calculus
posted by jeanie .
find the area of y=x sqrt(x^2+16), bound by the xaxis and the vertical line x=3
I got 22.5 is that correct?

The function crosses the origin
so I see it as
A = ∫ x(x^2+16)^(1/2) dx from 0 to 3
= [ (1/3)(x^2+16)^(3/2) ] from 0 to 3
= (1/3)(25)^(3/2)  0
= (1/3)(125)
= 125/3