math1B-Fall_06-mt1-Tataru-soln

math1B-Fall_06-mt1-Tataru-soln - Mathematics 1B. Fall...

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Unformatted text preview: Mathematics 1B. Fall Semester 2006 Professor: Daniel Tataru Midterm 1, Solutions 1 (20) . Evaluate the following (indefinite) integrals a ) Z e x dx Solution: Substitute x = u 2 , dx = 2 udu . The integral becomes Z 2 ue u du We integrate by parts to obtain Z 2 ue u du = 2 ue u- Z 2 e u du = 2 ue u- 2 e u + C = 2 xe x- 2 e x + C b ) Z x tan 2 x dx Solution: We rewrite the integral as Z x tan 2 x dx = Z x (sec 2 x- 1) dx = Z x sec 2 x dx- x 2 2 The first term is integrated by parts, Z x sec 2 x dx = x tan x- Z tan xdx = x tan x + ln | cos x | + C The final result is Z x tan 2 x dx = x tan x + ln | cos x | - x 2 2 + C 2 (20) . Evaluate the following (definite) integrals: a ) Z - 4 x 2 x 4 + 4 dx Solution: We use partial fractions. First we factor the denominator, x 4 + 4 = x 4 + 4 x 2 + 4- 4 x 2 = ( x 2 + 2) 2- (2 x ) 2 = ( x 2 + 2 x + 2)( x 2- 2 x + 2) Then we decompose into partial fractions, 4 x 2 x 4 + 4 = Ax + B x 2- 2 x + 2- Cx + D x 2 + 2 x + 2 This gives 4 x 2 = ( Ax + B )( x 2 + 2 x + 2) + ( Cx + D )( x 2- 2 x + 2) Identifying the coefficients we obtain the equations A + C = 0 , 2 A + B-...
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math1B-Fall_06-mt1-Tataru-soln - Mathematics 1B. Fall...

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