Question

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

Answer 1

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

Answer 2

yes

Answer 3

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

Answer 4

this would be:
\[x^4/4 + 4x^3/3\]

Answer 5

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

Answer 6

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

Answer 7

so your final result is 256/3 - 64

Answer 8

remember that you subtracted

Answer 9

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
\[\frac{0^{4}}{4} - \frac{40^{3}}{3} - (\frac{(-4)^{4}}{4} - \frac{4(-4)^{3}}{3})\]

Answer 10

compute that for your answer

Answer 11

I already did :P

Answer 12

but I would say that your answer is more formal

Answer 13

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