الأجوبة
∫_0^1√(x(1-x)) dx
I=∫_0^1√(x-x^2 ) dx=∫_0^1√(1/4-(x-1/2)^2 ) dx
Let x-1/2=1/2 sinθ→dx=1/2 cosθ dθ
x=0→θ=-π/2,x=1→θ=π/2
∴I=∫_(-π⁄2)^(π⁄2)√(1/4-1/4 〖sin〗^2 θ)*cosθ=∫_(-π⁄2)^(π⁄2)1/2 〖cos〗^2 θdθ
=1/2 [2∫_0^(π⁄2)1/2 (1+cos2θ )dθ]=1/2 [θ+1/2 sin2θ ]_0^(π⁄2)=1/2 (π/2)=π/4
معلومات ذات صلة