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| − | <span class="exam">Let <math>f(x)=1-x^2</math>.
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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.
| + | <span class="exam"> Evaluate the integral: |
| − | ::<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.
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| − | ::<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.
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| | + | ::<math>\int \sin^3x \cos^2x~dx</math> |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009B Sample Midterm 1, Problem 5 Solution|'''<u>Solution</u>''']] |
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| − | |See the page on [[Riemann_Sums|'''Riemann Sums''']].
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| − | |}
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| − | '''Solution:'''
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| − | '''(a)'''
| + | [[009B Sample Midterm 1, Problem 5 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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| − | |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
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| − | | <math style="vertical-align: 0px">1(f(0)+f(1)+f(2))</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-hand Riemann sum is
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| − | | <math style="vertical-align: -4px">1(f(0)+f(1)+f(2))=1+0+-3=-2</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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| − | |-
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| − | |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
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| − | | <math style="vertical-align: -5px">1(f(1)+f(2)+f(3))</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 right-hand Riemann sum is
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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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| − | |}
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| − | '''(c)'''
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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 right-hand Riemann sum for <math style="vertical-align: -5px">f(x)=1-x^2</math>.
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| − | |The width of each rectangle is <math style="vertical-align: -13px">\Delta x=\frac{3-0}{n}=\frac{3}{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 right-hand Riemann sum is
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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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| − | |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 <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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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | |'''(a)''' <math style="vertical-align: -2px">-2</math>
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| − | |'''(b)''' <math style="vertical-align: -2px">-11</math>
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| − | |'''(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>
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| − | |}
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| | [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] | | [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] |
