<s>[INST] Solving $\int_{-1}^{1}8x^3-5x^2+4dx$ I've done the following so far:
$$\left.\int_{-1}^{1}\left(8x^3-5x^2+4\right)dx=\left(\frac{8}{4}x^4-\frac{5}{3}x^3+4x\vphantom{\int}\right)\right|_{-1}^{1}$$
$$=\left(\frac{8}{4}-\frac{5}{3}+4\right)=\frac{13}{3}$$
However, I double-checked on wolfram alpha and the solution is actually $-\dfrac{14}{3}$. Would you know where I went wrong? I have no idea where the negative came from, in the solution, or how it's one value above mine.
 [/INST] The value of last limit must be $4-(-4)=8$
$$\int_{-1}^1(8x^3-5x^2+4)dx=\left(4x^4-\frac53x^3+4x\right)_{-1}^1$$
$$=\left(4(1)^4-\frac53(1)^3+4(1)-4(-1)^4+\frac53(-1)^3-4(-1)\right) $$
$$=\frac{14}{3}$$
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