How do I find integral of x sqrt(16-3x)dx
2007-02-22 15:48:51 · 5 answers · asked by atsbuy 2 in Science & Mathematics ➔ Mathematics
1) use TI 89 :)
2) let 16 - 3x = u
then x = (16-u)/3
dx = (-1/3) du
it should be manageable from here!
2007-02-22 15:57:31 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Find the integral â«{ x â(16 - 3x) } dx.
Integrate by parts.
â« u dv = uv - â«v du
___________
â«{ x â(16 - 3x) } dx
Let
u = x and dv = â(16 - 3x) dx
du = dx and v = (-2/9)(16 - 3x)^(3/2)
= x (-2/9)(16 - 3x)^(3/2) + â«{(2/9)(16 - 3x)^(3/2)} dx + c
= -(2/9)x(16 - 3x)^(3/2) - (4/135)(16 - 3x)^(5/2) + C
= -(2/9)(16 - 3x)^(3/2) [x + (2/15)(16 - 3x)] + C
= -(2/135)(16 - 3x)^(3/2) [15x + 2(16 - 3x)] + C
= -(2/135)(16 - 3x)^(3/2) [15x + 32 - 6x] + C
= -(2/135)(16 - 3x)^(3/2) [9x + 32] + C
2007-02-24 14:01:25 · answer #2 · answered by Northstar 7 · 0⤊ 0⤋
Let u = x, du = dx
dv = (16 - 3x)^(1/2)dx, v = (-2/9)(16 - 3x)^(3/2)
â«x(16 - 3x)^(1/2)dx =
(-2/9)x(16 - 3x)^(3/2) + â«(2/9)(16 - 3x)^(3/2)dx =
(-2/9)x(16 - 3x)^(3/2) - (4/135)(16 - 3x)^(5/2)
2007-02-22 16:24:30 · answer #3 · answered by Helmut 7 · 0⤊ 0⤋
what you need to do is take the antiderivative, which would be
F(x)=(2/3)*(16-3x)^(3/2)
then, if it's a definite integral, you evaluate it like so:
F(upper limit)-F(lower limit)
2007-02-22 15:56:09 · answer #4 · answered by dr_chee 1 · 0⤊ 1⤋
-(2/9)(16-3x)^[3/2]
2007-02-22 16:13:37 · answer #5 · answered by Anopheles 1 · 0⤊ 0⤋