To evaluate the double integral d = ˆ{x^2 + 6y} dA, where the region of integration is bounded by y = x, y = x^3 and x = 0, we first need to determine the limits of integration by finding the points of intersection of the curves.
1. **Finding Points of Intersection:** To find where y = x and y = x^3 intersect, set x = x^3. This implies that x^3 – x = 0, which factors to x(x^2 – 1) = 0. This means the solutions are x = 0, x = 1, and x = -1. However, since the bounds also require x = 0, the relevant points of integration in the first quadrant are x=0 and x=1 (as x = -1 is not in the limits). Accordingly, we find the corresponding y-values to be y = 0 and y = 1.
2. **Determining the Double Integral Limits:** We can express the integral based on a sketch of the area as follows:
– For every x from 0 to 1, y ranges from y = x^3 to y = x.
3. **Setting up the Double Integral:** Now we have our limits and can set up the double integral:
I = ∫∫ (x^2 + 6y) dy dx
where the outer integral limits for x are from 0 to 1 and the inner integral limits for y are from x^3 to x.
4. **Carrying out the Inner Integral:** We compute the inner integral:
I = ∫_0^1 (∫ (x^2 + 6y) dy) dx = ∫_0^1 [x^2y + 3y^2] | from (x^3 to x) dx
Evaluating this gives us:
I = ∫_0^1 [x^3 + 3x^2 – (x^2 * x^3 + 3(x^3)^2)] dx
This simplifies to:
I = ∫_0^1 [x^3 + 3x^2 – x^5 – 3x^6] dx
5. **Calculating the Outer Integral:** Now calculating this will yield:
I = [ rac{x^4}{4} + x^3 – rac{x^6}{6} – rac{x^7}{7}] |_0^1
Evaluating from 0 to 1 gives:
I = ( rac{1}{4} + 1 – rac{1}{6} – rac{1}{7}) – 0
This simplifies to:
I = 1.25 – 0.214286 = 1.035714
In conclusion, the value of the evaluated double integral is approximately 1.036.