∫(0->4)√(4x-x^2) dx4x-x^2=4-(x-2)^2letx-2=2sinudx =2cosu dux=0 , u= -π/2x=4, u=π/2 ∫(0->4)√(4x-x^2) dx=4∫(-π/2->π/2) (cosu)^2 du=2∫(-π/2->π/2) (1+cos2u) du=2[u+(1/2)sin2u] |(-π/2->π/2)=2π
