Помогите решить очень надо.

Решите задачу:

$\int _1^2\, \frac{7}{5}x^2\cdot dx=\frac{7}{5}\cdot \frac{x^3}{3}|_1^2=\frac{7}{15}(2^3-1^3)=\frac{49}{15}\\\\\\\int _1^2\, \frac{dx}{\sqrt{5x-1}}=[t=5x-1,dt=5dx,\to dx=\frac{dt}{5}]=\frac{1}{5}\int _4^9\frac{dt}{\sqrt{t}}=\\\\=\frac{1}{5}\cdot 2\sqrt{t}\, |_4^9=\frac{2}{5}\cdot (\sqrt9-\sqrt4)=\frac{2}{5}\cdot (3-2)=\frac{2}{5}\\\\\\S=\int_{-2}^3(\frac{1}{2}x^2+1)dx=(\frac{1}{2}\cdot \frac{x^3}{3}+x)|_{-2}^3=\frac{1}{2}\cdot \frac{3^3}{3}+3-(\frac{1}{2}\cdot \frac{(-2)^3}{3}-2)=$

$=\frac{9}{2}+3+\frac{4}{3}+2=5+\frac{35}{6}=10\frac{5}{6}$