Consider the function f(x) = 9/x^3 - (6/ x^(5) ).
Let F(x) be the antiderivative of f(x) with F(1) = 0.
2007-12-05 06:12:41 · 3 answers · asked by Mark K 1 in Science & Mathematics ➔ Mathematics
Consider the function f(x) = 9/x^3 - (6/ x^(5) ).
Let F(x) be the antiderivative of f(x) with F(1) = 0.
Then what does F(3)=
2007-12-05 06:14:03 · update #1
F(x) = 9 ∫ x^(-3) dx - 6 ∫ x^(-5) dx
F(x) = 9 x^(-2) / (-2) - 6 x^(-4) / (-4) + C
F(x) = (- 9/2) (1 / x²) + (3/2) (1 / x^4) + C
F(1) = (-9/2) + (3/2) = - 3 = C
F(3) = (-1/2) + (1/54) - 3
F(3) = -27/54 + 1/54 - 162/54
F(3) = - 188/54 = -94/27
2007-12-05 10:35:11 · answer #1 · answered by Como 7 · 1⤊ 0⤋
9â«x^-3 - 6â«x^-5
9(-1/2x^-2) - 6(-1/4x^-4)
F(x) = -9/(2x^2) + 3/(2x^4) + c
0 = -9/(2) + 3/(2) + c
6/2 = c
F(x) = -9/(2x^2) + 3/(2x^4) + 3
so F(3) = -9/(2(3)^2) + 3/(2(3)^4) + 3
= -9/(2(9)) + 3/(2(81)) + 3
= -1/2 + 1/54 + 3
= -26/54 + 1/54 + 162/54
= 137/54 same as 2 and 29/54
2007-12-05 06:32:29 · answer #2 · answered by J D 5 · 0⤊ 0⤋
F(x) = -9/ (2*x^2) + 3 / (2* x^4)
F(1) = - 3 not 0
F(3) = - 13 / 27
2007-12-05 06:30:22 · answer #3 · answered by Ash_Jx 4 · 0⤊ 0⤋