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

Revision as of 17:48, 4 November 2017

Let 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 f(x)=1-x^2} .

(a) Compute the left-hand Riemann sum approximation 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 \int_0^3 f(x)~dx} with 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 n=3} boxes.

(b) Compute the right-hand Riemann sum approximation 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 \int_0^3 f(x)~dx} with 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 n=3} boxes.

(c) Express 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_0^3 f(x)~dx} as a limit of right-hand Riemann sums (as in the definition of the definite integral). Do not evaluate the limit.

(insert picture of handwritten solution)

Detailed Solution for this Problem

Return to Sample Exam

@@ Line 1: / Line 1: @@
-<span class="exam">Evaluate the indefinite and definite integrals.
+<span class="exam"> Let &nbsp;<math style="vertical-align: -5px">f(x)=1-x^2</math>.
-<span class="exam">(a) &nbsp; <math>\int x^2\sqrt{1+x^3}~dx</math>
+<span class="exam">(a) Compute the left-hand Riemann sum approximation of &nbsp;<math style="vertical-align: -14px">\int_0^3 f(x)~dx</math>&nbsp; with &nbsp;<math style="vertical-align: 0px">n=3</math>&nbsp; boxes.
-<span class="exam">(b) &nbsp; <math>\int _{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\cos(x)}{\sin^2(x)}~dx</math>
+<span class="exam">(b) Compute the right-hand Riemann sum approximation of &nbsp;<math style="vertical-align: -14px">\int_0^3 f(x)~dx</math>&nbsp; with &nbsp;<math style="vertical-align: 0px">n=3</math>&nbsp; boxes.
+<span class="exam">(c) Express &nbsp;<math style="vertical-align: -14px">\int_0^3 f(x)~dx</math>&nbsp; as a limit of right-hand Riemann sums (as in the definition of the definite integral). Do not evaluate the limit.
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+(insert picture of handwritten solution)
-|-
-| How would you integrate &nbsp; <math style="vertical-align: -12px">\int \frac{\ln x}{x}~dx?</math>
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; You can use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -5px">u=\ln(x).</math>
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; Then, &nbsp;<math style="vertical-align: -13px">du=\frac{1}{x}dx.</math>
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; Thus,
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\int \frac{\ln x}{x}~dx} & = & \displaystyle{\int u~du}\\
-&&\\
-& = & \displaystyle{\frac{u^2}{2}+C}\\
-&&\\
-& = & \displaystyle{\frac{(\ln x)^2}{2}+C.}
-\end{array}</math>
-|}
+[[009B Sample Midterm 1, Problem 1 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']]
-'''Solution:'''
-'''(a)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|We use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
-|-
-|Let &nbsp;<math style="vertical-align: -2px">u=1+x^3.</math>
-|-
-|Then, &nbsp;<math style="vertical-align: 0px">du=3x^2dx</math>&nbsp; and &nbsp;<math style="vertical-align: -13px">\frac{du}{3}=x^2dx.</math>
-|-
-|Therefore, the integral becomes
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -13px">\frac{1}{3}\int \sqrt{u}~du.</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|We now have
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\int x^2\sqrt{1+x^3}~dx} & = & \displaystyle{\frac{1}{3}\int \sqrt{u}~du}\\
-&&\\
-& = & \displaystyle{\frac{2}{9}u^{\frac{3}{2}}+C}\\
-&&\\
-& = & \displaystyle{\frac{2}{9}(1+x^3)^{\frac{3}{2}}+C.}
-\end{array}</math>
-|}
-'''(b)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|We use &nbsp;<math>u</math>-substitution.
-|-
-|Let &nbsp;<math style="vertical-align: -5px">u=\sin(x).</math>
-|-
-|Then, &nbsp;<math style="vertical-align: -5px">du=\cos(x)dx.</math>
-|-
-|Also, we need to change the bounds of integration.
-|-
-|Plugging in our values into the equation &nbsp;<math style="vertical-align: -5px">u=\sin(x),</math>&nbsp; we get
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -15px">u_1=\sin\bigg(\frac{\pi}{4}\bigg)=\frac{\sqrt{2}}{2}</math>&nbsp; and &nbsp;<math style="vertical-align: -16px">u_2=\sin\bigg(\frac{\pi}{2}\bigg)=1.</math>&nbsp;
-|-
-|Therefore, the integral becomes
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -19px">\int_{\frac{\sqrt{2}}{2}}^1 \frac{1}{u^2}~du.</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|We now have
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\int _{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\cos(x)}{\sin^2(x)}~dx} & = & \displaystyle{\int_{\frac{\sqrt{2}}{2}}^1 \frac{1}{u^2}~du}\\
-&&\\
-& = & \displaystyle{\left.\frac{-1}{u}\right|_{\frac{\sqrt{2}}{2}}^1}\\
-&&\\
-& = & \displaystyle{-\frac{1}{1}-\frac{-1}{\frac{\sqrt{2}}{2}}}\\
-&&\\
-& = & \displaystyle{-1+\sqrt{2}.}
-\end{array}</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Final Answer: &nbsp;
-|-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>\frac{2}{9}(1+x^3)^{\frac{3}{2}}+C</math>
-|-
-|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>-1+\sqrt{2}</math>
-|}
 [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

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

Revision as of 17:48, 4 November 2017

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