Difference between revisions of "009B Sample Midterm 2, Problem 2"

Revision as of 10:17, 5 February 2017

Evaluate

a)

\int _{1}^{2}{\bigg (}2t+{\frac {3}{t^{2}}}{\bigg )}{\bigg (}4t^{2}-{\frac {5}{t}}{\bigg )}~dt

b)

\int _{0}^{2}(x^{3}+x){\sqrt {x^{4}+2x^{2}+4}}~dx

Foundations:
How would you integrate $\int (2x+1){\sqrt {x^{2}+x}}~dx?$
You could use $u$ -substitution. Let $u=x^{2}+x.$ Then, $du=(2x+1)~dx.$
Thus, $\int (2x+1){\sqrt {x^{2}+x}}~dx=\int {\sqrt {u}}={\frac {2}{3}}u^{3/2}+C={\frac {2}{3}}(x^{2}+x)^{3/2}+C.$

Solution:

(a)

Step 1:
We multiply the product inside the integral to get
$\int _{1}^{2}{\bigg (}2t+{\frac {3}{t^{2}}}{\bigg )}{\bigg (}4t^{2}-{\frac {5}{t}}{\bigg )}~dt=\int _{1}^{2}{\bigg (}8t^{3}-10+12-{\frac {15}{t^{3}}}{\bigg )}~dt=\int _{1}^{2}(8t^{3}+2-15t^{-3})~dt$ .

Step 2:
We integrate to get
$\int _{1}^{2}{\bigg (}2t+{\frac {3}{t^{2}}}{\bigg )}{\bigg (}4t^{2}-{\frac {5}{t}}{\bigg )}~dt=\left.2t^{4}+2t+{\frac {15}{2}}t^{-2}\right\|_{1}^{2}$ .
We now evaluate to get
$\int _{1}^{2}{\bigg (}2t+{\frac {3}{t^{2}}}{\bigg )}{\bigg (}4t^{2}-{\frac {5}{t}}{\bigg )}~dt=2(2)^{4}+2(2)+{\frac {15}{2(2)^{2}}}-{\bigg (}2+2+{\frac {15}{2}}{\bigg )}=36+{\frac {15}{8}}-4-{\frac {15}{2}}={\frac {211}{8}}$ .

(b)

Step 1:
We use $u$ -substitution. Let $u=x^{4}+2x^{2}+4$ . Then, $du=(4x^{3}+4x)dx$ and ${\frac {du}{4}}=(x^{3}+x)dx$ . Also, we need to change the bounds of integration.
Plugging in our values into the equation $u=x^{4}+2x^{2}+4$ , we get $u_{1}=0^{4}+2(0)^{2}+4=4$ and $u_{2}=2^{4}+2(2)^{2}+4=28$ .
Therefore, the integral becomes ${\frac {1}{4}}\int _{4}^{28}{\sqrt {u}}~du$ .

Step 2:
We now have:
$\int _{0}^{2}(x^{3}+x){\sqrt {x^{4}+2x^{2}+4}}~dx={\frac {1}{4}}\int _{4}^{28}{\sqrt {u}}~du=\left.{\frac {1}{6}}u^{\frac {3}{2}}\right\|_{4}^{28}={\frac {1}{6}}(28^{\frac {3}{2}}-4^{\frac {3}{2}})={\frac {1}{6}}(({\sqrt {28}})^{3}-({\sqrt {4}})^{3})={\frac {1}{6}}((2{\sqrt {7}})^{3}-2^{3})$ .
So, we have
$\int _{0}^{2}(x^{3}+x){\sqrt {x^{4}+2x^{2}+4}}~dx={\frac {28{\sqrt {7}}-4}{3}}$ .

Final Answer:
(a) ${\frac {211}{8}}$
(b) ${\frac {28{\sqrt {7}}-4}{3}}$

Return to Sample Exam

@@ Line 1: / Line 1: @@
-<span class="exam"> This problem has three parts:
+<span class="exam"> Evaluate
-::<span class="exam">a) State the Fundamental Theorem of Calculus.
+::<span class="exam">a) <math style="vertical-align: -14px">\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt</math>
-::<span class="exam">b) Compute &thinsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^{\cos (x)}\sin (t)~dt.</math>
+::<span class="exam">b) <math style="vertical-align: -14px">\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx</math>
-::<span class="exam">c) Evaluate <math style="vertical-align: -14px">\int_{0}^{\pi/4}\sec^2 x~dx.</math>
@@ Line 11: / Line 9: @@
 !Foundations: &nbsp;
 |-
-|'''1.''' What does Part 1 of the Fundamental Theorem of Calculus say about <math style="vertical-align: -15px">\frac{d}{dx}\int_0^x\sin(t)~dt?</math>
+|How would you integrate <math style="vertical-align: -12px">\int (2x+1)\sqrt{x^2+x}~dx?</math>
 |-
 |
-::Part 1 of the Fundamental Theorem of Calculus says that <math style="vertical-align: -15px">\frac{d}{dx}\int_0^x\sin(t)~dt=\sin(x).</math>
+::You could use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -2px">u=x^2+x.</math> Then, <math style="vertical-align: -4px">du=(2x+1)~dx.</math>
-|-
-|'''2.''' What does Part 2 of the Fundamental Theorem of Calculus say about <math style="vertical-align: -15px">\int_a^b\sec^2x~dx</math> where <math style="vertical-align: -5px">a,b</math> are constants?
 |-
 |
-::Part 2 of the Fundamental Theorem of Calculus says that <math style="vertical-align: -15px">\int_a^b\sec^2x~dx=F(b)-F(a)</math> where <math style="vertical-align: 0px">F</math> is any antiderivative of <math style="vertical-align: 0px">\sec^2x.</math>
+::Thus, <math style="vertical-align: -12px">\int (2x+1)\sqrt{x^2+x}~dx=\int \sqrt{u}=\frac{2}{3}u^{3/2}+C=\frac{2}{3}(x^2+x)^{3/2}+C.</math>
 |}
@@ Line 25: / Line 21: @@
 '''(a)'''
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 1: &nbsp;
 |-
-|The Fundamental Theorem of Calculus has two parts.
+|We multiply the product inside the integral to get
 |-
-|'''The Fundamental Theorem of Calculus, Part 1'''
+| &nbsp;&nbsp; <math style="vertical-align: -16px">\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt=\int_1^2 \bigg(8t^3-10+12-\frac{15}{t^3}\bigg)~dt=\int_1^2 (8t^3+2-15t^{-3})~dt</math>.
-|-
-|Let <math style="vertical-align: -4px">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>.
 |}
@@ Line 41: / Line 32: @@
 !Step 2: &nbsp;
 |-
-|'''The Fundamental Theorem of Calculus, Part 2'''
+|We integrate to get
 |-
-|Let <math style="vertical-align: -4px">f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: 0px">F</math> be any antiderivative of <math style="vertical-align: -4px">f</math>.
+| &nbsp;&nbsp; <math style="vertical-align: -16px">\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt=\left. 2t^4+2t+\frac{15}{2}t^{-2}\right|_1^2</math>.
 |-
-|Then, <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a)</math>.
+|We now evaluate to get
+|-
+| &nbsp;&nbsp; <math style="vertical-align: -16px">\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt=2(2)^4+2(2)+\frac{15}{2(2)^2}-\bigg(2+2+\frac{15}{2}\bigg)=36+\frac{15}{8}-4-\frac{15}{2}=\frac{211}{8}</math>.
 |}
 '''(b)'''
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 1: &nbsp;
 |-
-|Let <math style="vertical-align: -15px">F(x)=\int_0^{\cos (x)}\sin (t)~dt</math>. The problem is asking us to find <math style="vertical-align: -5px">F'(x)</math>.
+|We use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -2px">u=x^4+2x^2+4</math>. Then, <math style="vertical-align: -5px">du=(4x^3+4x)dx</math> and <math style="vertical-align: -14px">\frac{du}{4}=(x^3+x)dx</math>. Also, we need to change the bounds of integration.
 |-
-|Let <math style="vertical-align: -5px">g(x)=\cos(x)</math> and <math style="vertical-align: -14px">G(x)=\int_0^x \sin(t)~dt</math>.
+|Plugging in our values into the equation <math style="vertical-align: -2px">u=x^4+2x^2+4</math>, we get <math style="vertical-align: -5px">u_1=0^4+2(0)^2+4=4</math> and <math style="vertical-align: -4px">u_2=2^4+2(2)^2+4=28</math>.
 |-
-|Then, <math style="vertical-align: -5px">F(x)=G(g(x))</math>.
+|Therefore, the integral becomes&nbsp; <math style="vertical-align: -14px">\frac{1}{4}\int_4^{28}\sqrt{u}~du</math>.
+|-
+|
 |}
@@ Line 62: / Line 57: @@
 !Step 2: &nbsp;
 |-
-|If we take the derivative of both sides of the last equation, we get <math style="vertical-align: -5px">F'(x)=G'(g(x))g'(x)</math> by the Chain Rule.
+|We now have:
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 3: &nbsp;
-|-
-|Now, <math style="vertical-align: -5px">g'(x)=-\sin(x)</math> and <math style="vertical-align: -5px">G'(x)=\sin(x)</math> by the '''Fundamental Theorem of Calculus, Part 1'''.
-|-
-|Since <math style="vertical-align: -6px">G'(g(x))=\sin(g(x))=\sin(\cos(x))</math>, we have <math style="vertical-align: -5px">F'(x)=G'(g(x))\cdot g'(x)=\sin(\cos(x))\cdot(-\sin(x))</math>.
-|}
-'''(c)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-| Using the '''Fundamental Theorem of Calculus, Part 2''', we have
 |-
-| &nbsp;&nbsp; <math>\int_{0}^{\frac{\pi}{4}}\sec^2 x~dx=\tan(x)\biggr|_{0}^{\pi/4}</math>
+| &nbsp;&nbsp; <math style="vertical-align: -16px">\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx=\frac{1}{4}\int_4^{28}\sqrt{u}~du=\left.\frac{1}{6}u^{\frac{3}{2}}\right|_4^{28}=\frac{1}{6}(28^{\frac{3}{2}}-4^{\frac{3}{2}})=\frac{1}{6}((\sqrt{28})^3-(\sqrt{4})^3)=\frac{1}{6}((2\sqrt{7})^3-2^3)</math>.
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
 |-
-|So, we get
+|So, we have
 |-
-| &nbsp;&nbsp; <math style="vertical-align: -16px">\int_{0}^{\frac{\pi}{4}}\sec^2 x~dx=\tan \bigg(\frac{\pi}{4}\bigg)-\tan (0)=1</math>.
+| &nbsp;&nbsp; <math style="vertical-align: -16px">\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx=\frac{28\sqrt{7}-4}{3}</math>.
-|-
-|
 |}
@@ Line 96: / Line 69: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)'''
+|'''(a)''' &nbsp; <math>\frac{211}{8}</math>
-|-
-|'''The Fundamental Theorem of Calculus, Part 1'''
-|-
-|Let <math style="vertical-align: -4px">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 style="vertical-align: -4px">f</math> be continuous on <math style="vertical-align: -5px">[a,b]</math> and let <math style="vertical-align: 0px">F</math> be any antiderivative of <math style="vertical-align: -4px">f</math>.
-|-
-|Then, <math style="vertical-align: -14px">\int_a^b f(x)~dx=F(b)-F(a)</math>.
-|-
-|'''(b)''' &nbsp; <math style="vertical-align: -15px">\frac{d}{dx}\int_0^{\cos (x)}\sin (t)~dt\,=\,\sin(\cos(x))\cdot(-\sin(x))</math>.
 |-
-|'''(c)''' <math style="vertical-align: -14px">\int_{0}^{\pi/4}\sec^2 x~dx\,=\,1</math>.
+|'''(b)''' &nbsp; <math>\frac{28\sqrt{7}-4}{3}</math>
 |}
 [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

Difference between revisions of "009B Sample Midterm 2, Problem 2"

Revision as of 10:17, 5 February 2017

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