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beks73 [17]
3 years ago
15

So I have the following integral (Please note that the number below is -2 but for some reason LateX did not accept a negative nu

mber 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?
Mathematics
1 answer:
kramer3 years ago
6 0

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?

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