Question

Вычислите интегралы
4 и 5

Answer 1

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

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