080205quiz3solution

080205quiz3solution - Z 1 √ x 2 16 dx We use the...

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MATH 126 QUIZ #3 Solution 02/07/08 Mathias Knape Last Name : Circle discussion section First Name: 11 am 12 pm 1 pm Note: In order to receive full credit, please show all your work (conceptual steps, definitions you use, . ..) 1. Evaluate the integrals (a) Z 1 x 3 + 2 x 2 + x dx At first we have to find the partial fraction representation. Observe that 1 x 3 + 2 x 2 + x = 1 x ( x + 1) 2 = A x + B x + 1 + C ( x + 1) 2 Multiplying by x ( x + 1) 2 we obtain that 1 = A ( x + 1) 2 + Bx ( x + 1) + Cx. Plugging in x = 0 and x = - 1 gives us A = 1 and C = - 1 respectively. Now we can choose an arbitrary point x , e.g. x = 1, and insert it in the equation above to obtain B = - 1. Thus Z 1 x 3 + 2 x 2 + x dx = Z 1 x - 1 x + 1 - 1 ( x + 1) 2 dx = ln | x | - ln | x + 1 | + 1 x + 1 + D (b) Z x + 1 x 2 - 4 x + 6 dx Z x + 1 x 2 - 4 x + 6 dx = Z x + 1 ( x - 2) 2 + 2 dx = Z u + 3 u 2 + 2 du = Z u u 2 + 2 du + Z 3 u 2 + 2 du = 1 2 ln( u 2 + 2) + 3 2 arctan( u 2 ) + C = 1 2 ln( x 2 - 4 x + 6) + 3 2 arctan( x - 2 2 ) + C
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Unformatted text preview: Z 1 √ x 2 + 16 dx We use the trigonometric substitution x = 4 tan θ , then dx = 4 sec 2 θdθ . Thus Z 1 √ x 2 + 16 dx = Z 4 sec 2 θ 4 sec θ dθ = Z sec θdθ = ln | sec θ + tan θ | + C = ln ± ± ± ± ± √ x 2 + 16 4 + x 4 ± ± ± ± ± + C 2. Use Simpson’s Rule to approximate Z 3 1 1 + y 5 dy for n = 6. You do not have to com-pute a final numerical value. Define f ( y ) = 1 1+ y 5 and notice that Δ x = 3-6 = . 5. Using Simpson’s Rule we have that Z 3 1 1 + y 5 dy ≈ 1 3 1 2 [ f (0) + 4 f ( 1 2 ) + 2 f (1) + 4 f ( 3 2 ) + 2 f (2) + 4 f ( 5 2 ) + f (3)] ....
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This note was uploaded on 04/27/2008 for the course MATH 126 taught by Professor Mikulevicius during the Spring '07 term at USC.

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080205quiz3solution - Z 1 √ x 2 16 dx We use the...

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