To evaluate the double integral of 2x^2 over the region bounded by -1 ≤ x ≤ 1 and - ≤ y ≤ 1, we need to integrate with respect to y first and then with respect to x. ∫∫ 2x^2 dy dx

To evaluate the double integral of 2x^2 over the region bounded by -1 ≤ x ≤ 1 and -1 ≤ y ≤ 1, we need to integrate with respect to y first and then with respect to x. Here are the steps:

1. Integrate with respect to y: ∫ from -1 to 1 (2x^2) dy. Since there is no y in the integrand, this integral simplifies to 2x^2*y evaluated from -1 to 1, which is 2x^2.

2. Now, integrate the result with respect to x: ∫ from -1 to 1 (2x^2) dx.

3. The antiderivative of 2x^2 is (2/3)x^3. So, we evaluate (2/3)x^3 from -1 to 1.

4. Substituting the limits of integration, we get (2/3)1^3 - (2/3)(-1)^3 = 2/3 - (-2/3) = 4/3.

So, the double integral of 2x^2 over the region bounded by -1 ≤ x ≤ 1 and -1 ≤ y ≤ 1 is 4/3.