Answer:
[tex] A = \frac{74}{3} -\frac{7}{2} +36 = \frac{127}{6} +36 = \frac{343}{6}[/tex]
Step-by-step explanation:
For this case we have these two functions:
[tex] x+y^2 = 12[/tex] (1)
[tex] x+y=0[/tex] (2)
And as we can see we have the figure attached.
For this case we select the x axis in order to calculate the area.
If we solve y from equation (1) and (2) we got:
[tex] y = \pm \sqrt{12-x}[/tex]
[tex] y = -x[/tex]
Now we can solve for the intersection points:
[tex] \sqrt{12-x} = -\sqrt{12-x}[/tex]
[tex] 12-x = -12+x[/tex]
[tex] 2x=24 , x=12[/tex]
[tex] \sqrt{12-x} =-x[/tex]
[tex] 12-x = x^2[/tex]
[tex] x^2 +x -12=0 [/tex]
[tex] (x+4)*(x-3) =0[/tex]
And the solutions are [tex] x =-4, x=3[/tex]
So then we have in total 3 intersection point [tex] x=12, x=-4, x=3[/tex]
And we can find the area between the two curves separating the total area like this:
[tex] \int_{-4}^3 |\sqrt{12-x} - (-x)| dx +\int_{3}^{12}|-\sqrt{12-x} -\sqrt{12-x}|dx[/tex]
[tex] \int_{-4}^3 |\sqrt{12-x} + x| dx +\int_{3}^{12}|-2\sqrt{12-x}|dx[/tex]
We can separate the integrals like this:
[tex] \int_{-4}^3 |\sqrt{12-x} dx +\int_{-4}^3 x +2\int_{3}^{12}\sqrt{12-x} dx[/tex]
For this integral [tex] \int_{-4}^3 |\sqrt{12-x} dx [/tex] we can use the u substitution with [tex]u = 12-x[/tex] and after apply and solve the integral we got:
[tex] \int_{-4}^3 |\sqrt{12-x} dx =\frac{74}{3}[/tex]
The other integral:
[tex] \int_{-4}^3 x dx = \frac{3^2 -(-4)^2}{2} =-\frac{7}{2}[/tex]
And for the other integral:
[tex]2\int_{3}^{12}\sqrt{12-x} dx[/tex]
We can use the same substitution [tex] u = 12-x[/tex] and after replace and solve the integral we got:
[tex]2\int_{3}^{12}\sqrt{12-x} dx =36[/tex]
So then the final area would be given adding the 3 results as following:
[tex] A = \frac{74}{3} -\frac{7}{2} +36 = \frac{127}{6} +36 = \frac{343}{6}[/tex]
