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

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*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?


The first step should be simplifying the integral, like this:
$$\displaystyle \int -4x^{4} + 4x^{3} + 8x^2 \,dx$$
$$-4\int x^4dx + 4\int x^3 dx + 8\int x^2 dx$$
use the power rule, $$\int x^n \,dx = \frac{x^{n+1}}{n+1}$$
and apply to the problem
$$-4 \times\frac{x^5}{5} + 4\times \frac{x^4}{4} + 8 \times\frac{x^3}{3}$$
Now just plug in the values and you should get 12.6 :)
