the Fundamental Theorem of Calculus.
+<span class="exam">(a) State '''both parts''' of the Fundamental Theorem of Calculus.
-::<span class="exam">b) Evaluate the integral
+<span class="exam">(b) Evaluate the integral
-::::<math>\int_0^1 \frac{d}{dx} \bigg(e^{\tan^{-1}(x)}\bigg)dx</math>
+::<math>\int_0^1 \frac{d}{dx} \bigg(e^{\tan^{-1}(x)}\bigg)dx</math>
-::<span class="exam">c) Compute
+<span class="exam">(c) Compute
-::::<math>\frac{d}{dx}\int_1^{\frac{1}{x}} \sin t~dt</math>
+::<math>\frac{d}{dx}\int_1^{\frac{1}{x}} \sin t~dt</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 1: / Line 1: @@
-<span class="exam">Consider the region bounded by the following two functions:
+::<span class="exam">a) State '''both parts''' of the Fundamental Theorem of Calculus.
-<span class="exam">a) Using the lower sum with three rectangles having equal width, approximate the area.
+::<span class="exam">b) Evaluate the integral
-<span class="exam">b) Using the upper sum with three rectangles having equal width, approximate the area.
+::::<math>\int_0^1 \frac{d}{dx} \bigg(e^{\tan^{-1}(x)}\bigg)dx</math>
-<span class="exam">c) Find the actual area of the region.
+::<span class="exam">c) Compute
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 11: / Line 11: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Foundations: &nbsp;
 |-
+|
 |-
+|'''2.''' What does Part 1 of the Fundamental Theorem of Calculus say about &nbsp;<math style="vertical-align: -15px">\frac{d}{dx}\int_0^x\sin(t)~dt?</math>
 |-
+|
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^x\sin(t)~dt=\sin(x).</math>
 |}

@@ Line 28: / Line 28: @@
 !Step 1: &nbsp;
 |-
+|The Fundamental Theorem of Calculus has two parts.
 |-
+|'''The Fundamental Theorem of Calculus, Part 1'''
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;Let &nbsp;<math>f</math>&nbsp; be continuous on &nbsp;<math style="vertical-align: -5px">[a,b]</math>&nbsp; and let &nbsp;<math style="vertical-align: -14px">F(x)=\int_a^x f(t)~dt.</math>
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;Then, &nbsp;<math style="vertical-align: 0px">F</math>&nbsp; is a differentiable function on &nbsp;<math style="vertical-align: -5px">(a,b)</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">F'(x)=f(x).</math>
 |}
@@ Line 40: / Line 40: @@
 !Step 2: &nbsp;
 |-
+|'''The Fundamental Theorem of Calculus, Part 2'''
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;Let &nbsp;<math>f</math>&nbsp; be continuous on &nbsp;<math>[a,b]</math>&nbsp; and let &nbsp;<math style="vertical-align: 0px">F</math>&nbsp; be any antiderivative of &nbsp;<math>f.</math>
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;Then, &nbsp;<math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a).</math>
 |}
@@ Line 88: / Line 86: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)'''
+|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp; See above
 |-
 |'''(b)'''