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| − | <span class="exam">Evaluate the indefinite and definite integrals. | + | <span class="exam"> Let <math style="vertical-align: -5px">f(x)=1-x^2</math>. |
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| − | <span class="exam">(a) <math>\int x^2\sqrt{1+x^3}~dx</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. |
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| − | <span class="exam">(b) <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 <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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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | (insert picture of handwritten solution) |
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| − | | How would you integrate <math style="vertical-align: -12px">\int \frac{\ln x}{x}~dx?</math>
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| − | You can use <math style="vertical-align: 0px">u</math>-substitution.
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| − | | Let <math style="vertical-align: -5px">u=\ln(x).</math>
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| − | | Then, <math style="vertical-align: -13px">du=\frac{1}{x}dx.</math>
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| − | Thus,
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\int \frac{\ln x}{x}~dx} & = & \displaystyle{\int u~du}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{u^2}{2}+C}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{(\ln x)^2}{2}+C.}
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| − | \end{array}</math>
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| − | |}
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| | + | [[009B Sample Midterm 1, Problem 1 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']] |
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| − | '''Solution:'''
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| − | '''(a)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |We use <math style="vertical-align: 0px">u</math>-substitution.
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| − | |Let <math style="vertical-align: -2px">u=1+x^3.</math>
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| − | |Then, <math style="vertical-align: 0px">du=3x^2dx</math> and <math style="vertical-align: -13px">\frac{du}{3}=x^2dx.</math>
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| − | |Therefore, the integral becomes
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| − | | <math style="vertical-align: -13px">\frac{1}{3}\int \sqrt{u}~du.</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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| − | |We now have
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{\int x^2\sqrt{1+x^3}~dx} & = & \displaystyle{\frac{1}{3}\int \sqrt{u}~du}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{2}{9}u^{\frac{3}{2}}+C}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{2}{9}(1+x^3)^{\frac{3}{2}}+C.}
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| − | \end{array}</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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| − | |We use <math>u</math>-substitution.
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| − | |Let <math style="vertical-align: -5px">u=\sin(x).</math>
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| − | |Then, <math style="vertical-align: -5px">du=\cos(x)dx.</math>
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| − | |Also, we need to change the bounds of integration.
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| − | |Plugging in our values into the equation <math style="vertical-align: -5px">u=\sin(x),</math> we get
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| − | | <math style="vertical-align: -15px">u_1=\sin\bigg(\frac{\pi}{4}\bigg)=\frac{\sqrt{2}}{2}</math> and <math style="vertical-align: -16px">u_2=\sin\bigg(\frac{\pi}{2}\bigg)=1.</math>
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| − | |Therefore, the integral becomes
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| − | | <math style="vertical-align: -19px">\int_{\frac{\sqrt{2}}{2}}^1 \frac{1}{u^2}~du.</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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| − | |We now have
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| − | <math>\begin{array}{rcl}
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| − | \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}\\
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| − | &&\\
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| − | & = & \displaystyle{\left.\frac{-1}{u}\right|_{\frac{\sqrt{2}}{2}}^1}\\
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| − | &&\\
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| − | & = & \displaystyle{-\frac{1}{1}-\frac{-1}{\frac{\sqrt{2}}{2}}}\\
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| − | &&\\
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| − | & = & \displaystyle{-1+\sqrt{2}.}
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| − | \end{array}</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>\frac{2}{9}(1+x^3)^{\frac{3}{2}}+C</math>
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| − | | '''(b)''' <math>-1+\sqrt{2}</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>''']] |
Let 
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(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