Question

上限3下限-3,∫(根号(9-x^2)dx 用微积分基本定律求解

Answer 1

∫(-3->3) √(9-x^2) dx

let

x= 3sina

dx=3cosada

x=-3,a=-π/2

x=3,a=π/2

∫(-3->3) √(9-x^2) dx

=∫(-π/2->π/2) 3cosa (3cosa)da

=18∫(0->π/2) (cosa)^2da

=9∫(0->π/2)(1+cos2a) da

=9[a+ sin(2a)/2](0->π/2)

=9π/2