Integrate from 0 to 1 (x^2)(sqrt5x+6)dx
I am having trouble with u substitution, if anyone could help me it would be appreciated. Thanks
2007-12-11 05:40:57 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Take u = 5x + 6. Then du = 5 dx, x^2 = (u - 6)^2 / 25, and the integral becomes
Int(u from 6 to 11) 1/125 (u - 6)^2 sqrt u du.
Multiply out the integrand and the antiderivative is found just by using the power rule:
Int(u from 6 to 11) 1/125 [u^(5/2) - 12u^(3/2) + 36u^(1/2)] du.
2007-12-11 05:49:15 · answer #1 · answered by acafrao341 5 · 0⤊ 0⤋
Put sqrt(5x+6) = z
5/2 *1/sqrt(5x+6) dx = dz so
dx = 2/5 * z*dz
x^2= 1/25 *(z^2-6)^2 so you finally get
2/125 Int(0,1) (z^2-6)^2 * z^2dz = 2/125 Int(0,1)(z^6-12z^4+36z^2 )dz=(1/7-12/5+12)*2/125
2007-12-11 05:53:36 · answer #2 · answered by santmann2002 7 · 0⤊ 0⤋