The given equation of curves are, y=8x−x2 It can be written as, y=−(x2−8x+42−42) y=−((x−4)2−16) y−16=−(x−4)2 The above equation represents a parabola having vertex (4,16) . Consider the equation, 8x−4y+11=0 It represent a straight line passing through the points (0,
11
4
) and(−
11
8
,0). The intersection is calculated as, 8x−4(8x−x2)+11=0 4x2−24x+11=0 4x2−22x−2x+11=0 (2x−1)(2x−11)=0 Thus, x=
1
2
,
11
2
Consider the figure shown below.
The required area is the shaded region that is calculated as, A=