|
|
| (19 intermediate revisions by 2 users not shown) |
| 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 <math>\frac{d}{dx}\int_0^{\cos (x)}\sin (t)~dt</math>
| |
| − | ::<span class="exam">c) Evaluate <math>\int_{0}^{\frac{\pi}{4}}\sec^2 x~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> |
| | | | |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009B Sample Midterm 2, Problem 2 Solution|'''<u>Solution</u>''']] |
| − | |-
| |
| − | |
| |
| − | |} | |
| | | | |
| − | '''Solution:'''
| |
| | | | |
| − | '''(a)''' | + | [[009B Sample Midterm 2, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| | | | |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 1:
| |
| − | |-
| |
| − | |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 differential function on <math>(a,b)</math> and <math>F'(x)=f(x)</math>.
| |
| − | |}
| |
| | | | |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 2:
| |
| − | |-
| |
| − | |'''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>
| |
| − | |}
| |
| − |
| |
| − | '''(b)'''
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 1:
| |
| − | |-
| |
| − | |Let <math>F(x)=\int_0^{\cos (x)}\sin (t)~dt</math>. The problem is asking us to find <math>F'(x)</math>.
| |
| − | |-
| |
| − | |Let <math>g(x)=\cos(x)</math> and <math>G(x)=\int_0^x \sin(t)~dt</math>.
| |
| − | |-
| |
| − | |Then, <math>F(x)=G(g(x))</math>.
| |
| − | |}
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 2:
| |
| − | |-
| |
| − | |If we take the derivative of both sides of the last equation, we get <math>F'(x)=G'(g(x))g'(x)</math> by the Chain Rule.
| |
| − | |}
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 3:
| |
| − | |-
| |
| − | |Now, <math>g'(x)=-\sin(x)</math> and <math>G'(x)=\sin(x)</math> by the '''Fundamental Theorem of Calculus, Part 1'''.
| |
| − | |-
| |
| − | |Since <math>G'(g(x))=\sin(g(x))=\sin(\cos(x))</math>, we have <math>F'(x)=G'(g(x))g'(x)=\sin(\cos(x))(-\sin(x))</math>
| |
| − | |}
| |
| − |
| |
| − | '''(c)'''
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 1:
| |
| − | |-
| |
| − | | Using the '''Fundamental Theorem of Calculus, Part 2''', we have
| |
| − | |-
| |
| − | |<math>\int_{0}^{\frac{\pi}{4}}\sec^2 x~dx=\left.\tan(x)\right|_0^{\frac{\pi}{4}}</math>
| |
| − | |}
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 2:
| |
| − | |-
| |
| − | |So, we get
| |
| − | |-
| |
| − | |<math>\int_{0}^{\frac{\pi}{4}}\sec^2 x~dx=\tan \bigg(\frac{\pi}{4}\bigg)-\tan (0)=1</math>
| |
| − | |-
| |
| − | |
| |
| − | |}
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Final Answer:
| |
| − | |-
| |
| − | |'''(a)'''
| |
| − | |-
| |
| − | |'''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 differential 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>
| |
| − | |-
| |
| − | |'''(b)''' <math>\sin(\cos(x))(-\sin(x)</math>
| |
| − | |-
| |
| − | |'''(c)''' <math>1</math>
| |
| − | |}
| |
| | [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] | | [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] |
