math1a - quiz 10 - solns

# math1a - quiz 10 - solns - , 3] into n equal pieces, of...

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Math 1A Quiz 10 Solutions April 25th, 2008 1. Find the most general antiderivatives of each of the following functions: (a) f ( x ) = x 4 +3 x x 2 (b) g ( x ) = 4 1 - x 2 (c) h ( x ) = 2 cos x + sec 2 x , - π 2 < x < π 2 . Solution to a): Distribute the denominator. We get f ( x ) = x 2 + 3 x - 3 / 2 . Now use the power rules for antidiﬀerentiation to get F ( x ) = x 3 3 - 6 x - 1 / 2 + C. Solution to b): The antiderivative is G ( x ) = 4 sin - 1 ( x ) + C, by memorization. Solution to c): The antiderivative is H ( x ) = 2 sin x + tan x + C. 1

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2. (a) Write an expression for R 3 1 t 2 dt as a limit of Riemann sums. The Riemann sums you use should divide the interval [1 , 3] into n pieces of equal width, and you should use the right-hand end- points for your x * i . (b) Evaluate the Riemann sum you wrote down in part (a) when n = 4. Solution to a): We divide the interval [1
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Unformatted text preview: , 3] into n equal pieces, of length 3-1 n = 2 n . The i th piece of this partition has right-hand endpoint 1 + i 2 n , so according to the instructions in the problem we have Δ x = 2 n and x * i = 1 + i 2 n . The resulting Riemann sum is: n X i =1 f ( x * i ) · Δ x = n X i =1 ± 1 + i 2 n ² 2 · 2 n . And the integral is the limit of the Riemann sums: Z 3 1 t 2 dt = lim n →∞ n X i =1 ± 1 + i 2 n ² 2 · 2 n . Solution to b): Just plug in n = 4 into our Riemann sum. We get: 4 X i =1 ± 1 + i 2 ² 2 · 1 2 = 1 2 " ± 1 + 1 2 ² 2 + ± 1 + 2 2 ² 2 + ± 1 + 3 2 ² 2 + ± 1 + 4 2 ² 2 # = 1 2 ± 9 4 + 16 4 + 25 4 + 36 4 ² = 43 4 . 2...
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## This note was uploaded on 09/17/2011 for the course MATH 1A taught by Professor Wilkening during the Spring '08 term at Berkeley.

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math1a - quiz 10 - solns - , 3] into n equal pieces, of...

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