1873_solutions

1873_solutions

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Unformatted text preview: 11 t dt. 1. Apply the above exercise to the functions f and g defined by the equations f x  x  1 e x and g x  x  1 e x for all x  0. Deduce that   Ý Þ0 Þ0 x e x x y 1 1 y dydx. There really isn’t anything to do here. The desired equation   Ý Þ0 Þ0 x e x x y 1 1 y dydx follows at once from the earlier exercise. 2. By making the substitution y  ux in the inside integral in Exercise 1, deduce that       B ,  . The substitution y  ux gives us 388 Ý Þ0 Þ0 e x x Ý  1 1 y dydx y Þ0 Þ0  x Þ0  Ý 1 x e e xx x 1 ux 1  1 1 Þ0 Þ0 Ý nÝ lim n 1 u 1 1 u du  and deduce that n  1 n  1 xdudx Þ 1/n t  e t dt t  e t dt  n Ý lim ne 1    B ,  . 3. Apply the method of integration by parts to the integral that defines 1    . 1  ux Þ 1/n t  1 n 1/n e e t dt  . 4. Make the substitution t  sin 2 y in the definition of the beta function, deduce that B ,   2 Þ /2 sin 2 1  cos 2 1 d. 0 This exercise is a routine manipulation. Þ0 1 1 1 2 1 2 and deduce that 1 2 This exercise is a routine manipulation. 5. Use Exercise 4 to evaluate B , 11 t dt t  .  and deduce that 6. Make the substitution t  u 2 in the definition of   2Þ Ý u 2 1 exp u 2 du 0 and then use Exercise 5 to find another way of showing that Þ0 Ý exp x 2 dx   . 2 Recall that we obtained this identity in an earlier exercise. This exercise is a routine manipulation. 7. With an eye on the exercises in an earlier subsection, prove that if p  1 then Þ0  sin p d  2 Þ /2 0 p 1 2  sin p d  p 2 . 1 Suppose that p  1. Using Exercise 2 we obtain Þ0  sin p d  Þ0  sin 2 p1 2 1  cos 2 1/2 1 d p1 1  1B , 2 2 2 1 2 p 1 2 p 1 2 1 2  1 2  p 1 2  p 2 8. Prove that if   0 then B ,   2 1 2 B , 1 2 and deduce that  2  2 2 389 1   1 . 2 1 . dx Hint: Suppose that   0. First observe that B ,   2 Þ /2 cos 2 1  sin 2 1 d 0 /2 2 Þ0 2  Þ0 2 2 1  2 sin  cos  /2 2 2 1 2 1 sin 2 2 1 d d Then make the substitution u  2 and use Exercise 2. B ,   /2 Þ0 2 2 2 1 cos 2 1/2 1 d , 1 2 2 B  21 2 1 sin 2 and so    21  2 1 2 1 2  which gives us the desired result. 9. Prove that Þ0 /2 . 2 The intention of this exercise is to express the left side as Þ0 /2 sin 2 3/4 1  cos 2 1/4 tan x dx  1 d  1B 3, 1 2 44 3 4 1 2 Now if we put   1 4 1 4 3 4  1 2 1 4 3 4 1 4 in the identity  2  2 2 1  1 2  we see that  1 2 1 4  22 1 3 4 1 4 and so Þ0 /2  2 It is worth mentioning that this integral could have been evaluated with the substitution u  As a matter of fact, even the indefinite integral tan x dx  1 2 Þ 3 4 1 4  tan x . tan x dx can be evaluated in this way to give Þ 1 arccos cos x sin x 1 log cos x  2 cos x sin x  sin x 2 2 and you might want to challenge your students to come up with Þ 3 tan x dx  1 log 3 tan 2 x  1  1 log 3 tan 4 x 3 tan 2 x  1  23 arctan 1 2 3 tan 2 x 2 4 3 and that tan x dx  /2 Þ0 3 tan x dx ...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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