This preview shows page 1. Sign up to view the full content.
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 ...
View
Full
Document
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.
 Fall '08
 STAFF
 Math, Calculus

Click to edit the document details