Интегралы Помогите прошу

Решите задачу:

$\displaystyle 1) \int^2_{-1}dx=0$
$\displaystyle 5) \int^4_1 (3-2x)dx=3x-x^2 \big|^4_1=12-16-(3-1)=-4-2=-6$
$\displaystyle 32) \int^1_{-1} x^2dx= \frac{x^3}{3} \big|^1_{-1}= \frac{1}{3}-(- \frac{1}{3})= \frac{2}{3}$
$\displaystyle 18) \int ^{ \frac{\pi}{2}}_{ \frac{\pi}{6}} cosxdx=sinx\bigg|^{ \frac{\pi}{2}}_{ \frac{\pi}{6}}=sin \frac{\pi}{2}-sin \frac{\pi}{6}=1-0,5=0,5$
$\displaystyle 10) \int^7_2 \frac{dx}{x^2}=- \frac{1}{x} \big|^7_2=- \frac{1}{7}+ \frac{1}{2}= \frac{5}{14}$
$\displaystyle 22) \int^{ \frac{\pi}{4}}_0 \frac{dx}{cos^2x}=tgx \bigg|^{ \frac{\pi}{4}}_0=tg \frac{\pi}{4}-tg0=1$
$\displaystyle a) \int^3_12xdx=x^2 \big|^3_1=9-1=8$
$\displaystyle b) \int^4_1 (x^2-6x+9)dx= \frac{x^3}{3}-3x^2+9x \big|^4_1= \frac{64}{3}-48+36-( \frac{1}{3}-3+9)$
$\displaystyle = \frac{64}{3}-12- \frac{1}{3}-6= \frac{63}{3}-18=21-18=3$