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

Latest revision as of 15:05, 12 November 2017

Evaluate the integral:

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 \sin^3x \cos^2x~dx}

Solution

Detailed Solution

Return to Sample Exam

@@ 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: &nbsp;
-|-
-|'''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: &nbsp;
-|-
-|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
-|-
-| &nbsp;&nbsp; &nbsp; &nbsp; <math style="vertical-align: 0px">1(f(0)+f(1)+f(2)).</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Thus, the left-hand Riemann sum is
-|-
-|
-&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>
-|}
-'''(b)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|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
-|-
-| &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;"
-!Step 2: &nbsp;
-|-
-|Thus, the right-hand Riemann sum is
-|-
-|
-&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>
-|}
-'''(c)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|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
-|-
-| &nbsp; &nbsp; &nbsp; &nbsp; <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: &nbsp;
-|-
-|So, the right-hand Riemann sum is
-|-
-| &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>
-|-
-|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: &nbsp;
-|-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math style="vertical-align: -2px">-2</math>
-|-
-|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math style="vertical-align: -2px">-11</math>
-|-
-|&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>''']]

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

Latest revision as of 15:05, 12 November 2017

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