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| − | <span class="exam"> Consider the region <math style="vertical-align: 0px">S</math> bounded by <math style="vertical-align: -13px">x=1,x=5,y=\frac{1}{x^2}</math>  and the <math>x</math>-axis. | + | <span class="exam"> This problem has three parts: |
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| − | ::<span class="exam">a) Use four rectangles and a Riemann sum to approximate the area of the region <math style="vertical-align: 0px">S</math>. Sketch the region <math style="vertical-align: 0px">S</math> and the rectangles and
| + | <span class="exam">(a) State the both parts of the fundamental theorem of calculus. |
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| − | :::<span class="exam">indicate whether your rectangles overestimate or underestimate the area of <math style="vertical-align: 0px">S</math>.
| + | <span class="exam">(b) Compute <math style="vertical-align: -15px">\frac{d}{dx}\int_0^{\cos (x)}\sin (t)~dt</math>. |
| − | ::<span class="exam">b) Find an expression for the area of the region <math style="vertical-align: 0px">S</math> as a limit. Do not evaluate the limit.
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| | + | <span class="exam">(c) Evaluate <math style="vertical-align: -14px">\int_{0}^{\pi/4}\sec^2 x~dx</math>. |
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| − | [[File:9AMT2_1GP.png|right|400px|frame|Approximation of integral with left endpoints is an overestimate.]]
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009B Sample Midterm 2, Problem 1 Solution|'''<u>Solution</u>''']] |
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| − | |Recall:
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| − | ::'''1.''' The height of each rectangle in the left-hand Riemann sum is given by
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| − | :::choosing the left endpoint of the interval.
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| − | ::'''2.''' The height of each rectangle in the right-hand Riemann sum is given by
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| − | :::choosing the right endpoint of the interval.
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| − | ::'''3.''' See the page on [[Riemann_Sums|'''Riemann Sums''']] for more information.
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| − | |}
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| − | '''Solution:'''
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| − | '''(a)''' | + | [[009B Sample Midterm 2, Problem 1 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |Let <math style="vertical-align: -13px">f(x)=\frac{1}{x^2}.</math> Since our interval is <math style="vertical-align: -5px">[1,5]</math> and we are using <math style="vertical-align: -1px">4</math> rectangles, each rectangle has width <math style="vertical-align: -1px">1.</math> Since the problem doesn't specify, we can choose either right- or left-endpoints. Choosing left-endpoints, the Riemann sum is
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| − | ::<math>1\cdot (f(1)+f(2)+f(3)+f(4)).</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |Thus, the left-endpoint Riemann sum is
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| − | ::<math>\begin{array}{rcl}
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| − | \displaystyle{1\cdot (f(1)+f(2)+f(3)+f(4))} & = & \displaystyle{\bigg(1+\frac{1}{4}+\frac{1}{9}+{1}{16}\bigg)}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{205}{144}.}\\
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| − | \end{array}</math>
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| − | |The left-endpoint Riemann sum overestimates the area of <math style="vertical-align: 0px">S.</math>
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| − | |}
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| − | '''(b)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |Let <math style="vertical-align: 0px">n</math> be the number of rectangles used in the left-endpoint Riemann sum for <math style="vertical-align: -13px">f(x)=\frac{1}{x^2}.</math>
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| − | |The width of each rectangle is
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| − | ::<math style="vertical-align: -13px">\Delta x=\frac{5-1}{n}=\frac{4}{n}.</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |So, the left-endpoint Riemann sum is
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| − | ::<math>\Delta x \bigg(f(1)+f\bigg(1+\frac{4}{n}\bigg)+f\bigg(1+2\frac{4}{n}\bigg)+\ldots +f\bigg(1+(n-1)\frac{4}{n}\bigg)\bigg).</math>
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| − | |Now, we let <math style="vertical-align: 0px">n</math> go to infinity to get a limit.
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| − | |So, the area of <math style="vertical-align: 0px">S</math> is equal to
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| − | ::<math style="vertical-align: -20px">\lim_{n\to\infty} \frac{4}{n}\sum_{i=0}^{n-1}f\bigg(1+i\frac{4}{n}\bigg)\,=\,\lim_{n\to\infty} \frac{4}{n}\sum_{i=0}^{n-1}\bigg(1+i\frac{4}{n}\bigg)^{-2}.</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | |'''(a)''' The left-endpoint Riemann sum is <math style="vertical-align: -20px">\frac{205}{144}</math>, which overestimates the area of <math style="vertical-align: 0px">S</math>.
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| − | |'''(b)''' Using left-endpoint Riemann sums:
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| − | ::<math style="vertical-align: -20px">\lim_{n\to\infty} \frac{4}{n}\sum_{i=0}^{n-1}\bigg(1+i\frac{4}{n}\bigg)^{-2}</math>
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| − | |}
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| | [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] | | [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] |
