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practice-exam3-21b-sol

# practice-exam3-21b-sol - Math 21B UC Davis Winter 2010 Dan...

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Math 21B UC Davis, Winter 2010 Dan Romik Solutions to Sample Exam Question 1 Compute the following integrals: (a) Z π 0 ( x 2 + sin( x )) dx Z π 0 ( x 2 + sin( x )) dx = ± 1 3 x 3 - cos( x ) ² ³ ³ ³ ³ ³ π 0 = π 3 3 - ( - 1) - (0 - 1) = π 3 3 + 2 (b) Z sec 2 ( x ) dx Z sec 2 ( x ) dx = tan( x ) + C (standard integral) (c) Z cos(2 x - 1) dx Z cos(2 x - 1) dx = 1 2 sin(2 x - 1) + C (use substitution u = 2 x - 1) (d) Z (1 + x ) 7 x dx Z (1 + x ) 7 x dx = 2 Z (1 + x ) 7 2 x dx = 2 Z (1 + u ) 7 du = 1 4 (1 + u ) 8 + C = 1 4 ( 1 + x ) 8 + C (use the substitution u = x ) (e) Z b 1 ( e y + e - y ) dy (note: the answer is a function of b ) Z b 1 ( e y + e - y ) dy = ( e y - e - y ) ³ ³ ³ b 1 = e b - e - b - e + e - 1 (f) Z 2 0 xe x 2 dx Z 2 0 xe x 2 dx = Z 4 0 1 2 e u du = 1 2 ( e 4 - e 0 ) = 1 2 ( e 4 - 1 ) (use the substitution u = x 2 ) 1

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Question 2 Compute the arc length of the curve y = x 3 / 6+1 / (2 x ) from x = 2 to x = 3. Solution. The arc length is given by R 3 2 p 1 + ( y 0 ) 2 dx . First we simplify the integrand: y 0 = x 2 2 - 1 2 x 2 , 1 + ( y 0 ) 2 = 1 + ± x 2 2 - 1 2 x 2 ² 2 = 1 + ± x 4 4 - 1 2 + 1 4 x 4 ² = x 4 4 + 1 2 + 1 4 x 4 = ± x 2 2 + 1 2 x 2 ² 2 , p 1 + ( y 0 ) 2 = ³ ³ ³ ³ x 2 2 + 1 2 x 2 ³ ³ ³ ³ = x 2 2 + 1 2 x 2 (absolute value is redundant for 2 x 3). Now proceed to compute the arc length: Z 3 2 p 1 + ( y 0 ) 2 dx = Z 3 2 ± x 2 2 + 1 2 x 2 ² dx = ± x 3 6 - 1 2 x ²³ ³ ³ ³ 3 2 = ± 9 2 - 1 6 ² - ± 4 3 - 1 4 ² = 13 4 . Question 3 Compute the following integrals using integration by parts or any other technique. (a) Z ln xdx : Use integration by parts with u = ln x , v = x (so that dv = 1 · dx ): Z ln xdx = Z ln x · 1 dx = x · ln x - Z x d dx (ln x ) dx = x ln x - Z 1 dx = x ln x - x + C (b) Z x ln xdx : Use integration by parts with
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practice-exam3-21b-sol - Math 21B UC Davis Winter 2010 Dan...

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