Mathematics, 13.07.2019 17:00 jennsou

So i have the following integral ( 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: $2^4-\frac{2^2}{2}+3.)^4-(\frac{-2^2 }{2}+))$ the answer i get is 32 but the correct answer should be 12. i'm not seeing what i'm doing wrong. any ?

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So i have the following integral ( note that the number below is -2 but for some reason latex did no...

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