Помогите! Математика!

Решите задачу:

$\displaystyle\\1)\int\limits^2_1 {2x^2} \, dx =\frac{2x^3}{3}\mid^2_1=\frac{2*2^3}{3}-\frac{2*1}{3}=\frac{16}{3}-\frac{2}{3}=\frac{14}{3}\\\\\\2)\int\limits^{\frac{\pi}{2}} _{\frac{\pi}{6} } {\sin(x)} \, dx=-\cos(x)\mid^{\frac{\pi}{2} } _{\frac{\pi}{6} }=-\cos(\frac{\pi}{2})+\cos(\frac{\pi}{6})=0+\frac{\sqrt{3}}{2}= \frac{\sqrt{3}}{2}\\\\\\3)\int\limits^2_{-1} {x+2} \, dx=(\frac{x^2}{2}+2x)\mid^2_{-1}=\frac{2^2}{2}+2*2-(\frac{1}{2}-2)=2+4-\frac{1}{2}+2=\\\\\\ =8-\frac{1}{2}=7,5$

$\displaystyle\\ 2)S=\int\limits^1_{-1} {x^2} \, dx =\frac{x^3}{3}\mid^1_{-1}=\frac{1}{3}-(-\frac{1}{3})=\frac{2}{3}$