009B Sample Midterm 1, Problem 5 - Revision history

Kayla Murray at 23:05, 12 November 2017

2017-11-12T23:05:14Z

Kayla Murray: Replaced content with " Evaluate the integral: :: $\int \sin^3x \cos^2x~dx$
(insert picture of handwritten solution) 009B Sample Midterm 1, Problem 5 Deta..."

2017-11-05T01:54:44Z

Replaced content with " <span class="exam"> Evaluate the integral: ::<math>\int \sin^3x \cos^2x~dx</math> <hr> (insert picture of handwritten solution) 009B Sample Midterm 1, Problem 5 Deta..."

Kayla Murray at 01:01, 27 February 2017

2017-02-27T01:01:36Z

Kayla Murray at 00:10, 19 February 2017

2017-02-19T00:10:43Z

Kayla Murray at 00:16, 8 February 2017

2017-02-08T00:16:14Z

Kayla Murray at 00:09, 8 February 2017

2017-02-08T00:09:41Z

Kayla Murray at 16:23, 7 February 2017

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Kayla Murray at 16:21, 7 February 2017

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Kayla Murray at 16:13, 6 February 2017

2017-02-06T16:13:41Z

Kayla Murray at 16:00, 6 February 2017

2017-02-06T16:00:41Z

@@ Line 5: / Line 5: @@
 <hr>
-(insert picture of handwritten solution)
+[[009B Sample Midterm 1, Problem 5 Solution|'''<u>Solution</u>''']]
 [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
-<span class="exam">(a) Compute the left-hand Riemann sum approximation of &nbsp;<math style="vertical-align: -14px">\int_0^3 f(x)~dx</math>&nbsp; with &nbsp;<math style="vertical-align: 0px">n=3</math>&nbsp; boxes.
+<span class="exam"> Evaluate the integral:
-<span class="exam">(b) Compute the right-hand Riemann sum approximation of &nbsp;<math style="vertical-align: -14px">\int_0^3 f(x)~dx</math>&nbsp; with &nbsp;<math style="vertical-align: 0px">n=3</math>&nbsp; boxes.
+::<math>\int \sin^3x \cos^2x~dx</math>
-<span class="exam">(c) Express &nbsp;<math style="vertical-align: -14px">\int_0^3 f(x)~dx</math>&nbsp; as a limit of right-hand Riemann sums (as in the definition of the definite integral). Do not evaluate the limit.
+<hr>
 [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
-<span class="exam">Let <math>f(x)=1-x^2</math>.
+<span class="exam">Let &nbsp;<math>f(x)=1-x^2</math>.
-<span class="exam">(a) Compute the left-hand Riemann sum approximation of <math style="vertical-align: -14px">\int_0^3 f(x)~dx</math> with <math style="vertical-align: 0px">n=3</math> boxes.
+<span class="exam">(a) Compute the left-hand Riemann sum approximation of &nbsp;<math style="vertical-align: -14px">\int_0^3 f(x)~dx</math>&nbsp; with &nbsp;<math style="vertical-align: 0px">n=3</math>&nbsp; boxes.
-<span class="exam">(b) Compute the right-hand Riemann sum approximation of <math style="vertical-align: -14px">\int_0^3 f(x)~dx</math> with <math style="vertical-align: 0px">n=3</math> boxes.
+<span class="exam">(b) Compute the right-hand Riemann sum approximation of &nbsp;<math style="vertical-align: -14px">\int_0^3 f(x)~dx</math>&nbsp; with &nbsp;<math style="vertical-align: 0px">n=3</math>&nbsp; boxes.
-<span class="exam">(c) Express <math style="vertical-align: -14px">\int_0^3 f(x)~dx</math> as a limit of right-hand Riemann sums (as in the definition of the definite integral). Do not evaluate the limit.
+<span class="exam">(c) Express &nbsp;<math style="vertical-align: -14px">\int_0^3 f(x)~dx</math>&nbsp; as a limit of right-hand Riemann sums (as in the definition of the definite integral). Do not evaluate the limit.
@@ Line 25: / Line 25: @@
 !Step 1: &nbsp;
 |-
-|Since our interval is <math style="vertical-align: -5px">[0,3]</math> and we are using 3 rectangles, each rectangle has width 1.
+|Since our interval is &nbsp;<math style="vertical-align: -5px">[0,3]</math>&nbsp; and we are using 3 rectangles, each rectangle has width 1.
 |-
 |So, the left-hand Riemann sum is
@@ Line 49: / Line 49: @@
 !Step 1: &nbsp;
 |-
-|Since our interval is <math style="vertical-align: -5px">[0,3]</math> and we are using 3 rectangles, each rectangle has width 1.
+|Since our interval is &nbsp;<math style="vertical-align: -5px">[0,3]</math>&nbsp; and we are using 3 rectangles, each rectangle has width 1.
 |-
 |So, the right-hand Riemann sum is
@@ Line 72: / Line 72: @@
 !Step 1: &nbsp;
 |-
-|Let <math style="vertical-align: 0px">n</math> be the number of rectangles used in the right-hand Riemann sum for <math style="vertical-align: -5px">f(x)=1-x^2.</math>
+|Let &nbsp;<math style="vertical-align: 0px">n</math>&nbsp; be the number of rectangles used in the right-hand Riemann sum for &nbsp;<math style="vertical-align: -5px">f(x)=1-x^2.</math>
 |-
 |The width of each rectangle is
@@ Line 86: / Line 86: @@
 | &nbsp;&nbsp; &nbsp; &nbsp; <math style="vertical-align: -14px">\Delta x \bigg(f\bigg(1\cdot \frac{3}{n}\bigg)+f\bigg(2\cdot \frac{3}{n}\bigg)+f\bigg(3\cdot \frac{3}{n}\bigg)+\ldots +f(3)\bigg).</math>
 |-
-|Finally, we let <math style="vertical-align: 0px">n</math> go to infinity to get a limit.
+|Finally, we let &nbsp;<math style="vertical-align: 0px">n</math>&nbsp; go to infinity to get a limit.
 |-
-|Thus, <math style="vertical-align: -14px">\int_0^3 f(x)~dx</math> is equal to <math style="vertical-align: -21px">\lim_{n\to\infty} \frac{3}{n}\sum_{i=1}^{n}f\bigg(i\frac{3}{n}\bigg).</math>
+|Thus, &nbsp;<math style="vertical-align: -14px">\int_0^3 f(x)~dx</math>&nbsp; is equal to &nbsp;<math style="vertical-align: -21px">\lim_{n\to\infty} \frac{3}{n}\sum_{i=1}^{n}f\bigg(i\frac{3}{n}\bigg).</math>
 |}

@@ Line 1: / Line 1: @@
 <span class="exam">Let <math>f(x)=1-x^2</math>.
-::<span class="exam">a) Compute the left-hand Riemann sum approximation of <math style="vertical-align: -14px">\int_0^3 f(x)~dx</math> with <math style="vertical-align: 0px">n=3</math> boxes.
+<span class="exam">(a) Compute the left-hand Riemann sum approximation of <math style="vertical-align: -14px">\int_0^3 f(x)~dx</math> with <math style="vertical-align: 0px">n=3</math> boxes.
-::<span class="exam">b) Compute the right-hand Riemann sum approximation of <math style="vertical-align: -14px">\int_0^3 f(x)~dx</math> with <math style="vertical-align: 0px">n=3</math> boxes.
-::<span class="exam">c) Express <math style="vertical-align: -14px">\int_0^3 f(x)~dx</math> as a limit of right-hand Riemann sums (as in the definition of the definite integral). Do not evaluate the limit.
+<span class="exam">(b) Compute the right-hand Riemann sum approximation of <math style="vertical-align: -14px">\int_0^3 f(x)~dx</math> with <math style="vertical-align: 0px">n=3</math> boxes.

@@ Line 35: / Line 35: @@
 |Thus, the left-hand Riemann sum is
 |-
-| &nbsp;&nbsp; &nbsp; &nbsp; <math style="vertical-align: -4px">1(f(0)+f(1)+f(2))=1+0+-3=-2.</math>
 |}
@@ Line 53: / Line 58: @@
 |Thus, the right-hand Riemann sum is
 |-
-| &nbsp;&nbsp; &nbsp; &nbsp; <math style="vertical-align: -5px">1(f(1)+f(2)+f(3))=0+-3+-8=-11.</math>
 |}

@@ Line 27: / Line 27: @@
 |So, the left-hand Riemann sum is
 |-
-| &nbsp;&nbsp; <math style="vertical-align: 0px">1(f(0)+f(1)+f(2)).</math>
+| &nbsp;&nbsp; &nbsp; &nbsp; <math style="vertical-align: 0px">1(f(0)+f(1)+f(2)).</math>
 |}
@@ Line 35: / Line 35: @@
 |Thus, the left-hand Riemann sum is
 |-
-| &nbsp;&nbsp; <math style="vertical-align: -4px">1(f(0)+f(1)+f(2))=1+0+-3=-2.</math>
+| &nbsp;&nbsp; &nbsp; &nbsp; <math style="vertical-align: -4px">1(f(0)+f(1)+f(2))=1+0+-3=-2.</math>
 |}
@@ Line 46: / Line 46: @@
 |So, the right-hand Riemann sum is
 |-
-| &nbsp;&nbsp; <math style="vertical-align: -5px">1(f(1)+f(2)+f(3)).</math>
+| &nbsp;&nbsp; &nbsp; &nbsp; <math style="vertical-align: -5px">1(f(1)+f(2)+f(3)).</math>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 53: / Line 53: @@
 |Thus, the right-hand Riemann sum is
 |-
-| &nbsp;&nbsp; <math style="vertical-align: -5px">1(f(1)+f(2)+f(3))=0+-3+-8=-11.</math>
+| &nbsp;&nbsp; &nbsp; &nbsp; <math style="vertical-align: -5px">1(f(1)+f(2)+f(3))=0+-3+-8=-11.</math>
 |}
@@ Line 64: / Line 64: @@
 |The width of each rectangle is
 |-
-| &nbsp; &nbsp; <math style="vertical-align: -13px">\Delta x=\frac{3-0}{n}=\frac{3}{n}.</math>
+| &nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -13px">\Delta x=\frac{3-0}{n}=\frac{3}{n}.</math>
 |}
@@ Line 72: / Line 72: @@
 |So, the right-hand Riemann sum is
 |-
-| &nbsp;&nbsp; <math style="vertical-align: -14px">\Delta x \bigg(f\bigg(1\cdot \frac{3}{n}\bigg)+f\bigg(2\cdot \frac{3}{n}\bigg)+f\bigg(3\cdot \frac{3}{n}\bigg)+\ldots +f(3)\bigg).</math>
+| &nbsp;&nbsp; &nbsp; &nbsp; <math style="vertical-align: -14px">\Delta x \bigg(f\bigg(1\cdot \frac{3}{n}\bigg)+f\bigg(2\cdot \frac{3}{n}\bigg)+f\bigg(3\cdot \frac{3}{n}\bigg)+\ldots +f(3)\bigg).</math>
 |-
 |Finally, we let <math style="vertical-align: 0px">n</math> go to infinity to get a limit.
@@ Line 83: / Line 83: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)''' &nbsp;<math style="vertical-align: -2px">-2</math>
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math style="vertical-align: -2px">-2</math>
 |-
-|'''(b)''' &nbsp;<math style="vertical-align: -2px">-11</math>
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math style="vertical-align: -2px">-11</math>
 |-
-|'''(c)''' &nbsp;<math style="vertical-align: -22px">\lim_{n\to\infty} \frac{3}{n}\sum_{i=1}^{n}f\bigg(i\frac{3}{n}\bigg)</math>
+|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp; <math style="vertical-align: -22px">\lim_{n\to\infty} \frac{3}{n}\sum_{i=1}^{n}f\bigg(i\frac{3}{n}\bigg)</math>
 |}
 [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 8: / Line 8: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Foundations: &nbsp;
 |-
 |'''1.''' The height of each rectangle in the left-hand Riemann sum is given by choosing the left endpoint of the interval.

@@ Line 17: / Line 17: @@
 |'''3.''' See the Riemann sums (insert link) for more information.
 |}
 '''Solution:'''
@@ Line 81: / Line 82: @@
 |Thus, <math style="vertical-align: -14px">\int_0^3 f(x)~dx</math> is equal to <math style="vertical-align: -21px">\lim_{n\to\infty} \frac{3}{n}\sum_{i=1}^{n}f\bigg(i\frac{3}{n}\bigg).</math>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 26: / Line 26: @@
 |Since our interval is <math style="vertical-align: -5px">[0,3]</math> and we are using 3 rectangles, each rectangle has width 1. So, the left-hand Riemann sum is
 |-
-| &nbsp;&nbsp; <math style="vertical-align: 0px">1(f(0)+f(1)+f(2))</math>.
+| &nbsp;&nbsp; <math style="vertical-align: 0px">1(f(0)+f(1)+f(2)).</math>
 |-
+|
@@ Line 36: / Line 36: @@
 |Thus, the left-hand Riemann sum is
 |-
-| &nbsp;&nbsp; <math style="vertical-align: -4px">1(f(0)+f(1)+f(2))=1+0+-3=-2</math>.
+| &nbsp;&nbsp; <math style="vertical-align: -4px">1(f(0)+f(1)+f(2))=1+0+-3=-2.</math>
 |}
@@ Line 45: / Line 45: @@
 |Since our interval is <math style="vertical-align: -5px">[0,3]</math> and we are using 3 rectangles, each rectangle has width 1. So, the right-hand Riemann sum is
 |-
-| &nbsp;&nbsp; <math style="vertical-align: -5px">1(f(1)+f(2)+f(3))</math>.
+| &nbsp;&nbsp; <math style="vertical-align: -5px">1(f(1)+f(2)+f(3)).</math>
 |-
+|
@@ Line 54: / Line 54: @@
 |Thus, the right-hand Riemann sum is
 |-
-| &nbsp;&nbsp; <math style="vertical-align: -5px">1(f(1)+f(2)+f(3))=0+-3+-8=-11</math>.
+| &nbsp;&nbsp; <math style="vertical-align: -5px">1(f(1)+f(2)+f(3))=0+-3+-8=-11.</math>
 |}
@@ Line 61: / Line 61: @@
 !Step 1: &nbsp;
 |-
-|Let <math style="vertical-align: 0px">n</math> be the number of rectangles used in the right-hand Riemann sum for <math style="vertical-align: -5px">f(x)=1-x^2</math>.
+|Let <math style="vertical-align: 0px">n</math> be the number of rectangles used in the right-hand Riemann sum for <math style="vertical-align: -5px">f(x)=1-x^2.</math>
 |-
-|The width of each rectangle is <math style="vertical-align: -13px">\Delta x=\frac{3-0}{n}=\frac{3}{n}</math>.
+|The width of each rectangle is <math style="vertical-align: -13px">\Delta x=\frac{3-0}{n}=\frac{3}{n}.</math>
 |-
+|
@@ Line 75: / Line 75: @@
 |So, the right-hand Riemann sum is
 |-
-| &nbsp;&nbsp; <math style="vertical-align: -14px">\Delta x \bigg(f\bigg(1\cdot \frac{3}{n}\bigg)+f\bigg(2\cdot \frac{3}{n}\bigg)+f\bigg(3\cdot \frac{3}{n}\bigg)+\ldots +f(3)\bigg)</math>.
+| &nbsp;&nbsp; <math style="vertical-align: -14px">\Delta x \bigg(f\bigg(1\cdot \frac{3}{n}\bigg)+f\bigg(2\cdot \frac{3}{n}\bigg)+f\bigg(3\cdot \frac{3}{n}\bigg)+\ldots +f(3)\bigg).</math>
 |-
 |Finally, we let <math style="vertical-align: 0px">n</math> go to infinity to get a limit.
 |-
-|Thus, <math style="vertical-align: -14px">\int_0^3 f(x)~dx</math> is equal to <math style="vertical-align: -21px">\lim_{n\to\infty} \frac{3}{n}\sum_{i=1}^{n}f\bigg(i\frac{3}{n}\bigg)</math>.
+|Thus, <math style="vertical-align: -14px">\int_0^3 f(x)~dx</math> is equal to <math style="vertical-align: -21px">\lim_{n\to\infty} \frac{3}{n}\sum_{i=1}^{n}f\bigg(i\frac{3}{n}\bigg).</math>
 |}