Question

срочно 30 балооввв!!!!!!!!!!!!!!!!!!!!!

Answer 1

$1); Tochki; peresecheniya:; left { {{y=frac{x^2}{2}} atop {y=2-frac{3}{2}x}} right. ; ,\\ frac{x^2}{2}=2-frac{3}{2}x; ,; ; x^2+3x-4=0; ,; ; x_1=-4,; x_2=1\\S=int _{-4}^1, (2-frac{3}{2}x-frac{x^2}{2})dx=(2x-frac{3x^2}{4}-frac{x^3}{6})|_{-4}^1=\\=2-frac{3}{4}-frac{1}{6}_(-8-12+frac{32}{3})=22-frac{139}{12}=frac{125}{12}\\2); int , sqrt[3]{(4-3t)^2}dt=int (4-3t)^{frac{2}{3}}dt=-frac{1}{3}cdot frac{(4-3t)^{frac{5}{3}}}{frac{5}{3}}+C=$

$=-frac{1}{5}cdot sqrt[3]{(4-3t)^5}+C;\\Formyla:; ; ; int (ax+b)^{k}dx=frac{1}{a}cdot frac{(ax+b)^{k+1}}{k+1}+C$