Question

can someone help me set this up?

Answer 1

\[\int\limits_{0}^{2} (2x-3)(4x^2+1)dx\]

Answer 2

Just in the integrating format... x^n+1/n+1

Answer 3

multiply the terms in the brackets first

Answer 4

soo... 8x^3-12x^2+2x-3?

Answer 5

right so

\[\int\limits_{0}^{2} (2x-3)(4x^2+1)\text dx\]
\[=\int\limits_{0}^{2}\left(8x^3-12x^2+2x-3\right)\text dx\]

Answer 6

(8)*(x^4/4)-(12)*(x^3/3)+(2)*(x^2/2)-(3)*(x^2/2)?

Answer 7

check that last term,


\(3\) is really the same as \(3x^0\)

Answer 8

x/1?

Answer 9

yes

Answer 10

So now you have found a function who's derivative is the function inside the integral. But you are integrating over a specific interval, so you need to get a number out, rather than a function.

Answer 11

...? huh

Answer 12

\[=\int\limits_{0}^{2}\left(8x^3-12x^2+2x-3\right)\text dx\]\[=(8)\times(\frac{x^4}4)-(12)\times(\frac{x^3}3)+(2)\times(\frac{x^2}2)-(3)\times(\frac{x}1)\Big|_0^2\]

Answer 13

I got 2 :) is that right?