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hw3sol

# hw3sol - y = sin 2 x 2 tan x and y = sin 2 x 2 tan x e 2...

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Fall 2010, MAT21B, Paper Homework 3 (Solution) 1. Evaluate Z e 1 ln x x dx. Solution: Let u = ln x , then du = 1 x · dx 2 x = dx 2 x . Thus, Z e 1 ln x x dx = Z 1 2 0 2 udu = u 2 | u = 1 2 u =0 = 1 4 . Method 2: Note that Z e 1 ln x x dx = Z e 1 1 2 ln x x dx Let u = ln x , du = dx x we have Z e 1 ln x x dx = Z e 1 1 2 ln x x dx = 1 2 Z 1 0 udu = u 2 4 | 1 0 = 1 4 . 2. Solve the differential equation dy dx = 12 x 2 - 10 x (4 x 3 - 5 x 2 + 8) 2 / 3 with the initial value y (0) = 7. Solution: y ( x ) = Z 12 x 2 - 10 x (4 x 3 - 5 x 2 + 8) 2 / 3 dx. Let u = 4 x 3 - 5 x 2 + 8, then du = (12 x 2 - 10 x ) dx. Thus, y ( x ) = Z du u 2 / 3 = 3 u 1 3 + C = 3(4 x 3 - 5 x 2 + 8) 1 3 + C Since y (0) = 7, we have 7 = 3 · 8 1 / 3 + C = 6 + C So, C = 1 and the solution is y = 3(4 x 3 - 5 x 2 + 8) 1 3 + 1 . 1

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3. Find the area of the region enclosed by the curves
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Unformatted text preview: y = sin 2 ( x 2 + tan x ) and y = sin 2 ( x 2 + tan x ) + e 2 √ x √ x over 1 ≤ x ≤ 4. Solution: The area is A = Z 4 1 ± ± ± ± ² sin 2 ( x 2 + tan x ) + e 2 √ x √ x ³-sin 2 ( x 2 + tan x ) ± ± ± ± dx = Z 4 1 e 2 √ x √ x dx Let u = 2 √ x , we have du = dx √ x . Therefore, A = Z 4 1 e 2 √ x √ x dx = Z 4 2 e u du = e 4-e 2 . 2...
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hw3sol - y = sin 2 x 2 tan x and y = sin 2 x 2 tan x e 2...

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