Найдите определенный интеграл: ₁³∫x^3dx/4-₃⁵∫x^3dx/4(20б)

Решите задачу:

$\displaystyle \int\limits^3_1 {\frac{x^3}4} \, dx - \int\limits^5_3 {\frac{x^3}4} \, dx =\frac{x^4}{16}\bigg|^3_1-\frac{x^4}{16}\bigg|^5_3=\frac{1}{16}-\frac{3^4}{16}-\bigg(\frac{3^4}{16}-\frac{3^5}{16}\bigg)=\\\\\\=\frac{1}{16}+\frac{243}{16}=\frac{244}{16}=\boxed{\frac{61}4}$