Difference between revisions of "009B Sample Final 1, Problem 2"

Latest revision as of 17:15, 2 December 2017

We would like to evaluate

{\frac {d}{dx}}{\bigg (}\int _{-1}^{x}\sin(t^{2})2t\,dt{\bigg )}.

(a) Compute $f(x)=\int _{-1}^{x}\sin(t^{2})2t\,dt.$

(b) Find $f'(x).$

(c) State the Fundamental Theorem of Calculus.

(d) Use the Fundamental Theorem of Calculus to compute ${\frac {d}{dx}}{\bigg (}\int _{-1}^{x}\sin(t^{2})2t\,dt{\bigg )}$ without first computing the integral.

Solution

Detailed Solution

Return to Sample Exam

@@ Line 1: / Line 1: @@
 <span class="exam"> We would like to evaluate
-:::::<math>\frac{d}{dx}\bigg(\int_{-1}^{x} \sin(t^2)2tdt\bigg)</math>.
+::<math>\frac{d}{dx}\bigg(\int_{-1}^{x} \sin(t^2)2t\,dt\bigg).</math>
-<span class="exam">a) Compute <math>f(x)=\int_{-1}^{x} \sin(t^2)2t~dt</math>.
+<span class="exam">(a) Compute &nbsp;<math style="vertical-align: -15px">f(x)=\int_{-1}^{x} \sin(t^2)2t\,dt.</math>
-<span class="exam">b) Find <math>f'(x)</math>.
+<span class="exam">(b) Find &nbsp;<math style="vertical-align: -5px">f'(x).</math>
-<span class="exam">c) State the fundamental theorem of calculus.
+<span class="exam">(c) State the Fundamental Theorem of Calculus.
-<span class="exam">d) Use the fundamental theorem of calculus to compute <math>\frac{d}{dx}\bigg(\int_{-1}^{x} \sin(t^2)2t~dt\bigg)</math> without first computing the integral.
+<span class="exam">(d) Use the Fundamental Theorem of Calculus to compute &nbsp;<math style="vertical-align: -15px">\frac{d}{dx}\bigg(\int_{-1}^{x} \sin(t^2)2t\,dt\bigg)</math>&nbsp; without first computing the integral.
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009B Sample Final 1, Problem 2 Solution|'''<u>Solution</u>''']]
-|-
-|How would you integrate <math>\int e^{x^2}2x~dx</math>?
-|-
-|
-::You could use <math style="vertical-align: -1px">u</math>-substitution. Let <math style="vertical-align: 0px">u=x^2</math>. Then, <math style="vertical-align: 0px">du=2xdx</math>.
-|-
-|
-::So, we get <math style="vertical-align: -14px">\int e^u~du=e^u+C=e^{x^2}+C</math>.
-|}
-'''Solution:'''
-'''(a)'''
+[[009B Sample Final 1, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']]
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|We proceed using <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: 0px">u=t^2</math>. Then, <math style="vertical-align: 0px">du=2tdt</math>.
-|-
-|Since this is a definite integral, we need to change the bounds of integration.
-|-
-|Plugging in our values into the equation <math style="vertical-align: 0px">u=t^2</math>, we get <math style="vertical-align: -5px">u_1=(-1)^2=1</math> and <math style="vertical-align: -3px">u_2=x^2</math>.
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-||So, we have
-|-
-|
-::<math>\begin{array}{rcl}
-f(x) & = & \displaystyle{\int_{-1}^{x} \sin(t^2)2t~dt}\\
-&&\\
-& = & \displaystyle{\int_{1}^{x^2} \sin(u)~du}\\
-&&\\
-& = & \displaystyle{-\cos(u)\bigg|_{1}^{x^2}}\\
-&&\\
-& = & \displaystyle{-\cos(x^2)+\cos(1)}\\
-\end{array}</math>
-|}
-'''(b)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|From part (a), we have <math style="vertical-align: -5px">f(x)=-\cos(x^2)+\cos(1)</math>.
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|If we take the derivative, we get <math style="vertical-align: -5px">f'(x)=\sin(x^2)2x</math>.
-|}
-'''(c)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|The Fundamental Theorem of Calculus has two parts.
-|-
-|'''The Fundamental Theorem of Calculus, Part 1'''
-|-
-|Let <math>f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: -14px">F(x)=\int_a^x f(t)~dt</math>.
-|-
-|Then, <math style="vertical-align: 0px">F</math> is a differentiable function on <math style="vertical-align: -5px">(a,b)</math> and <math style="vertical-align: -5px">F'(x)=f(x)</math>.
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|'''The Fundamental Theorem of Calculus, Part 2'''
-|-
-|Let <math>f</math> be continuous on <math>[a,b]</math> and let <math style="vertical-align: 0px">F</math> be any antiderivative of <math>f</math>.
-|-
-|Then, <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a)</math>.
-|}
-'''(d)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|By the '''Fundamental Theorem of Calculus, Part 1''',
-|-
-|
-::<math>\frac{d}{dx}\bigg(\int_{-1}^{x} \sin(t^2)2t~dt\bigg)=\sin(x^2)2x</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Final Answer: &nbsp;
-|-
-|'''(a)''' <math>f(x)=-\cos(x^2)+\cos(1)</math>
-|-
-|'''(b)''' <math>f'(x)=\sin(x^2)2x</math>
-|-
-|'''(c)''' '''The Fundamental Theorem of Calculus, Part 1'''
-|-
-|Let <math>f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: -14px">F(x)=\int_a^x f(t)~dt</math>.
-|-
-|Then, <math style="vertical-align: 0px">F</math> is a differentiable function on <math style="vertical-align: -5px">(a,b)</math> and <math style="vertical-align: -5px">F'(x)=f(x)</math>.
-|-
-|'''The Fundamental Theorem of Calculus, Part 2'''
-|-
-|Let <math>f</math> be continuous on <math>[a,b]</math> and let <math style="vertical-align: 0px">F</math> be any antiderivative of <math>f</math>.
-|-
-|Then, <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a)</math>.
-|-
-|'''(d)''' <math style="vertical-align: -5px">\sin(x^2)2x</math>
-|}
 [[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Difference between revisions of "009B Sample Final 1, Problem 2"

Latest revision as of 17:15, 2 December 2017

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