31bm1sol

31bm1sol - 31B MIDTERM 1-SOLUTION Integration &...

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Unformatted text preview: 31B MIDTERM 1-SOLUTION Integration & Infinite Series 262-186-200 Monday April 21, 2008 FOWLER A103B 09:00AM-09:50AM Exercise 1. (10 points) In each of the following, calculate the indefinite integral: (1) Z x 2 e x 3 dx Z x 2 e x 3 dx = 1 3 e x 3 + C. (2) Z e- 2 x sin( e- 2 x ) dx Z e- 2 x sin( e- 2 x ) dx = 1 2 cos( e- 2 x ) + C. (3) Z ln 2 ( x ) dx x , use u = ln( x ). Set u = ln( x ), so that du = 1 x dx . Thus we have, Z ln 2 ( x ) dx x = Z u 2 du = 1 3 u 3 + C = 1 3 ln 3 ( x ) + C. Exercise 2. (10 points) In each of the following, calculate the derivative: (1) f ( x ) = ln(4 x 2 + 1) f ( x ) = 8 x 4 x 2 + 1 (2) f ( x ) = e sin 2 ( x ) f ( x ) = 2 cos( x ) sin( x ) e sin 2 ( x ) (3) f ( x ) = cos- 1 1 x . (Note that cos- 1 means the inverse function of cos). Remind that if g is an invertible function, the derivative of g- 1 is given by the following formula: ( g- 1 ) ( b ) = 1 g ( g- 1 ( b )) 1 2 Apply this for g = cos. We get using the Chain Rule d dx cos- 1 1 x =- 1 x 2 (cos- 1 )...
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31bm1sol - 31B MIDTERM 1-SOLUTION Integration &...

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