Question

Помгите вычислить интегралы, : а) б)

Answer 1

7/3   x^3 = 7/3 *216 - 7/3  *27 = 504- 63 =441

sin 6x  = 1-0=0

Answer 2

$a)\int\limits^6_3 {7x^2} \, dx = (\frac{7x^3}{3}) |\limits^6_3 = (\frac{7*216}{3})-(\frac{7*27}{3})=441\\ b) \int\limits^{\frac{\pi}{12}}_0 {6cos6x} \, dx =sin6x |\limits^{\frac{\pi}{12}}_0 = sin\frac{\pi}{2}-sin0 = 1 - 0 = 1$