Applications of Definite Integration

Find the area of the region bounded by the curve $y = 2x - x^2$ and the line $y = -x$.


Solving the equation of the curve $y = 2x - x^2$ and the equation of the line $y = - x$ $\;$ simultaneously gives the points of intersection of the two as $\left(0,0\right)$ and $\left(3,-3\right)$

The curve $y = 2x - x^2$ cuts the X axis at the point $\left(2,0\right)$ in the first quadrant.

Let the area enclosed by the curve $y = 2x - x^2$ and the X axis in the first quadrant $= A_1 = \left|I_1\right|$ where

$\begin{aligned} I_1 & = \int \limits_{0}^{2} y \; dx \\\\ & = \int \limits_{0}^{2} \left(2x - x^2\right) \; dx \\\\ & = 2 \times \left[\dfrac{x^2}{2}\right]_{0}^{2} - \left[\dfrac{x^3}{3}\right]_{0}^{2} \\\\ & = 4 - \dfrac{8}{3} = \dfrac{4}{3} \end{aligned}$

$\therefore$ $\;$ $A_1 = \left|I_1\right| = \dfrac{4}{3}$ sq units $\;\;\; \cdots \; (1)$

Let the area enclosed by the line $y = -x$, the X axis and the line $x = 3$ be $A_2 = \left|I_2\right|$ where

$\begin{aligned} I_2 & = \int \limits_{0}^{3} y \; dx \\\\ & = \int \limits_{0}^{3} -x \; dx \\\\ & = \left[\dfrac{- x^2}{2}\right]_{0}^{3} = \dfrac{-9}{2} \end{aligned}$

$\therefore$ $\;$ $A_2 = \left|I_2\right| = \dfrac{9}{2}$ sq units $\;\;\; \cdots \; (2)$

Let the area enclosed by the curve $y = 2x - x^2$, the X axis and the line $x = 3$ be $A_3 = \left|I_3\right|$ where

$\begin{aligned} I_3 & = \int \limits_{2}^{3} y \; dx \\\\ & = \int \limits_{2}^{3} \left(2x - x^2\right) \; dx \\\\ & = 2 \times \left[\dfrac{x^2}{2}\right]_{2}^{3} - \left[\dfrac{x^3}{3}\right]_{2}^{3} \\\\ & = 9 - 4 - \dfrac{1}{3} \left(27 - 8\right) = \dfrac{-4}{3} \end{aligned}$

$\therefore$ $\;$ $A_3 = \left|I_3\right| = \dfrac{4}{3}$ sq units $\;\;\; \cdots \; (3)$

$\therefore$ $\;$ Area of the region bounded by the curve $y = 2x - x^2$ and the line $y = -x$ $\;$ (shaded region) is

$\begin{aligned} A & = A_1 + A_2 - A_3 \\\\ & = \dfrac{4}{3} + \dfrac{9}{2} - \dfrac{4}{3} \hspace{3em} \left[\text{From equations } (1), (2) \text{ and } (3)\right] \\\\ & = \dfrac{9}{2} \; \text{sq units} \end{aligned}$