009B Sample Midterm 1, Problem 5 - Revision history

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Created page with "Let <math>f(x)=1-x^2</math>. ::a) Compute the left-hand Riemann sum approximation of <math style="vertical-align: -14px">\int_0^3 f(x)~d..."

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Let <math>f(x)=1-x^2</math>.

::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.
::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.
::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.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Link to Riemann sums page
|}

'''Solution:'''

'''(a)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|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
|-
|    <math style="vertical-align: 0px">1(f(0)+f(1)+f(2))</math>.
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Thus, the left-hand Riemann sum is
|-
|    <math style="vertical-align: -4px">1(f(0)+f(1)+f(2))=1+0+-3=-2</math>.
|}

'''(b)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|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
|-
|    <math style="vertical-align: -5px">1(f(1)+f(2)+f(3))</math>.
|-
|
|}
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Thus, the right-hand Riemann sum is
|-
|    <math style="vertical-align: -5px">1(f(1)+f(2)+f(3))=0+-3+-8=-11</math>.
|}

'''(c)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|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>.
|-
|
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|So, the right-hand Riemann sum is
|-
|    <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>.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)'''  <math style="vertical-align: -2px">-2</math>
|-
|'''(b)'''  <math style="vertical-align: -2px">-11</math>
|-
|'''(c)'''  <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|'''Return to Sample Exam''']]

@@ 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>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+[[009B Sample Midterm 1, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[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>.
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Foundations: &nbsp;
 |-
-|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 Riemann sums (insert link) for more information.
 |}
 '''Solution:'''
@@ Line 26: / Line 25: @@
 !Step 1: &nbsp;
 |-
-|Since our interval is <math style="vertical-align: -5px">[0,3]</math> and we are using <math style="vertical-align: -1px">3</math> rectangles, each rectangle has width <math style="vertical-align: -1px">1.</math> So, the left-hand Riemann sum is
+|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.
 |-
+| &nbsp;&nbsp; &nbsp; &nbsp; <math style="vertical-align: 0px">1(f(0)+f(1)+f(2)).</math>
 |}
@@ Line 37: / Line 37: @@
 |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>
 |}
@@ Line 44: / Line 49: @@
 !Step 1: &nbsp;
 |-
-|Since our interval is <math style="vertical-align: -5px">[0,3]</math> and we are using <math style="vertical-align: -1px">3</math> rectangles, each rectangle has width <math style="vertical-align: -1px">1.</math> So, the right-hand Riemann sum is
+|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
 |-
+| &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 56: / Line 60: @@
 |Thus, the right-hand Riemann sum is
 |-
-−
+|
-::<math style="vertical-align: -5px">1(f(1)+f(2)+f(3))=0+-3+-8=-11.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |}
@@ Line 64: / 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
 |-
+| &nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -13px">\Delta x=\frac{3-0}{n}=\frac{3}{n}.</math>
 |}
@@ Line 77: / Line 84: @@
 |So, the right-hand Riemann sum is
 |-
+| &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>
 |-
-|Thus, <math style="vertical-align: -14px">\int_0^3 f(x)~dx</math> is equal to
+|Finally, we let &nbsp;<math style="vertical-align: 0px">n</math>&nbsp; go to infinity to get a limit.
 |-
+|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>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; '''(a)''' &nbsp;<math style="vertical-align: -2px">-2</math>
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math style="vertical-align: -2px">-2</math>
 |-
-|&nbsp;&nbsp; '''(b)''' &nbsp;<math style="vertical-align: -2px">-11</math>
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math style="vertical-align: -2px">-11</math>
 |-
-|&nbsp;&nbsp; '''(c)''' &nbsp;<math style="vertical-align: -22px">\lim_{n\to\infty} \frac{3}{n}\sum_{i=1}^{n}\bigg(1-\bigg(i\frac{3}{n}\bigg)^2\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 91: / Line 91: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)''' &nbsp;<math style="vertical-align: -2px">-2</math>
+|&nbsp;&nbsp; '''(a)''' &nbsp;<math style="vertical-align: -2px">-2</math>
 |-
-|'''(b)''' &nbsp;<math style="vertical-align: -2px">-11</math>
+|&nbsp;&nbsp; '''(b)''' &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}\bigg(1-\bigg(i\frac{3}{n}\bigg)^2\bigg)</math>
+|&nbsp;&nbsp; '''(c)''' &nbsp;<math style="vertical-align: -22px">\lim_{n\to\infty} \frac{3}{n}\sum_{i=1}^{n}\bigg(1-\bigg(i\frac{3}{n}\bigg)^2\bigg)</math>
 |}
 [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 4: / Line 4: @@
 ::<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.
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Foundations: &nbsp;
 |-
-|See the page on [[Riemann_Sums|'''Riemann Sums''']].
+|Recall:
 |}

@@ Line 9: / Line 9: @@
 !Foundations: &nbsp;
 |-
-|Link to Riemann sums page
+|See the page on [[Riemann_Sums|'''Riemann Sums''']].
 |}

@@ Line 10: / Line 10: @@
 |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 23: / Line 26: @@
 !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. So, the left-hand Riemann sum is
+|Since our interval is <math style="vertical-align: -5px">[0,3]</math> and we are using <math style="vertical-align: -1px">3</math> rectangles, each rectangle has width <math style="vertical-align: -1px">1.</math> So, the left-hand Riemann sum is
 |-
+|
 |}
@@ Line 35: / Line 37: @@
 |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 42: / Line 44: @@
 !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. So, the right-hand Riemann sum is
+|Since our interval is <math style="vertical-align: -5px">[0,3]</math> and we are using <math style="vertical-align: -1px">3</math> rectangles, each rectangle has width <math style="vertical-align: -1px">1.</math> So, the right-hand Riemann sum is
 |-
-| &nbsp;&nbsp; <math style="vertical-align: -5px">1(f(1)+f(2)+f(3))</math>.
 |-
+|
@@ Line 53: / Line 56: @@
 |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>.
 |}
@@ Line 60: / Line 64: @@
 !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
 |-
+|
 |}
@@ Line 74: / Line 77: @@
 |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>.
 |-
 |Finally, we let <math style="vertical-align: 0px">n</math> go to infinity to get a limit.
@@ Line 80: / Line 84: @@
 |Thus, <math style="vertical-align: -14px">\int_0^3 f(x)~dx</math> is equal to
 |-
-| &nbsp;&nbsp; <math style="vertical-align: -21px">\lim_{n\to\infty} \frac{3}{n}\sum_{i=1}^{n}f\bigg(i\frac{3}{n}\bigg)\,=\,\lim_{n\to\infty} \frac{3}{n}\sum_{i=1}^{n}\bigg(1-\bigg(i\frac{3}{n}\bigg)^2\bigg)</math>&thinsp;.
 |}