Question

Can someone help me evaluate this integral? I will post.

Answer 1

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

Answer 2

the 4x^2 is supposed to be 4x^-2 I couldn't figure out how to get the - to work with the equation function.

Answer 3

\[\int\limits_{2}^{3}(4x^{-2} - 3x)dx = \int\limits_{2}^{3}4x^{-2}dx - \int\limits_{2}^{3}3dx\]Remember the power rule:
cx^n = (cx^n+1)/(n+1). So, 4x^(-2) becomes: 4x^(-1)/-1 = -4x^(-1)

Answer 4

So we get:\[-4x^{-1} - 3x\]evaluated at 3 and 2. :-)

Answer 5

That is 7/3, if I did my arithmetic correctly.

Answer 6

I got the answer 17/3

Answer 7

I meant a -7/3.

Answer 8

Remember that it's
F(a) - F(b). Maybe you are messing up with the signs? F(b) will be negative here, so it will become F(a) + F(b)

Answer 9

Isn't it:\[(4/x) - 3x\]So, we get:\[-4/3 - 9 - (-2 -6) = -4/3 - 9 + 8 = -4/3 - 1 = -7/3\]

Answer 10

Thank you. I figured the question out with your help!

Answer 11

The answer was -7/3

Answer 12

You are welcome, mate :-)