9514 1404 393
Answer:
121.5
Step-by-step explanation:
The region of integration appears to be that area bounded by the parabola and the line. The line intersects the parabola at (-5, 25) and (4, 16).
Integrating in the y-direction requires the region be divided into two parts. One part is for y-values between 0 and 16, where the differential of area is bounded by the parabola on both ends. The other part is for y-values between 16 and 25, where the line is one boundary and the parabola is the other. Then the integral is ...
[tex]\displaystyle A=\int_0^{16}{2\sqrt{y}}\,dy+\int_{16}^{25}{(20-y+\sqrt{y})}\,dy\\\\=\dfrac{4}{3}16^{\frac{3}{2}}+20(25-16)-\dfrac{25^2-16^2}{2}+\dfrac{2}{3}(25^{\frac{3}{2}}-16^{\frac{3}{2}})\\\\=\dfrac{256}{3}+180-\dfrac{369}{2}+\dfrac{122}{3}\\\\\boxed{A=121.5}[/tex]
