Spring 2004 - Nordgren's Class - Quiz 1

Spring 2004 - Nordgren's Class - Quiz 1 - x 2. Answer: If...

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Solutions to Quiz 1 By H˚ akan Nordgren Question 1: Find the most general anti-derivative of f ( x ) = x 3 +2 x 2 x 1 2 . Answer: It is easier to see how to proceed if we simplify the f a little: f ( x ) = x 3 + 2 x 2 x 1 2 = x 5 2 + 2 x 3 2 . Thus from this we can see that the function F ( x ) = 2 7 x 7 2 + 4 5 x 5 2 + C , where C is a constant will have derivative f . Remember to check your answer by differentiating it, to see that its derivative really is the function you started with. Question 2: Find a function f such that f 00 ( x ) = 3 e - x , with f (0) = 1 and f 0 (0) = 2. Answer: So we need to find an anti-derivative of f 00 . The function - 3 e - x + C 1 works (check it). We also need that this function when evaluated at 0 gives 2. Thus we need - 3 .e 0 + C 1 = 2. Therefore, C 1 = 5. So f 0 ( x ) = - 3 e - x + 5. Now we need to find an anti-derivative of f 0 . The function 3 e - x +5 x + C 2 works. We also need 3 e 0 + 5 . 0 + C 2 = 1. Thus C 2 = - 2. Therefore, f ( x ) = 3 e - x + 5 x - 2. Question 3: Using right end-points, find R n and an expression for the area under the curve y = x 2 on the interval 0
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Unformatted text preview: x 2. Answer: If we split up the interval 0 x 2 into n pieces, each piece will be 2-n = 2 n long. The end-points of the rst sub-interval will be 0 and 0 + 2 n = 2 n . The end-points of the second sub-interval will be 2 n and (0+ 2 n )+ 2 n = 2 2 n .The end-points of the third sub-interval will be 2 2 n and (2 2 n ) + 2 n = 3 2 n . If we label the end-points x ,x 1 ,...,x n , then we have x i = i 2 n . The point on the curve y = x 2 corresponding to x = 0 is 0. The point cor-responding to x 1 = 2 n is 2 n 2 . And the point corresponding to x 2 = 2 2 n is 2 2 n 2 . Thus R n = n X i =1 i 2 n 2 2 n = 2 n 3 n X i =1 i 3 . Thus an expression for the area under the curve is Area = lim n 2 n 3 n X i =1 i 3 . 1...
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