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

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<span class="exam">Let <math>f(x)=1-x^2</math>.
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<span class="exam">Let &nbsp;<math>f(x)=1-x^2</math>.
 +
 
 +
<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 &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 &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.
  
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::<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.
 
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::<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;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Foundations: &nbsp;  
 
!Foundations: &nbsp;  
 
|-
 
|-
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|Recall:
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|'''1.''' The height of each rectangle in the left-hand Riemann sum is given by choosing the left endpoint of the interval.
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|-
 
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|
 
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::'''1.''' The height of each rectangle in the left-hand Riemann sum is given by choosing the left endpoint of the interval.
 
 
|-
 
|-
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|
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|'''2.''' The height of each rectangle in the right-hand Riemann sum is given by choosing the right endpoint of the interval.
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::'''2.''' The height of each rectangle in the right-hand Riemann sum is given by choosing the right endpoint of the interval.
 
 
|-
 
|-
−
|
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|'''3.''' See the Riemann sums (insert link) for more information.
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::'''3.''' See the page on [[Riemann_Sums|'''Riemann Sums''']] for more information.
 
 
|}
 
|}
 +
  
 
'''Solution:'''
 
'''Solution:'''
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
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|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  
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|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  
 
|-
 
|-
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|
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| &nbsp;&nbsp; &nbsp; &nbsp; <math style="vertical-align: 0px">1(f(0)+f(1)+f(2)).</math>
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::<math style="vertical-align: 0px">1(f(0)+f(1)+f(2)).</math>
 
 
|}
 
|}
  
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|Thus, the left-hand Riemann sum is  
 
|Thus, the left-hand Riemann sum is  
 
|-
 
|-
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| &nbsp;&nbsp; <math style="vertical-align: -4px">1(f(0)+f(1)+f(2))=1+0+-3=-2.</math>
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|  
 +
&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 +
\displaystyle{1(f(0)+f(1)+f(2))} & = & \displaystyle{1+0+-3}\\
 +
&&\\
 +
& = & \displaystyle{-2.}
 +
\end{array}</math>
 
|}
 
|}
  
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
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|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
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::<math style="vertical-align: -5px">1(f(1)+f(2)+f(3)).</math> 
 
 
|-
 
|-
−
|
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| &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;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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|Thus, the right-hand Riemann sum is  
 
|Thus, the right-hand Riemann sum is  
 
|-
 
|-
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|
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|  
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::<math style="vertical-align: -5px">1(f(1)+f(2)+f(3))=0+-3+-8=-11.</math>
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&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 +
\displaystyle{1(f(1)+f(2)+f(3))} & = & \displaystyle{0+-3+-8}\\
 +
&&\\
 +
& = & \displaystyle{-11.}
 +
\end{array}</math>
 
|}
 
|}
  
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
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|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>
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|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  
 
|The width of each rectangle is  
 
|-
 
|-
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|
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| &nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -13px">\Delta x=\frac{3-0}{n}=\frac{3}{n}.</math>
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::<math style="vertical-align: -13px">\Delta x=\frac{3-0}{n}=\frac{3}{n}.</math>
 
 
|}
 
|}
  
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|So, the right-hand Riemann sum is  
 
|So, the right-hand Riemann sum is  
 
|-
 
|-
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|
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| &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>
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::<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>
 
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|-
 
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|Finally, we let <math style="vertical-align: 0px">n</math> go to infinity to get a limit. 
 
 
|-
 
|-
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|Thus, <math style="vertical-align: -14px">\int_0^3 f(x)~dx</math> is equal to
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|Finally, we let &nbsp;<math style="vertical-align: 0px">n</math>&nbsp; go to infinity to get a limit. 
 
|-
 
|-
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|
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|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>
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::<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>
 
 
|}
 
|}
 +
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
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|&nbsp;&nbsp; '''(a)''' &nbsp;<math style="vertical-align: -2px">-2</math>  
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|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math style="vertical-align: -2px">-2</math>  
 
|-
 
|-
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|&nbsp;&nbsp; '''(b)''' &nbsp;<math style="vertical-align: -2px">-11</math>
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|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math style="vertical-align: -2px">-11</math>
 
|-
 
|-
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|&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>
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|&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>''']]
 
[[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

Revision as of 11:01, 9 April 2017

Let  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)=1-x^2} .

(a) Compute the left-hand Riemann sum approximation of  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_0^3 f(x)~dx}   with  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n=3}   boxes.

(b) Compute the right-hand Riemann sum approximation of  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_0^3 f(x)~dx}   with  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n=3}   boxes.

(c) Express  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_0^3 f(x)~dx}   as a limit of right-hand Riemann sums (as in the definition of the definite integral). Do not evaluate the limit.


Foundations:  
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:

(a)

Step 1:  
Since our interval is  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [0,3]}   and we are using 3 rectangles, each rectangle has width 1.
So, the left-hand Riemann sum is
       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1(f(0)+f(1)+f(2)).}
Step 2:  
Thus, the left-hand Riemann sum is

        Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{1(f(0)+f(1)+f(2))} & = & \displaystyle{1+0+-3}\\ &&\\ & = & \displaystyle{-2.} \end{array}}

(b)

Step 1:  
Since our interval is  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [0,3]}   and we are using 3 rectangles, each rectangle has width 1.
So, the right-hand Riemann sum is
       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1(f(1)+f(2)+f(3)).}
Step 2:  
Thus, the right-hand Riemann sum is

        Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{1(f(1)+f(2)+f(3))} & = & \displaystyle{0+-3+-8}\\ &&\\ & = & \displaystyle{-11.} \end{array}}

(c)

Step 1:  
Let  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n}   be the number of rectangles used in the right-hand Riemann sum for  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)=1-x^2.}
The width of each rectangle is
        Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta x=\frac{3-0}{n}=\frac{3}{n}.}
Step 2:  
So, the right-hand Riemann sum is
       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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).}
Finally, we let  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n}   go to infinity to get a limit.
Thus,  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_0^3 f(x)~dx}   is equal to  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n\to\infty} \frac{3}{n}\sum_{i=1}^{n}f\bigg(i\frac{3}{n}\bigg).}


Final Answer:  
    (a)     Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -2}
    (b)     Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -11}
    (c)     Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n\to\infty} \frac{3}{n}\sum_{i=1}^{n}f\bigg(i\frac{3}{n}\bigg)}

Return to Sample Exam

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