|
|
| 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 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.
| + | ::<math>\int \sin^3x \cos^2x~dx</math> |
| | | | |
| − | <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. | + | <hr> |
| | + | [[009B Sample Midterm 1, Problem 5 Solution|'''<u>Solution</u>''']] |
| | | | |
| | | | |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | [[009B Sample Midterm 1, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| − | !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)'''
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 1:
| |
| − | |-
| |
| − | |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
| |
| − | |-
| |
| − | | <math style="vertical-align: 0px">1(f(0)+f(1)+f(2)).</math>
| |
| − | |}
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 2:
| |
| − | |-
| |
| − | |Thus, the left-hand Riemann sum is
| |
| − | |-
| |
| − | |
| |
| − | <math>\begin{array}{rcl}
| |
| − | \displaystyle{1(f(0)+f(1)+f(2))} & = & \displaystyle{1+0+-3}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{-2.}
| |
| − | \end{array}</math>
| |
| − | |}
| |
| − |
| |
| − | '''(b)'''
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 1:
| |
| − | |-
| |
| − | |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
| |
| − | |-
| |
| − | | <math style="vertical-align: -5px">1(f(1)+f(2)+f(3)).</math>
| |
| − | |}
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 2:
| |
| − | |-
| |
| − | |Thus, the right-hand Riemann sum is
| |
| − | |-
| |
| − | |
| |
| − | <math>\begin{array}{rcl}
| |
| − | \displaystyle{1(f(1)+f(2)+f(3))} & = & \displaystyle{0+-3+-8}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{-11.}
| |
| − | \end{array}</math>
| |
| − | |}
| |
| − |
| |
| − | '''(c)'''
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 1:
| |
| − | |-
| |
| − | |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>
| |
| − | |-
| |
| − | |The width of each rectangle is
| |
| − | |-
| |
| − | | <math style="vertical-align: -13px">\Delta x=\frac{3-0}{n}=\frac{3}{n}.</math>
| |
| − | |}
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 2:
| |
| − | |-
| |
| − | |So, the right-hand Riemann sum is
| |
| − | |-
| |
| − | | <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>
| |
| − | |-
| |
| − | |Finally, we let <math style="vertical-align: 0px">n</math> go to infinity to get a limit.
| |
| − | |-
| |
| − | |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>
| |
| − | |}
| |
| − |
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Final Answer:
| |
| − | |-
| |
| − | | '''(a)''' <math style="vertical-align: -2px">-2</math>
| |
| − | |-
| |
| − | | '''(b)''' <math style="vertical-align: -2px">-11</math>
| |
| − | |-
| |
| − | | '''(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>
| |
| − | |}
| |
| | [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] | | [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] |
