Question

очень прошу помочь!

Answer 1

1) =(3x³/3-4x²/2-x)=(x³-2x²-x) ₂|³=3³-2*3²-3-(2³-2*2²-2)=27-18-3-8+8+2=8.
2) 

Answer 2

$1)\; \; \int\limits^3_2\, (3x^2-4x-1)dx=(x^3-2x^2-x)\Big |_2^3=\\\\=(27-18-3)-(8-8-2)=8\\\\2)\; \; \int\limits^{\frac{\pi}{2}}_{\frac{\pi}{6}}\, (cosx+\frac{1}{sin^2x})\, dx=(sinx-ctgx)\Big |_{\frac{\pi}{6}}^{\frac{\pi}{2}}=\\\\=sin\frac{\pi }{2}-ctg\frac{\pi }{2}-(sin\frac{\pi}{6}-ctg \frac{\pi}{6})=1-0-\frac{1}{2}+\sqrt3=\\\\=\frac{1}{2}+\sqrt3=\frac{1+2\sqrt3}{2}\\\\3)\; \; \int\limits^1_0 \frac{dx}{\sqrt{1-x^2}}=arcsinx\Big |_0^1=arcsin1-arcsin0= \frac{\pi }{2}-0=\frac{\pi}{2}$