Evaluate $\int_{-1}^{2} (4-x^2) dx$
I know the function is increasing on the interval $[-1,0]$ and decreasing on the interval $(0,2]$. So,
$\phi_n(x) = [4-(\displaystyle\frac{-1+3(k-1)}{2^n})^2]$ for $x\in [\displaystyle\frac{-1+3(k-1)}{2^n}, \frac{-1+3k}{2^n})$ for $k\in 1,2,\dots,2^{n-1}$ on the interval $[-1,0]$.
$\phi_n(x) = \big[4-\displaystyle\frac{-1+3k}{2^n})\big]$ for $x\in \big(\displaystyle\frac{-1+3(k-1)}{2^n}, \frac{-1+3k}{2^n}\big]$ for $k\in 2^{n-1}+1, 2^{n-1}+2, \dots, 2^n$ on the interval $(0,2]$.
Then $\int\phi_n(x) = \sum_{k=1}^{2^{n-1}} \big[4-\big(\displaystyle\frac{-1+3(k-1)}{2^n}\big)^2 \cdot \frac{3}{2^n} + \sum_{2^{n-1}+1}^{2^n} \big[4-\big(\displaystyle\frac{-1+3k}{2^n}\big)^2\big] \cdot \frac{3}{2^n}$
$\int f =$ lim$_{n\rightarrow\infty} \int \phi_n(x)$
Is there an easy way to simplify this integral? I am having trouble evaluating from here.