integrate with respect to x. [x^3+4x^2] dx b=0 a=-4

Question

anonymous · Answer

is this a definite integral from -4 to 0, then?

anonymous · Answer

yes

anonymous · Answer

alright, so with any definite integral, first you have to find the indefinite integral (or the antiderivative)

anonymous · Answer

this would be:
$$x^4/4 + 4x^3/3$$

anonymous · Answer

now you plug in the higher value of x into the equation (you get 0)

anonymous · Answer

you the subtract the indefinite integral evaluated at the lower value of x.
this happens to be:
64-256/3

anonymous · Answer

so your final result is 256/3 - 64

anonymous · Answer

remember that you subtracted

australopithecus · Answer

Split the integral
$$\int\limits_{-4}^{0} x^{3} +\int\limits_{-4}^{0} 4x^{2}$$
you will get:
$$x^{4}/4 + 4x^{3}/3 $$
then use 

b - a

so
$$\frac{0^{4}}{4} - \frac{40^{3}}{3} - (\frac{(-4)^{4}}{4} - \frac{4(-4)^{3}}{3})$$

australopithecus · Answer

compute that for your answer

anonymous · Answer

I already did :P

anonymous · Answer

but I would say that your answer is more formal

anonymous · Answer

you may have wanted to put a parenthesis around the 0 in the second term, though. (its not 40^3)