APPM 1350 Review Solutions 3

A find the average value of f t t 1 t2 over the

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Unformatted text preview: 5]. (b) What does the Mean Value Theorem for integrals say? Find a c in [2, 5] such that fave = f (c) where f (x) = (x 3)2 Solution: (a) Note that Z5p Z 1 1 1 26 p 1 fave = t 1 + t2 dt |{z} = · u du = 5 00 521 10 2 2 3/2 u 3 u=1+t 26 1 ! = 1 (263/2 15 1) (b) Mean Value Theorem for Integrals: If f is continuous on [a, b], then there exists a number c in [a, b] such that Zb 1 f (c) = fave = f (x)dx. b aa Now note that Z5 Z 1 1 22 1 u3 2 9 fave = (x 3)2 dx |{z} = u du = · = =1 5 22 31 3319 u= x 3 3)2 = 1 and so c = 3 ± 1 = 2, 4. Z cos(y) p Zx cos(t) 3 2 dt (b) y = 4. Find the derivative of : (a) g (y ) = 1t dt p t 1 x Now f (c) = fave implies (c Solution: (a) g 0 (y ) = (b) p 3 1 cos2 (y ) · dy d = dx dx Z x p sin(y ) 3 1 sin(y ) = cos(t) dt t Z 0 cos(t) dt + t Z x cos(t) dt t x x 0 " Zp # Zx x d cos(t) cos(t) = dt + dt dx t t 0 0 p p cos( x) 1 cos(x) cos( x) cos(x) p p+ = = + x 2x x x 2x Zx 1 5. Find the intervals where the function f (x) = dt is concave up and concave down. 2...
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This note was uploaded on 02/04/2014 for the course APPM 1350 taught by Professor Segur,harv during the Fall '07 term at Colorado.

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