3.5.1 Integration as the Summation of Areas – Examples


Example 1
Find the area of the shaded region.


Solution:
Area of the shaded region=baydx=40(6xx2)dx=[6x22x33]40=[3(4)2(4)33]0=2623unit2


Example 2
Find the area of the shaded region.


Solution:
y = x -----(1)
x = 8yy2-----(2)
Substitute (1) into (2),
y = 8yy2
y2 – 7y = 0
y (y – 7) = 0
y = 0 or 7
From (1), x = 0 or 7
Therefore the intersection points of the curve and the straight line is (0, 0) and (7, 7).

Intersection point of the curve and y-axis is,
x = 8yy2
At y-axis, x = 0
0 = 8yy2
y (y – 8) = 0
y = 0, 8

Area of shaded region = (A1) Area of triangle + (A2) Area under the curve from y = 7 to y = 8.
=12×base×height +87xdy=12×(7)(7)+87(8yy2)dy=492+[8y22y33]87=2412+[4(8)2(8)33][4(7)2(7)33]=2412+85138123=2816unit2