31B(F07)_Midterm2_Solutions

# 31B(F07)_Midterm2_Solutions - MATH 31B Lecture 4 Fall 2007...

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MATH 31B - Lecture 4 - Fall 2007 Solutions to Midterm 2 - November 16, 2007 This is a closed-book and closed-note examination. Calculators are not allowed. Please show all your work. Use only the paper provided. You may write on the back if you need more space, but clearly indicate this on the front. There are 6 problems for a total of 100 points. POINTS: 1. 2. 3. 4. 5. 6. TOTAL: 1

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2 1. (15 points) Evaluate the integral Z 3 x + 1 x 4 + x 3 + x 2 dx Solution. Here we do a partial fraction decomposition and write: 3 x + 1 x 4 + x 3 + x 2 = 3 x + 1 x 2 ( x 2 + x + 1) = A x + B x 2 + Cx + D x 2 + x + 1 Comparing the numerators of both sides gives us: Ax ( x 2 + x + 1) + B ( x 2 + x + 1) + ( Cx + D )( x 2 ) = 3 x + 1 Comparing the coefficients for x 3 , x 2 , x , and the constants gives us the following equations: A + C = 0 A + B + D = 0 A + B = 3 B = 1 Solving the equations gives us A = 2, B = 1, C = - 2, D = - 3 Now we can try to solve our integral. Z 3 x + 1 x 4 + x 3 + x 2 dx = Z 2 x dx + Z 1 x 2 dx + Z - 2 - 3 x x 2 + x + 1 dx = 2 ln | x | - 1 /x + Z - 2 x - 3 x 2 + x + 1 dx Let us complete the square to evaluate the last integral. We let u = ( x +1 / 2) then Z - 2 x - 3 x 2 + x + 1 = Z - 2 u - 2 u 2 + 3 / 4 du = Z - 2 u u 2 + 3 / 4 + Z - 2 u 2 + 3 / 4 du = - ln | u 2 + 3 / 4 | - 4 3 3 tan - 1 ( u/ p 3 / 4) .
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