Let x 2 = u, x = u (1/2), dx = 1/2 u (- 1/2),

When x=0, u=0, when x=a, u = a 2.

Therefore, the formula on the left is = ∫ (A 2 above the interval and 0 below the interval) u (3/2) f (u) *1/2u (-1/2) du.

= 1/2∫ (2 in interval, 0 in interval) uf (u) du

= 1/2∫ (2 in interval, 0 in interval) xf (x) dx

= correct type

The last step is because the integral is only related to the integrand expression and interval, and has nothing to do with the specific variable code, so it can be changed to any code.