Final answer:
The question concerns evaluating a double integral over a specific region. We set up the integral with respect to y first, then x, and found the integral over the triangluar region bounded by the curves x² and x. The result of the integral is 1/12.
Step-by-step explanation:
The question asks us to evaluate the double integral ∧∧xdA over the region bounded by x2 ≤ y ≤ x. To approach this problem, we first need to sketch the region of integration and then set up the integral with appropriate limits.
Given the inequalities, we can see that the region is a triangular area in the xy-plane. The curve y = x2 is an upwards opening parabola, and y = x is a straight line with a 45-degree slope. The region of interest is between these two curves.
To set up the double integral, we choose to integrate with respect to y first and then x, as the boundaries are functions of x. The limits for y are from x2 to x, and the limits for x are from 0 to 1, where the curve y = x and the parabola intersect.
The double integral becomes:
∧01∧x2x x dy dx
First, we integrate with respect to y:
∧01 [xy] x2x dx
This simplifies to:
∧01 (x2 - x3) dx
We finally integrate with respect to x:
∧01 (x2 - x3) dx = [1/3 x3 - 1/4 x4] 01
Evaluating the definite integral, we find:
1/3 - 1/4 = 1/12
So, the result of the double integral ∧∧xdA over the given region is 1/12.