Difference between revisions of "009B Sample Midterm 1, Problem 5"

From Grad Wiki
Jump to navigation Jump to search
 
(15 intermediate revisions by 2 users not shown)
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>\int_0^3 f(x)dx</math> with <math>n=3</math> boxes.
+
<span class="exam"> Evaluate the integral:
::<span class="exam">b) Compute the right-hand Riemann sum approximation of <math>\int_0^3 f(x)dx</math> with <math>n=3</math> boxes.
 
::<span class="exam">c) Express <math>\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.
 
  
 +
::<math>\int \sin^3x \cos^2x~dx</math>
  
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+
<hr>
!Foundations: &nbsp;
+
[[009B Sample Midterm 1, Problem 5 Solution|'''<u>Solution</u>''']]
|-
 
|Link to Riemann sums page
 
|}
 
  
'''Solution:'''
 
  
'''(a)'''
+
[[009B Sample Midterm 1, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']]
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 1: &nbsp;
 
|-
 
|Since our interval is <math>[0,3]</math> and we are using 3 rectangles, each rectangle has width 1. So, the left-hand Riemann sum is
 
|-
 
|<math>1(f(0)+f(1)+f(2))</math>
 
|-
 
|
 
|}
 
  
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 2: &nbsp;
 
|-
 
|Thus, the left-hand Riemann sum is
 
|-
 
|<math>1(f(0)+f(1)+f(2))=1+0+-3=-2</math>. 
 
|}
 
  
'''(b)'''
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 1: &nbsp;
 
|-
 
|Since our interval is <math>[0,3]</math> and we are using 3 rectangles, each rectangle has width 1. So, the right-hand Riemann sum is
 
|-
 
|<math>1(f(1)+f(2)+f(3))</math>. 
 
|-
 
|
 
|}
 
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 2: &nbsp;
 
|-
 
|Thus, the right-hand Riemann sum is
 
|-
 
|<math>1(f(1)+f(2)+f(3))=0+-3+-8=-11</math>. 
 
|}
 
 
'''(c)'''
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 1: &nbsp;
 
|-
 
|Let <math>n</math> be the number of rectangles used in the right-hand Riemann sum for <math>f(x)=1-x^2</math>.
 
|-
 
|The width of each rectangle is <math>\Delta x=\frac{3-0}{n}=\frac{3}{n}</math>.
 
|-
 
|
 
|-
 
|
 
|}
 
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 2: &nbsp;
 
|-
 
|So, the right-hand Riemann sum is
 
|-
 
|<math>\Delta x \bigg(f\bigg(\frac{3}{n}\bigg)+f\bigg(2\frac{3}{n}\bigg)+f\bigg(3\frac{3}{n}\bigg)+\ldots +f(3)\bigg)</math>.
 
|-
 
|Now, we let <math>n</math> go to infinity to get a limit. 
 
|-
 
|So, the area of <math>S</math> is equal to <math>\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;
 
|-
 
|'''(a)''' <math>-2</math>
 
|-
 
|'''(b)''' <math>-11</math>
 
|-
 
|'''(c)''' <math>\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>''']]
 
[[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 16:05, 12 November 2017

Evaluate the integral:


Solution


Detailed Solution


Return to Sample Exam