solutions_11_23

# Solutions_11_23 - Solution This integral represents the amount of oil leaked from time t = 0 to t = 120 ± 4 In this problem we will determine Z e

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EVALUATING DEFINITE INTEGRALS (SOLUTIONS) 1. Evaluate the following integrals. (1) Z 2 0 (6 x 2 - 4 x + 5) dx Solution. An antiderivative of 6 x 2 - 4 x + 5 is 2 x 3 - 2 x 2 + 5 x . Hence the integral evaluates to (2 · 2 3 - 2 · 2 2 + 5 · 2) - 0 = 18. ± (2) Z 2 π π cos θdθ Solution. An antiderivative of cos θ is - sin θ . Hence the integral evaluates to - sin(2 π ) + sin( π ) = 0. ± (3) Z 1 0 4 1 + x 2 dx Solution. An antideriative of 4 1+ x 2 is 4 arctan( x ). Hence the integral evaluates to 4 arctan(1) - 4 arctan(0) = π . ± (4) Z π/ 4 0 ± 1 + cos 2 θ cos 2 θ ² Solution. We have Z π/ 4 0 1 + cos 2 θ cos 2 θ = Z π/ 4 0 sec 2 θdθ + Z π/ 4 0 1 dθ. Now, use that tan θ is an antiderivative of sec 2 θ . ± 2. What is wrong with the following equation? Z 3 - 1 1 x 2 dx = - 1 x ³ ³ ³ ³ 3 - 1 = - 4 3 . Solution. The fundamental theorem of calculus only applies to integrands that are continuous on the interval of integration. The function 1 x 2 is not continuous on [ - 1 , 3], since it is not deﬁned at x = 0. ± 3. If oil leaks from a tank at a rate of r ( t ) gallons per minute at time t , what does Z 120 0 r ( t ) dt represent?

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Unformatted text preview: Solution. This integral represents the amount of oil leaked from time t = 0 to t = 120. ± 4. In this problem, we will determine Z e 1 log( y ) dy . (1) Compute d dy ( y (log y-1). Solution. By the product and sum rules, we get d dy ( y log y-y ) = y · 1 y + 1 · log y-y = log y. ± (2) Hence determine Z e 1 log( y ) dy . Date : November 23, 2009. 1 Solution. From the previous part, we see that y log y-y is an antiderivative of log y . Hence, the integral is ( e log e-e )-(1 log 1-1) = 1. ± (3) Also, ﬁnd Z 1 e x dx . Solution. Since an antiderivative of e x is e x , we ﬁnd the integral to be e 1-e = e-1. ± (4) Your answers to parts (b) and (c) should add up to e . Can you explain this graphically? Solution. I will leave this to you. ± 2...
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## This note was uploaded on 03/27/2012 for the course MATH 1a taught by Professor Benedictgross during the Fall '09 term at Harvard.

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Solutions_11_23 - Solution This integral represents the amount of oil leaked from time t = 0 to t = 120 ± 4 In this problem we will determine Z e

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