009B Sample Midterm 2, Problem 1

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Consider the region 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 S} bounded by 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 x=1,x=5,y=\frac{1}{x^2}}   and the 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 x} -axis.

a) Use four rectangles and a Riemann sum to approximate the area of the region 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 S} . Sketch the region and the rectangles and
indicate whether your rectangles overestimate or underestimate the area of .
b) Find an expression for the area of the region as a limit. Do not evaluate the limit.


Approximation of integral with left endpoints is an overestimate.
Foundations:  
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 for more information.

Solution:

(a)

Step 1:  
Let Since our interval is and we are using rectangles, each rectangle has width Since the problem doesn't specify, we can choose either right- or left-endpoints. Choosing left-endpoints, the Riemann sum is
Step 2:  
Thus, the left-endpoint Riemann sum is
The left-endpoint Riemann sum overestimates the area of

(b)

Step 1:  
Let be the number of rectangles used in the left-endpoint Riemann sum for
The width of each rectangle is
Step 2:  
So, the left-endpoint Riemann sum is
Now, we let go to infinity to get a limit.
So, the area of is equal to
Final Answer:  
(a) The left-endpoint Riemann sum is , which overestimates the area 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 S} .
(b) Using left-endpoint Riemann sums:
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{4}{n}\sum_{i=0}^{n-1}\bigg(1+i\frac{4}{n}\bigg)^{-2}}

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