151reviewfsolF09

151reviewfsolF09 - Partial solutions to the final exam...

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Unformatted text preview: Partial solutions to the final exam review problems (1) The area is Z 4 / 3 / 3 sin x- 3cos xdx = 4. (2) We find R 4 1 f ( x ) dx = R 10 1 f ( x ) dx- R 10 4 f ( x ) dx = 2- 7 =- 5. Similarly, we get R 10 8 f ( x ) dx = R 10 1 f ( x ) dx- R 8 1 f ( x ) dx = 2- 14 =- 12. Finally, R 8 4 f ( x ) dx = R 10 1 f ( x ) dx- R 4 1 f ( x ) dx- R 10 8 f ( x ) dx = 2- (- 5)- (- 12) = 19. (3) The integrals are 2 x + 5 x 2 2 + x 3 + C , 2ln 2 3- 15 2 and 5 2 . (4) The integrals are e x 3 +4 3 + C , sin 2 x 2 + C and tan 2 x 2 + C . (5) It takes 7ln3 ln4 days. (6) It is continuous because lim x 1- 2 x +1 = lim x 1+ 4 x- 1 = 3 = f (1). If it were differentiable at 1 then the derivative of 2 x + 1 at 1 would be the same as the derivative of 4 x- 1 at 1, but this would imply 2 = 4, which is a contradiction. (7) Continuity at 0 implies a = lim x 0+ f ( x ) = lim x 0+ x ln x = lim x 0+ ln x 1 /x = lim x 0+ 1 /x- 1 /x 2 = 0, where we used lH opitals Rule at the end. If the function were differentiable at 0 then we would have f (0) = lim x 0+ f ( x )- f (0) x- = lim x 0+ x ln x x = lim x 0+ ln x =- , and f (0) would not be a number. We conclude that the function is not differentiable at 0. The functionnot be a number....
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151reviewfsolF09 - Partial solutions to the final exam...

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