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APPM 1350 Review Solutions 3

# APPM 1350 Review Solutions 3 - Solution APPM 1350 Review#3...

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Solution: APPM 1350 Review #3 Fall 2013 1. Evaluate the following integrals: (a) Z sin( p x ) p x dx (b) Z 1 (5 t + 4) 2 . 7 dt (c) Z sec 3 ( x ) tan( x ) dx (d) Z 1 0 x 2 (1 + 2 x 3 ) 5 dx (e) Z 2 - 2 ( x + 3) p 4 - x 2 dx (f) Z 2 - 1 ( x - 2 | x | ) dx (g) 10 Z 0 - 1 (1 + x 5 ) 3 / 2 x 9 dx (h) Z 1 0 4 x p 1 - x 4 dx Solution: (a) Let u = p x then du = dx/ 2 p x and, Z sin( p x ) p x dx = 2 Z sin( u ) du = - 2 cos( u ) + C = - 2 cos( p x ) + C (b) Let u = 5 t + 4 then du = 5 dt and so Z 1 (5 t + 4) 2 . 7 dt = 1 5 Z u - 2 . 7 du = 1 5 u - 1 . 7 - 1 . 7 + C = - (5 t + 4) 8 . 5 - 1 . 7 + C = - 1 8 . 5(5 t + 4) 1 . 7 + C (c) Z sec 3 ( x ) tan( x ) dx = Z sec 2 ( x ) sec( x ) tan( x ) dx = |{z} u =sec( x ) Z u 2 du = u 3 3 + C = sec 3 ( x ) 3 + C (d) Z 1 0 x 2 (1 + 2 x 3 ) 5 dx = |{z} u =1+2 x 3 1 6 Z 3 1 u 5 du = u 6 36 3 1 = 728 / 36 = 182 / 9 (e) Note that Z 2 - 2 ( x +3) p 4 - x 2 dx = Z 2 - 2 x p 4 - x 2 dx + Z 2 - 2 3 p 4 - x 2 dx , and note that f ( x ) = x p 4 - x 2 is an odd function so R 2 - 2 x p 4 - x 2 dx = 0 and if we interpret Z 2 - 2 p 4 - x 2 dx geometrically as the area of a semicircle of radius 2 then Z 2 - 2 p 4 - x 2 dx = 2 and thus, Z 2 - 2 ( x + 3) p 4 - x 2 dx = Z 2 - 2 x p 4 - x 2 dx + 3 Z 2 - 2 p 4 - x 2 dx = 0 + 3(2 ) = 6 (f) Using the fact that | x | = x if x 0 and | x | = - x if x < 0, we have Z 2 - 1 ( x - 2 | x | ) dx = Z 0 - 1 3 x dx + Z 2 0 - x dx = 3 x 2 2 0 - 1 - x 2 2 2 0 = - 7 / 2 (g) If u = 1 + x 5 , then du = 5 x 4 dx (and note x 5 = u - 1), so 10 Z 0 - 1 (1 + x 5 ) 3 / 2 x 9 dx = 2 Z 1 0 u 3 / 2 ( u - 1) du = 2 Z 1 0 ( u 5 / 2 - u 3 / 2 ) du = 2 2 7 u 7 / 2 - 2 5 u 5 / 2 1 0 = 2 2 7 - 2 5 = - 8 35 (h) If we use the change of variables u = x 2 , then du = 2 xdx and so we have Z 1 0 4 x p 1 - x 4 dx = 2 Z 1 0 p 1 - u 2 du now note that the integral Z 1 0 p 1 - u 2 du represents the area of one quarter of a unit circle, since the graph of y = p 1 - u 2 is the top half of a circle with radius equal to one. Thus, using the formula A = r 2 we have, Z 1 0 4 x p 1 - x 4 dx = 2 Z 1 0 p 1 - u 2 du = 2 · 4 = 2

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2. (a) If f is continuous and Z 4 0 f ( x ) dx = 10, find Z 2 0 f (2 x ) dx . (b) If f 0 is continuous on [ a, b ], show that 2 Z b a f ( x ) f 0 ( x ) dx = [ f ( b )] 2 - [ f ( a )] 2 . Solution: (a) Let u = 2 x , then du = 2 dx and so Z 2 0 f (2 x ) dx = 1 2 Z 4 0 f ( u ) du = 5 . (b) Note that since f 0 is continuous that implies the product f ( x ) f 0 ( x ) is continuous and therefore it is integrable , so the integral makes sense. Now if we let u = f ( x ) then du = f 0 ( x ) dx and so 2 Z b a f ( x ) f 0 ( x ) dx = 2 Z f ( b ) f ( a ) u du = 2 · u 2 2 f ( b ) f ( a ) = [ f ( b )] 2 - [ f ( a )] 2 3. (a) Find the average value of f ( t ) = t p 1 + t 2 over the interval [0 , 5].
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