Question

So I have the following integral (Please note that the number below is -2 but for some reason LateX did not accept a negative number as the parameter
$\int\limits^2_2 {4x^3-x+1} \, dx$

Which I have integrated to this:

$\int\limits^2_2 {x^4-\frac{x^2}{2}+3x} \, dx$

And now I need to find the integrated value which I conclude this would be it:

$2^4-\frac{2^2}{2}+3.2-((-2)^4-(\frac{-2^2}{2}+3.(-2))$

The answer I get is 32 but the correct answer should be 12. I'm not seeing what I'm doing wrong. Any help?

Answer 1

$f(x)=4x^3-4x+1\\ F(x)=x^4-\frac{x^2}{2}+x\\ \int_{-2}^{2} f(x)=F(2)-F(-2) \implies\\ (2)^4-\frac{2^2}{2}+2)-((-2)^2-\frac{(-2)^2}{2}-2)= \\ 16-2+2-(16-2-2)=16-(16-4)=4$

I've done this a million times and I'm still getting 4 instead of 32. Are you sure you posted the coefficients?