Integral????? need help

3
∫ 4x(x-6)dx=
0

multiply it out
∫ 4x(x-6) dx
becomes
∫ 4x^2 - 24x dx
which becomes
4/3x^3 - 12x^2 | from 0 to 3
plug in upper and lower limits, and you get
(36 - 108) - (0 - 0)
= -72

first u have to distribute the 4x so the equation becomes 4x^2-24xdx. the u do the actual integral so it becomes 4/3x^3 -12x^2. then all u have to do is plug in 3. 4/3(3^3)-12(3^2) which gets u -72. normally in an integral u would also have to minus -72 by the number u get if u plug in the bottom number but since it is zero its unnecessary

The integral can be rewritten as:

3
4 ∫ (x^2 - 6x)dx
0

Now the integral of x^2 - 6x is just a simple "reverse" power rule. Add a power to the exponent and divide. You get: x^3 / 3 - 3x^2

Now plug in your limits. (9-27) - (0-0) = -18
Don't forget to multiply by 4 since we took it out
-18 x 4 = -72