__3.5 Integration as the Summation of Areas__**(A) Area of the region between a Curve and the**

*x*-axis.

Area of the shaded region;
$A={\displaystyle {\int}_{a}^{b}y\text{}dx}$

**(B) Area of the region between a curve and the**

*y*-axis.

Area of the shaded region;
$$A={\displaystyle {\int}_{a}^{b}x\text{}dy}$$

**Example 1**

Find the area of the shaded region.

*Solution:*$\begin{array}{l}\text{Area of the shaded region}\\ ={\displaystyle {\int}_{a}^{b}y\text{}dx}\\ ={\displaystyle {\int}_{0}^{4}\left(6x-{x}^{2}\right)dx}\\ ={\left[\frac{6{x}^{2}}{2}-\frac{{x}^{3}}{3}\right]}_{0}^{4}\\ =\left[3{(4)}^{2}-\frac{{\left(4\right)}^{3}}{3}\right]-0\\ =26\frac{2}{3}{\text{unit}}^{2}\end{array}$

**Example 2**

Find the area of the shaded region.

*Solution:**y*=

*x*—–(1)

*x*= 8

*y*–

*y*

^{2}—–(2)

Substitute (1) into (2),

*y*= 8

*y*–

*y*

^{2}

*y*

^{2 }– 7y = 0

*y*(

*y*– 7) = 0

*y*= 0 or 7

From (1),

*x*= 0 or 7Therefore 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*= 8

*y*–

*y*

^{2}

At

*y*-axis,*x*= 00 = 8

*y*–*y*^{2}*y*(

*y*– 8) = 0

*y*= 0, 8

**Area of shaded region = (**

*A*_{1}) Area of triangle + (*A*_{2}) Area under the curve from*y*= 7 to*y*= 8.