Question

f(x)=x+√(1-x^2)tf(t)dt1到0的定积分

Answer 1

∫_0^1t(f(t)dt=

=∫_0^1 [t(t+√(1-t^2))]dt

=∫_0^1 t^2dt + (1/2)∫_0^1 √(1-t^2)d(t^2)

= [t^3/3]_0^1 - [(1/2)*(2/3)*(1-t^2)^(3/2)]_0^1

= 1/3 + 1/3

= 2/3