Интегралы,решите пожалуйста!!))

Решите задачу:

$\int _1^2x^4\, dx=\frac{x^5}{5}|_1^2=\frac{1}{5}(32-1)=\frac{31}{5}\\\\\int _{-1}^0(x^3+2x)dx=(\frac{x^4}{4}+x^2)|_{-1}^0=0-(\frac{1}{4}+1)=-1,25\\\\\int _0^4\sqrt{x}dx=\frac{x^{\frac{3}{2}}}{\frac{3}{2}}|_0^4=\frac{2}{3}(\sqrt{4^3}-0)=\frac{2}{3}\cdot 2^3=\frac{16}{3}\\\\\int _{-\frac{\pi}{4}}^{\frac{\pi}{4}}}\, cosx\, dx=sinx|_{-\frac{\pi}{4}}^{\frac{\pi}{4}}=sin \frac{\pi}{4}-sin(-\frac{\pi}{4})=sin\frac{\pi}{4}+sin\frac{\pi}{4}=\\\\=2\cdot \frac{\sqrt2}{2}=\sqrt2\\\\\int _{-2}^5(4x^3-3x^2+2x+1)dx=$

$=(x^4-x^3+x^2+x)|_{-2}^5=(5^4-5^3+5^2+5)-(2^4+2^3+2^2-2)=\\\\=530-26=504$