9514 1404 393
Answer:
121.5
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 ...
