solutions for b

# X0 we know this equals from above consequently 2 ex

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Unformatted text preview: s information and evaluate the integral 2 e−x dx x=0 follows: In the integral in (#) above, do the substitution t = x2 , and hence dt = 2 x dx The interval correspondence: t = +∞ x = +∞ t= 0 x= 0 Accordingly, the integral in (#) is rewritten as +∞ 1 t− 2 e−t dt +∞ = x=0 t=0 = +∞ 2 1 x 2 · e−x · 2 x dx 2 e−x dx. x=0 We know this equals √ π from (#) above. Consequently, +∞ 2 e−x dx = x=0 11 1√ π. 2 . as √ 1 √ 2 [IV] (30pts) 3 2 x=0 3 − x2 dx 1 − x2 · 1 = 2 − 3 2 2 arctan 3 2 ·x 2· − 3 2 3 2 −1 √2 1 − x2 x2 x=0 1 = 9 3 − ·x 4 2 3 − 3−1 2 2 arctan √ 1 − x2 2 x2 x=0 1 = = = = 3 4 2 arctan arctan √ arctan arctan √ √ 3 = − 2 x2 2 x=0 √1 2 3 ·x 1− 3 − 2 x2 2 1 ·√ 2 3 2 3 1 − x2 ·x 3 2 √ · x2 x=0 1− 1 √ 2 −2· π . 3 12 1 √ 2 2 2 − arctan 0 [V] (30pts) Recall √ 1−x = ∞ 2 m − 3 !! m=1 1− 2 m !! Note: For m = 1, the quantity (1) (2) 1 − tx x with t x, then you obtain = 1− ∞ 2 m − 3 !! m=1 2 m !! Divide the both sides of the above by √ 1 − tx √√ t 1−t = √ − (3) 2 m − 3 !! is understood as 1. Agree that, in the above, if you substitute √ xm . √ t √ tm x m . 1 − t , and obtain 1 √ t 1−t ∞ 2 m − 3 !! m=1 2 m !! √ The Taylor series expression for 1 Ex = t=0 √ 1 − tx √√ t 1−t is found as follows: 13 dt tm √ t 1−t xm . Integrate the both sides of the...
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## This note was uploaded on 01/12/2014 for the course MATH 116 taught by Professor Scholle,minho during the Fall '08 term at Kansas.

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