The parabola y = x ² and the line x + y = 12 intersect for
x ² = 12 - x
x ² + x - 12 = 0
(x - 3) (x + 4) = 0
===> x = 3
so you can compute the area by using two integrals,
[tex]\displaystyle \int_0^3 x^2\,\mathrm dx + \int_3^{12}(12-x)\,\mathrm dx[/tex]
Then the area you want is
[tex]\displaystyle \frac{x^3}3\bigg|_0^3 + \left(12x-\frac{x^2}2\right)\bigg|_3^{12} = \left(\frac{3^3}3-\frac{0^3}3\right) + \left(12^2-\frac{12^2}2 - 12\times3 + \frac{3^2}2\right) \\\\ = \boxed{\frac{99}2}[/tex]
Alternatively, you can subtract the area bounded by y = x ², x + y = 12, and the y-axis in the first quadrant from the area of a triangle with height 12 (the y-intercept of the line) and length 12 (the x-intercept).
Such a triangle has area
1/2 × 12 × 12 = 72
and the area you want to cut away from this is given by a single integral,
[tex]\displaystyle \int_0^3 ((12-x)-x^2)\,\mathrm dx = \int_0^3(12-x-x^2)\,\mathrm dx[/tex]
The integral has a value of
[tex]\displaystyle \left(12x-\frac{x^2}2-\frac{x^3}3\right)\bigg|_0^3 = 12\times3 - \frac{3^2}2 - \frac{3^3}3 \\\\ = \frac{45}2[/tex]
and so the area of the shaded region is again 72 - 45/2 = 99/2.
