|
|
| (20 intermediate revisions by 2 users not shown) |
| Line 1: |
Line 1: |
| − | <span class="exam">Evaluate the indefinite and definite integrals. | + | <span class="exam"> Let <math style="vertical-align: -5px">f(x)=1-x^2</math>. |
| | | | |
| − | ::<span class="exam">a) <math>\int x^2\sqrt{1+x^3}dx</math>
| + | <span class="exam">(a) Compute the left-hand Riemann sum approximation of <math style="vertical-align: -14px">\int_0^3 f(x)~dx</math> with <math style="vertical-align: 0px">n=3</math> boxes. |
| − | ::<span class="exam">b) <math>\int _{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\cos(x)}{\sin^2(x)}dx</math>
| |
| | | | |
| | + | <span class="exam">(b) Compute the right-hand Riemann sum approximation of <math style="vertical-align: -14px">\int_0^3 f(x)~dx</math> with <math style="vertical-align: 0px">n=3</math> boxes. |
| | | | |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <span class="exam">(c) Express <math style="vertical-align: -14px">\int_0^3 f(x)~dx</math> as a limit of right-hand Riemann sums (as in the definition of the definite integral). Do not evaluate the limit. |
| − | !Foundations:
| |
| − | |-
| |
| − | | Review u substitution
| |
| − | |}
| |
| | | | |
| − | '''Solution:''' | + | <hr> |
| | + | [[009B Sample Midterm 1, Problem 1 Solution|'''<u>Solution</u>''']] |
| | | | |
| − | '''(a)'''
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 1:
| |
| − | |-
| |
| − | |We need to use substitution. Let <math>u=1+x^3</math>. Then, <math>du=3x^2dx</math> and <math>\frac{du}{3}=x^2dx</math>.
| |
| − | |-
| |
| − | |Therefore, the integral becomes <math>\frac{1}{3}\int \sqrt{u}du</math>.
| |
| − | |}
| |
| | | | |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | [[009B Sample Midterm 1, Problem 1 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| − | !Step 2:
| |
| − | |-
| |
| − | |We now have:
| |
| − | |-
| |
| − | |<math>\int x^2\sqrt{1+x^3}dx=\frac{1}{3}\int \sqrt{u}du=\frac{2}{9}u^{\frac{3}{2}}+C=\frac{2}{9}(1+x^3)^{\frac{3}{2}}+C</math>. | |
| − | |}
| |
| | | | |
| − | '''(b)'''
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 1:
| |
| − | |-
| |
| − | |Again, we need to use substitution. Let <math>u=\sin(x)</math>. Then, <math>du=\cos(x)dx</math>. Also, we need to change the bounds of integration.
| |
| − | |-
| |
| − | |Plugging in our values into the equation <math>u=\sin(x)</math>, we get <math>u_1=\sin\bigg(\frac{\pi}{4}\bigg)=\frac{\sqrt{2}}{2}</math> and <math>u_2=\sin\bigg(\frac{\pi}{2}\bigg)=1</math>.
| |
| − | |-
| |
| − | |Therefore, the integral becomes <math>\int_{\frac{\sqrt{2}}{2}}^1 \frac{1}{u^2}du</math>.
| |
| − | |-
| |
| − | |
| |
| − | |}
| |
| | | | |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 2:
| |
| − | |-
| |
| − | |We now have:
| |
| − | |-
| |
| − | |<math>\int _{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\cos(x)}{\sin^2(x)}dx=\int_{\frac{\sqrt{2}}{2}}^1 \frac{1}{u^2}du=\left.\frac{-1}{u}\right|_{\frac{\sqrt{2}}{2}}^1=-\frac{1}{1}-\frac{-1}{\frac{\sqrt{2}}{2}}=-1+\sqrt{2}</math>.
| |
| − | |-
| |
| − | |
| |
| − | |-
| |
| − | |
| |
| − | |}
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Final Answer:
| |
| − | |-
| |
| − | |'''(a)''' <math>\frac{2}{9}(1+x^3)^{\frac{3}{2}}+C</math>
| |
| − | |-
| |
| − | |'''(b)''' <math>-1+\sqrt{2}</math>
| |
| − | |}
| |
| | [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] | | [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] |
Let 
.
(a) Compute the left-hand Riemann sum approximation of 
with 
boxes.
(b) Compute the right-hand Riemann sum approximation of with boxes.
(c) Express as a limit of right-hand Riemann sums (as in the definition of the definite integral). Do not evaluate the limit.
Solution
Detailed Solution
Return to Sample Exam