integral problem...

[jnimagine]

suppose that f is continuous and (integral]0->2 f(x)dx = 6. Then (integral]0 -> pi/2 f(2sinb)cosbdb=?

ok.... i'm totally lost here... how do u solve this without knowing what f(x) is..????

[dx]

Do you know how to change variables in integrals? In the second integral, change variables to y = 2sinb.

[jnimagine]

Do you know how to change variables in integrals? In the second integral, change variables to y = 2sinb.

so if y = 2sinb then dy = 2cosbdb hence 1/2(integral]0->2f(y)dy) = 1/2(6) = 3
Is this the answer?

[dx]

Yes.