Срочно 30 балооввв!!!!!!!!!!!!!!!!!!!!!

Решите задачу:

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

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