Evaluate
(From 0 to 65) 1/(t +1)^1/3 dt
A) 765/4
B) 51/2
C) 24
D) 45/2
2006-07-23 14:03:09 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
This is also same as the before one.
let x = t + 1
then dx = dt
Limits.
When t = 0. x = 0 +1 = 1.
when t = 65. x = 66.
then the integral gives, (integral, 1 to 66) 1/x^1/3 dx
gives, (integral, 1 to 66) x^ (-1/3) dx
As integral y^n dy = [y^(n+1)] / n+1,
then we get, 3/2 x^2/3 (from 1 to 66 )
substituting upper and lower limit and subtrating.
= 3/2 { 66^2/3 - 1}
I think you can do this from now as I gave you the working in the previous problem.
Hope you got it.
2006-07-23 14:37:23 · answer #1 · answered by Sherlock Holmes 6 · 1⤊ 0⤋
Hmm well A)
2006-07-23 14:08:04 · answer #2 · answered by Anonymous · 0⤊ 0⤋
e
2006-07-23 14:06:30 · answer #3 · answered by quizqueen 4 · 0⤊ 0⤋
C bAbY
2006-07-23 14:07:49 · answer #4 · answered by Anonymous · 0⤊ 0⤋
C
2006-07-23 14:12:56 · answer #5 · answered by Ron DMC 2 · 0⤊ 0⤋
i have no idea
2006-07-23 14:08:15 · answer #6 · answered by crazi8red 6 · 0⤊ 0⤋
C
2006-07-23 14:05:10 · answer #7 · answered by Anonymous · 0⤊ 0⤋
eh, sorry...
2006-07-23 14:05:52 · answer #8 · answered by kiki Dee 5 · 0⤊ 0⤋