What is the Integral of
1/ ( (x-2)^(3/2) ) from 0 to 4.
Thanks!
2007-06-25 20:08:53 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
I = ∫(x-2)^(-3/2) dx
I = [(x-2)^(-3/2+1)/(-1/2)] limits
I = -2 [(x-2)^(-1/2)] limits o to 4
I = -2 [1/√(x-2)] limits o to 4
I = -2 [1/(√2) - 1/(i√2)]
I = -(2/√2) [1 - 1/(i)]
I = -(2/√2) [1 + i^2/(i)] >>>> i = √(-1) and -1 = i^2
I = - √2 [1 + i]
answer
2007-06-25 20:18:25 · answer #1 · answered by anil bakshi 7 · 1⤊ 0⤋
4
â« ((x^2 - 2)^-(3/2))dx =
0
- x/(2â(x^2 - 2)) from 0 to 4 =
- 2/â(4^2 - 2) =
- 2/â(16 - 2) =
- (1/7)â14 â - 0.5345225
If y = -(1/2)x(x^2 - 2)^-(1/2)
y' = (1/2)x^2(x^2 - 2)^-(3/2) - (1/2)(x^2 - 2)^-(1/2)
y' = ((1/2)x^2 - (1/2)(x^2 - 2))/(x^2 - 2)^(3/2)
y' = ((1/2)x^2 - (1/2)x^2 + 1))/(x^2 - 2)^(3/2)
y' = 1/(x^2 - 2)^(3/2)
2007-06-26 04:16:01 · answer #2 · answered by Helmut 7 · 0⤊ 1⤋
2
- â¯â¯â¯â¯â¯â¯â¯â¯â¯â¯
â(x - 2)
evaluated in (0,4)
-1.414213562 - 1.414213562·î
2007-06-26 03:27:58 · answer #3 · answered by LENNONLNX0618 2 · 0⤊ 0⤋