<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] As $x^3$ is odd, the integral of $\int_{-1}^{1}8x^3\,dx$ is zero. Therefore
\begin{align}\int_{-1}^{1}\left(8x^3-5x^2+4\right)dx&=\int_{-1}^{1}(-5x^2+4)\,dx\\&=\left(\frac{-5x^3}{3}+4x\right)\Bigg|_{-1}^{1}\\&=\left(\frac{-5}{3}+4\right)-\left(\frac{5}{3}-4\right)\\&=\frac{24-10}{3}\\&=\frac{14}{3}.\end{align}
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