Question

Вычислите интегралы
4 и 5

Answer 1

4.
$\int\limits^{2}_{-2} { \frac{1}{ \sqrt{2x+5} } } \, dx = \frac{1}{2} \int\limits^{2}_{-2} { \frac{1}{ \sqrt{2x+5} } } \, d(2x) = \frac{1}{2} \int\limits^{2}_{-2} { \frac{1}{ \sqrt{2x+5} } } \, d(2x+5) = \\ \\ =\frac{1}{2} \int\limits^{2}_{-2} {(2x+5)^{- \frac{1}{2} } } \, d(2x+5) = \frac{1}{2} \frac{1}{- \frac{1}{2}+1 } (2x+5)^{- \frac{1}{2} +1} |_{-2}^{2} = \\ \\ =(2x+5)^{ \frac{1}{2} } |_{-2}^{2} = \sqrt{2x+5} |_{-2}^{2} = \sqrt{2*2+5} - \sqrt{2*(-2)+5} = 2$

5.
$\int\limits^{ \frac{ \pi }{3} }_0 { \frac{1}{3} sin(x- \frac{ \pi }{3} )} \, dx = \frac{1}{3} \int\limits^{ \frac{ \pi }{3} }_0 {sin(x- \frac{ \pi }{3} )} \, d(x- \frac{ \pi }{3}) = \\ \\ = - \frac{1}{3} cos(x- \frac{ \pi }{3})|_0^{ \frac{ \pi }{3} } = - \frac{1}{3} (cos(\frac{ \pi }{3} -\frac{ \pi }{3} ) -cos(0-\frac{ \pi }{3} ) ) = \\ \\ = - \frac{1}{3} (1 - \frac{1}{2} ) = -\frac{1}{6}$