Question

what is \[\int\limits_{-\infty}^{0} dx/(8-3x)^{5/3}\] ?

Answer 1

please help..

Answer 2

rewrite it as \[\lim_{t\rightarrow_-\infty}{\int_t^0{\frac{dx}{(8-3x)^{\frac{5}{3}}}}}\]

Answer 3

use substitution :-D

Answer 4

ok.=) sorry. thanks..

Answer 5

so i'll use 8-3x as u?

Answer 6

1/36

Answer 7

Yes

Answer 8

yes

Answer 9

so when i solve, what is the value of the negative infinity?

Answer 10

Just find the limit of the expression, as t goes to - infinity

Answer 11

let 8-3x = u

du/dx = -3
du/-3 = dx

the integral becomes ::

-1/3 integral (u^(-5/3)du)
the answer to integral (u^(-5/3)du) -> u^(-2/3) / (-2/3)
-1/3 * ( -3u^(-2/3)/2)
u^(-2/3)/2

1/(2* (8-3x)^(2/3))
at 0 its : 1/(2*4) = 1/8

at -infinity its :

0.5lim(1/(8-3x)^(2/3)) = 0

so its 1/8

i hope didnt make any mistakes

Answer 12

why is it at 0.5?

Answer 13

i took the 1/2 out of the limit

Answer 14

Coolsector is right.

he just took out the 0.5 factor from the expression :-D

Answer 15

ah. i see.=) thank you guys.