Spring 2004 - Nordgren's Class - Quiz 2

# Spring 2004 - Nordgren's Class - Quiz 2 - u = t and dv =...

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Solutions to quiz 2 By H˚ akan Nordgren Question 1: Evaluate the following indeﬁnite integrals 1. R x 2 x 3 + 1 dx and 2. R te t 2 dt by changing variables. Answer: 1. Let us make the substitution u = x 3 + 1. Then x = ( u - 1) 1 3 and dx = 1 3 ( u - 1) - 2 3 du . Thus Z x 2 x 3 + 1 dx = Z ( u - 1) 2 3 u 1 3 ( u - 1) - 2 3 du = 1 3 Z udu = 2 9 ( x 3 + 1) 3 2 + constant 2. In this case, let us make that change of variable u = t 2 . Then t = u and dt = 1 2 u - 1 2 du . Thus Z te t 2 dt = Z ue u 1 2 u - 1 2 du = Z 1 2 e u du = 1 2 e t 2 + constant Question 2: Use integration by parts to determine R 2 0 te - t dt . Answer: Integration by parts is easy, because you will eventually make the appro- priate choice of u and dv , and therefore you will eventually be able to solve problem. Remember that if your choice of u and dv does not make the problem easier, then you should go back and try again with new u and dv . For this problem, I think that it will make sense to pick

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Unformatted text preview: u = t and dv = e-t dt . Then du = dt and v =-e-t . Thus, Z 2 te-t dt = h-e-t t i 2-Z 2-e-t dt = h-e-t t i 2-h e-t i 2 =-e-2 2 + e-2-1 . 1 Question 3: Sketch 2 x 2 and sin x from-1 to 1. Hence ﬁnd R 1-1 (2 x 2 +sin x ) dx using symmetry. Check your answer by an explicit calculation. Answer: You will have to provide the sketch yourselves, because I have not yet worked out how to do that in Latex. From your diagram you will hopefully be able to tell that the integral of sin x is going to cancel out, since sin is odd. So R 1-1 (2 x 2 + sin x ) dx = R 1-1 2 x 2 dx = 4 R 1 x 2 = 4 3 2...
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## This note was uploaded on 04/29/2008 for the course MATH 20B taught by Professor Justin during the Spring '08 term at UCSD.

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Spring 2004 - Nordgren's Class - Quiz 2 - u = t and dv =...

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