APPM 1350 Review Solutions 3

Find the intervals where the function f x dt is

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Unformatted text preview: 0 1+t+t Solution: 1 ( 1 + 2x) Note that f 0 (x) = = (1 + x + x2 ) 1 and f 00 (x) = and the only potential 1 + x + x2 (1 + x + x2 )2 00 (x) > 0 if x < 1/2 and f 00 (x) < 0 if x > 1/2 and so f (x) is point of inflection is x = 1/2 and f concave up on ( 1, 1/2) and concave down on ( 1/2, +1). p = d dx cos2 (y ) p Concavity Test ^ _ | -1/2 6. The velocity function (in meters per second) of a particle in motion is given by v (t) = 3t 5, find (a) the displacement and (b) the total distance traveled by the particle in the first 3 seconds. Solution: Z (a) Here, 3 v (t) dt = 0 Z 3 3t 5 dt = 0 3 3t 2 2 5t = 0 27 15 = 2 3/2. So the displacement is -1.5 meters. (b) Note, total distance is given by, Z 3 |v (t)| dt = 0 Z 3 0 | 3t 5| dt = = Z 5/3 (5 0 5t 3t 2 2 3t) dt + ! 5 /3 Z + 0 3 (3t 5/3 3t2 2 5) dt ! 3 5t 5/3 = 41 6 So the total distance travelled is 41/6 meters. Z2 7. (a) Find an approximation to the integral (2 x2 ) dx using a Riemann sum with a regular partition 0...
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