What is the value of $\int_{-1}^{2} 4 \, x^2 \, (2+x-x^2) \, dx$? $$\int_{-1}^{2} 4 \, x^2 \, (2+x-x^2) \, dx$$

*

*When I solved this took the constant out of the integral and then multiplied $x^2$
by the bracket and evaluated the integral to get:
$$4 \, \left[ \frac{x^4}{4} +\frac{2x^3}{3}-\frac{x^5}{5} \right]^2_{-1}=\frac{63}{20}$$

*But when I solve this integral on my calculator on (Casio 991ERX) or Wolfram I get $ \frac{63}{5}$. Why is that?


Let us do it step by step:
\begin{align*}
\int_{-1}^{2}4x^{2}(x + 2 - x^{2})\mathrm{d}x & = \int_{-1}^{2}(4x^{3} + 8x^{2} - 4x^{4})\mathrm{d}x\\\\
& = \left(x^{4} + \frac{8x^{3}}{3} - \frac{4x^{5}}{5}\right)\bigg\rvert_{-1}^{+2}\\\\
& = \left(16 + \frac{64}{3} - \frac{128}{5}\right) - \left(1 - \frac{8}{3} + \frac{4}{5}\right)\\\\
& = \frac{176}{15} + \frac{13}{15} = \frac{189}{15} = \frac{63}{5}
\end{align*}
Hopefully this helps!
