Winter 2006 - Wavrik's Class - Exam 1

# Winter 2006 - Wavrik's Class - Exam 1 - Prof Wavrik Math...

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Prof Wavrik Math 20B Winter 2006 Midterm_solutions Part I - Short Answers Evaluate the following integrals: 1. 3 1 0 1 2 0 2 | 33 1 x x xdx −= = 2. 3 1 1 22 2 0 0 0 1 12 | 3 (1 3 ) x xx d x xd x −+ =−+ = 1 3. x x e C ed x π = + 4. 2 2 1 ln(1 ) 2 1 x x C dx x + = + + 5. 2 sin( )cos( ( ) sin ) 2 d x x C = +

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Part II - Applications 1_a. The shaded region lies between the curves y=sin(x) and y=cos(x). Find the area of the shaded region. The intersection is where sin(x)=cos(x) which is x= π /4. The area is /4 0 cos( ) sin( ) 2 1 xx d x π −= 1_b. The shaded region lies between the curves y=sin(x) and y=cos(x). Find the area of the shaded region. In this case the area is 5/ 4 sin( ) cos( ) 2 2 d x
2_a . ..Find the volume of a right circular cone of height h and base radius r. It is easiest to see this by laying the cone on its side and treating it as a solid obtained by revolving the line rx h y = 2 () b Af x π for x = 0 to h around the x-axis. The general formula is . In this case we get a = d x 2 2 0 1 3 h rx dx r h h ππ  =   2_b. Find the volume of a sphere of radius r.

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Winter 2006 - Wavrik's Class - Exam 1 - Prof Wavrik Math...

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