009B Sample Midterm 2, Problem 1 - Revision history

MathAdmin at 01:08, 13 November 2017

2017-11-13T01:08:05Z

MathAdmin at 19:14, 9 April 2017

2017-04-09T19:14:40Z

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MathAdmin at 22:06, 18 April 2016

2016-04-18T22:06:11Z

MathAdmin at 21:36, 18 April 2016

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MathAdmin at 22:11, 8 April 2016

2016-04-08T22:11:36Z

MathAdmin at 21:38, 8 February 2016

2016-02-08T21:38:17Z

MathAdmin at 21:37, 8 February 2016

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MathAdmin at 18:49, 3 February 2016

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MathAdmin at 16:16, 3 February 2016

2016-02-03T16:16:50Z

MathAdmin at 07:19, 3 February 2016

2016-02-03T07:19:03Z

@@ Line 1: / Line 1: @@
 <span class="exam"> This problem has three parts:
-<span class="exam">(a) State the Fundamental Theorem of Calculus.
+<span class="exam">(a) State the both parts of the fundamental theorem of calculus.
-<span class="exam">(b) Compute &nbsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^{\cos (x)}\sin (t)~dt.</math>
+<span class="exam">(b) Compute &nbsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^{\cos (x)}\sin (t)~dt</math>.
-<span class="exam">(c) Evaluate &nbsp;<math style="vertical-align: -14px">\int_{0}^{\pi/4}\sec^2 x~dx.</math>
+<span class="exam">(c) Evaluate &nbsp;<math style="vertical-align: -14px">\int_{0}^{\pi/4}\sec^2 x~dx</math>.
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009B Sample Midterm 2, Problem 1 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009B Sample Midterm 2, Problem 1 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 88: / Line 88: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)''' The left-endpoint Riemann sum is <math style="vertical-align: -20px">\frac{205}{144}</math>, which overestimates the area of <math style="vertical-align: 0px">S</math>.
+|&nbsp;&nbsp; '''(a)''' The left-endpoint Riemann sum is <math style="vertical-align: -20px">\frac{205}{144}</math>, which overestimates the area of <math style="vertical-align: 0px">S</math>.
 |-
-|'''(b)''' Using left-endpoint Riemann sums:
+|&nbsp;&nbsp; '''(b)''' Using left-endpoint Riemann sums:
 |-
+|

@@ Line 1: / Line 1: @@
 <span class="exam"> Consider the region <math style="vertical-align: 0px">S</math> bounded by <math style="vertical-align: -13px">x=1,x=5,y=\frac{1}{x^2}</math>&thinsp; and the <math>x</math>-axis.
-::<span class="exam">a) Use four rectangles and a Riemann sum to approximate the area of the region <math style="vertical-align: 0px">S</math>. Sketch the region <math style="vertical-align: 0px">S</math> and the rectangles and indicate whether your rectangles overestimate or underestimate the area of <math style="vertical-align: 0px">S</math>.
+::<span class="exam">a) Use four rectangles and a Riemann sum to approximate the area of the region <math style="vertical-align: 0px">S</math>. Sketch the region <math style="vertical-align: 0px">S</math> and the rectangles and
 ::<span class="exam">b) Find an expression for the area of the region <math style="vertical-align: 0px">S</math> as a limit. Do not evaluate the limit.
@@ Line 12: / Line 14: @@
 |Recall:
 |-
 |'''1.''' The height of each rectangle in the left-hand Riemann sum is given by choosing the left endpoint of the interval.
 |-
-|'''2.''' The height of each rectangle in the right-hand Riemann sum is given by choosing the right endpoint of the interval.
 |-
 |'''3.''' See the page on [[Riemann_Sums|'''Riemann Sums''']] for more information.
 |}
@@ Line 25: / Line 36: @@
 !Step 1: &nbsp;
 |-
-|Let <math style="vertical-align: -13px">f(x)=\frac{1}{x^2}</math>.  Since our interval is <math style="vertical-align: -5px">[1,5]</math> and we are using 4 rectangles, each rectangle has width 1. Since the problem doesn't specify, we can choose either right- or left-endpoints.  Choosing left-endpoints, the Riemann sum is
+|Let <math style="vertical-align: -13px">f(x)=\frac{1}{x^2}.</math>  Since our interval is <math style="vertical-align: -5px">[1,5]</math> and we are using <math style="vertical-align: -1px">4</math> rectangles, each rectangle has width <math style="vertical-align: -1px">1.</math> Since the problem doesn't specify, we can choose either right- or left-endpoints.  Choosing left-endpoints, the Riemann sum is
 |-
+|
 |}
@@ Line 37: / Line 47: @@
 |Thus, the left-endpoint Riemann sum is
 |-
-| &nbsp;&nbsp; <math>1\cdot (f(1)+f(2)+f(3)+f(4))=\bigg(1+\frac{1}{4}+\frac{1}{9}+{1}{16}\bigg)=\frac{205}{144}</math>.
 |-
-|The left-endpoint Riemann sum overestimates the area of <math style="vertical-align: 0px">S</math>.
+|The left-endpoint Riemann sum overestimates the area of <math style="vertical-align: 0px">S.</math>
 |}
@@ Line 46: / Line 61: @@
 !Step 1: &nbsp;
 |-
-|Let <math style="vertical-align: 0px">n</math> be the number of rectangles used in the left-endpoint Riemann sum for <math style="vertical-align: -13px">f(x)=\frac{1}{x^2}</math>.
+|Let <math style="vertical-align: 0px">n</math> be the number of rectangles used in the left-endpoint Riemann sum for <math style="vertical-align: -13px">f(x)=\frac{1}{x^2}.</math>
 |-
+|The width of each rectangle is
 |-
+|
 |}
@@ Line 60: / Line 74: @@
 |So, the left-endpoint Riemann sum is
 |-
-| &nbsp;&nbsp; <math>\Delta x \bigg(f(1)+f\bigg(1+\frac{4}{n}\bigg)+f\bigg(1+2\frac{4}{n}\bigg)+\ldots +f\bigg(1+(n-1)\frac{4}{n}\bigg)\bigg)</math>.
 |-
 |Now, we let <math style="vertical-align: 0px">n</math> go to infinity to get a limit.
 |-
-|So, the area of <math style="vertical-align: 0px">S</math> is equal to <math style="vertical-align: -20px">\lim_{n\to\infty} \frac{4}{n}\sum_{i=0}^{n-1}f\bigg(1+i\frac{4}{n}\bigg)\,=\,\lim_{n\to\infty} \frac{4}{n}\sum_{i=0}^{n-1}\bigg(1+i\frac{4}{n}\bigg)^{-2}</math>.
+|So, the area of <math style="vertical-align: 0px">S</math> is equal to
 |}
@@ Line 74: / Line 92: @@
 |'''(b)''' Using left-endpoint Riemann sums:
 |-
-| &nbsp;&nbsp; <math style="vertical-align: -20px">\lim_{n\to\infty} \frac{4}{n}\sum_{i=0}^{n-1}\bigg(1+i\frac{4}{n}\bigg)^{-2}</math>
 |}
 [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 10: / Line 10: @@
 !Foundations: &nbsp;
 |-
-|See the page on [[Riemann_Sums|'''Riemann Sums''']].
+|Recall:
 |}

@@ Line 25: / Line 25: @@
+|
 |}
-== Temp 1 ==
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 2: &nbsp;

← Older revision		Revision as of 16:16, 3 February 2016
Line 1:		Line 1:
−	<span class="exam"> Consider the region <math style="vertical-align: 0px">S</math> bounded by <math style="vertical-align: -13px">x=1,x=5,y=\frac{1}{x^2}</math> and the <math>x</math>-axis.	+	<span class="exam"> Consider the region <math style="vertical-align: 0px">S</math> bounded by <math style="vertical-align: -13px">x=1,x=5,y=\frac{1}{x^2}</math>  and the <math>x</math>-axis.

	::<span class="exam">a) Use four rectangles and a Riemann sum to approximate the area of the region <math style="vertical-align: 0px">S</math>. Sketch the region <math style="vertical-align: 0px">S</math> and the rectangles and indicate whether your rectangles overestimate or underestimate the area of <math style="vertical-align: 0px">S</math>.		::<span class="exam">a) Use four rectangles and a Riemann sum to approximate the area of the region <math style="vertical-align: 0px">S</math>. Sketch the region <math style="vertical-align: 0px">S</math> and the rectangles and indicate whether your rectangles overestimate or underestimate the area of <math style="vertical-align: 0px">S</math>.

← Older revision		Revision as of 07:19, 3 February 2016
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−	[[File:9AMT2_1GP.png\|right\|400px\|frame\|Approximation with left endpoints is an overestimate.]]	+	[[File:9AMT2_1GP.png\|right\|400px\|frame\|Approximation of integral with left endpoints is an overestimate.]]

	{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"		{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"