Tích phân

Question

$\int\limits_{-2}^{2}(x^{5}+x^{2})\sqrt{4-x^{2}}$ dx

score 1 · Answer 1

$I = \int\limits_{ - 2}^2 {({x^5} + {x^2})} \sqrt {4 - {x^2}} dx = \int\limits_{ - 2}^2 {{x^5}} \sqrt {4 - {x^2}} dx + \int\limits_{ - 2}^2 {{x^2}} \sqrt {4 - {x^2}} dx$

*** Ta tính

${I_1} = \int\limits_{ - 2}^2 {{x^5}} \sqrt {4 - {x^2}} dx$

Đặt

$t = - x \Rightarrow dt = - dx$

${I_1} = - \int\limits_{ - 2}^2 {{t^5}} \sqrt {4 - {t^2}} dt = - {I_1} \Rightarrow {I_1} = 0.$

*** Ta tính

${I_2} = \int\limits_{ - 2}^2 {{x^2}} \sqrt {4 - {x^2}} dx$

Đặt

$x = 2\sin u,u \in \left[ { - \frac{\pi }{2};\frac{\pi }{2}} \right]$

$dx = 2\cos udu$

${I_2} = \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {4{{\sin }^2}u} .2\cos u.2\cos udu = 16\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {{{\sin }^2}u} .{\cos ^2}udu = 4\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {{{\sin }^2}2u} du$

$= 2\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {(1 - c{\rm{os4u}})} du = 2(u - \frac{1}{4}\sin 4u)\begin{cases}{\frac{\pi }{2}} \\ { - \frac{\pi }{2}} \end{cases} = 2\pi .$

Vậy

$I = {I_1} + {I_2} = 2\pi .$