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

Latest revision as of 17:15, 2 December 2017

We would like to evaluate

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

(a) Compute 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)=\int_{-1}^{x} \sin(t^2)2t\,dt.}

(b) Find 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).}

(c) State the Fundamental Theorem of Calculus.

(d) Use the Fundamental Theorem of Calculus to compute 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 \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>''']]
-|-
-|Review <math>u</math>-substitution
-|}
-'''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>u</math>-substitution. Let <math>u=t^2</math>. Then, <math>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>u=t^2</math>, we get <math>u_1=(-1)^2=1</math> and <math>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>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>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>[a,b]</math> and let <math>F(x)=\int_a^x f(t)~dt</math>.
-|-
-|Then, <math>F</math> is a differentiable function on <math>(a,b)</math> and <math>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>F</math> be any antiderivative of <math>f</math>.
-|-
-|Then, <math>\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>[a,b]</math> and let <math>F(x)=\int_a^x f(t)~dt</math>.
-|-
-|Then, <math>F</math> is a differentiable function on <math>(a,b)</math> and <math>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>F</math> be any antiderivative of <math>f</math>.
-|-
-|Then, <math>\int_a^b f(x)~dx=F(b)-F(a)</math>.
-|-
-|'''(d)''' <math>\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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