Example 4 | Plane Areas in Rectangular Coordinates

$y = 4x - x^2$

$(x - 2)^2 = -(y - 4)$   →   downward parabola, vertex at (2, 4), LR = 1
 

$A = \displaystyle{{\int_{x_1}}^{x_2}} y \, dx$

$A = \displaystyle{{\int_{x_1}}^{x_2}} (y_U - y_L) \, dx$

$A = \displaystyle{{\int_{-2}}^2} [ \, 4 - (4x - x^2) \, ] \, dx$

$A = 2\displaystyle{{\int_0}^2} (4 + x^2) \, dx$

$A = 2\left[ 4x + \dfrac{x^3}{3} \right]_0^2$

$A = 2 \left[ 4(2 - 0) + \dfrac{(2 - 0)^3}{3} \right]$

$A = \frac{64}{3} \, \text{ unit}^2$           answer