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| | <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">(b) <math style="vertical-align: -14px">\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx</math> |
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| | + | (insert picture of handwritten solution) |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | [[009B Sample Midterm 2, Problem 2 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']] |
| − | !Foundations:
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| − | |-
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| − | |How would you integrate <math style="vertical-align: -12px">\int (2x+1)\sqrt{x^2+x}~dx?</math>
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| − | You can use <math style="vertical-align: 0px">u</math>-substitution.
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| − | |-
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| − | | Let <math style="vertical-align: -2px">u=x^2+x.</math> | |
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| − | | Then, <math style="vertical-align: -4px">du=(2x+1)~dx.</math>
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| − | Thus,
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| − | |-
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\int (2x+1)\sqrt{x^2+x}~dx} & = & \displaystyle{\int \sqrt{u}~du}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{2}{3}u^{3/2}+C}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{2}{3}(x^2+x)^{3/2}+C.}
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| − | \end{array}</math>
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| − | |}
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| − |
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| − | '''Solution:'''
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| − |
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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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| − | |-
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| − | |We multiply the product inside the integral to get
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| − | |-
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| − | |
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt} & = & \displaystyle{\int_1^2 \bigg(8t^3-10+12-\frac{15}{t^3}\bigg)~dt}\\
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| − | &&\\
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| − | & = & \displaystyle{\int_1^2 (8t^3+2-15t^{-3})~dt.}
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| − | \end{array}</math>
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| − | |}
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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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| − | |-
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| − | |We integrate to get
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| − | |-
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| − | | <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>
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| − | |-
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| − | |We now evaluate to get
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| − | |-
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt} & = & \displaystyle{2(2)^4+2(2)+\frac{15}{2(2)^2}-\bigg(2+2+\frac{15}{2}\bigg)}\\
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| − | &&\\
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| − | & = & \displaystyle{36+\frac{15}{8}-4-\frac{15}{2}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{211}{8}.}
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| − | \end{array}</math>
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| − | |}
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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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| − | |-
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| − | |We use <math style="vertical-align: 0px">u</math>-substitution.
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| − | |-
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| − | |Let <math style="vertical-align: -2px">u=x^4+2x^2+4.</math>
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| − | |-
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| − | |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>
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| − | |-
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| − | |Also, we need to change the bounds of integration.
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| − | |-
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| − | |Plugging in our values into the equation <math style="vertical-align: -4px">u=x^4+2x^2+4,</math> we get
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| − | |-
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| − | | <math style="vertical-align: -5px">u_1=0^4+2(0)^2+4=4</math> and <math style="vertical-align: -5px">u_2=2^4+2(2)^2+4=28.</math>
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| − | |-
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| − | |Therefore, the integral becomes
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| − | | <math style="vertical-align: -14px">\frac{1}{4}\int_4^{28}\sqrt{u}~du.</math>
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| − | |}
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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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| − | |-
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| − | |We now have
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| − | |-
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| − | |
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx} & = & \displaystyle{\frac{1}{4}\int_4^{28}\sqrt{u}~du}\\
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| − | &&\\
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| − | & = & \displaystyle{\left.\frac{1}{6}u^{\frac{3}{2}}\right|_4^{28}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{1}{6}(28^{\frac{3}{2}}-4^{\frac{3}{2}})}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{1}{6}((\sqrt{28})^3-(\sqrt{4})^3)}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{1}{6}((2\sqrt{7})^3-2^3).}
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| − | \end{array}</math>
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| − | |-
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| − | |Therefore,
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| − | |-
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| − | | <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>
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| − | |}
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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{211}{8}</math>
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| − | |-
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| − | | '''(b)''' <math>\frac{28\sqrt{7}-4}{3}</math>
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| − | |}
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| | [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] | | [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] |
