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| − | <span class="exam">Evaluate the integral: | + | <span class="exam"> Evaluate the indefinite and definite integrals. |
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| − | ::<math>\int \sin^3x \cos^2x~dx</math>
| + | <span class="exam">(a) <math>\int x^2 e^x~dx</math> |
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| | + | <span class="exam">(b) <math>\int_{1}^{e} x^3\ln x~dx</math> |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Foundations:
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| − | |-
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| − | |'''1.''' Recall the trig identity: <math style="vertical-align: -2px">\sin^2x+\cos^2x=1.</math>
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| − | |'''2.''' How would you integrate <math style="vertical-align: -14px">\int \sin^2x\cos x~dx?</math>
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| − | ::You could use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -2px">u=\sin x.</math> Then, <math style="vertical-align: -1px">du=\cos x~dx.</math> Thus,
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| − | ::<math>\begin{array}{rcl}
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| − | \displaystyle{\int \sin^2x\cos x~dx} & = & \displaystyle{\int u^2~du}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{u^3}{3}+C}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{\sin^3x}{3}+C.} \\
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| − | \end{array}</math>
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| − | |}
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| − | '''Solution:'''
| + | <hr> |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | [[009B Sample Midterm 1, Problem 4 Solution|'''<u>Solution</u>''']] |
| − | !Step 1:
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| + | |
| − | |First, we write
| + | [[009B Sample Midterm 1, Problem 4 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | ::<math style="vertical-align: -13px">\int\sin^3x\cos^2x~dx=\int (\sin x) \sin^2x\cos^2x~dx.</math>
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| − | |-
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| − | |Using the identity <math style="vertical-align: -2px">\sin^2x+\cos^2x=1,</math> we get
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| − | ::<math style="vertical-align: -1px">\sin^2x=1-\cos^2x.</math>
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| − | |-
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| − | |If we use this identity, we have
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| − | ::<math>\begin{array}{rcl}
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| − | \displaystyle{\int\sin^3x\cos^2x~dx} & = & \displaystyle{\int (\sin x) (1-\cos^2x)\cos^2x~dx}\\
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| − | &&\\
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| − | & = & \displaystyle{\int (\cos^2x-\cos^4x)\sin(x)~dx.}\\
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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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| − | !Step 2:
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| − | |-
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| − | |Now, we use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -5px">u=\cos(x).</math> Then, <math style="vertical-align: -5px">du=-\sin(x)dx.</math> Therefore,
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| − | ::<math>\begin{array}{rcl}
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| − | \displaystyle{\int\sin^3x\cos^2x~dx} & = & \displaystyle{\int -(u^2-u^4)~du}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{-u^3}{3}+\frac{u^5}{5}+C}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{\cos^5x}{5}-\frac{\cos^3x}{3}+C.}\\
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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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| − | |-
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| − | | <math>\frac{\cos^5x}{5}-\frac{\cos^3x}{3}+C</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>''']] |
