Answer to Question #277195 in Calculus for Air

Question

Find by double integration the area of the region in 𝑥𝑦 plane bounded by the curves 𝑦 = 𝑥

2 and

𝑦 = 4𝑥 − 𝑥

2

.

Answer 1

Let us find by double integration the area of the region in 𝑥𝑦 plane bounded by the curves $𝑦 = 𝑥^2$ and

$𝑦 = 4𝑥 − 𝑥^2.$ Taking into account that the equation $x^2=4x-x^2$ is equivalent to $2x^2-4x=0,$ and hence has the roots $x_1=0$ and $x_2=2,$ we conclude that the area $A$ of the region is equal to

$A=\int\limits_0^2dx\int\limits_{x^2}^{4x-x^2}dy =\int\limits_0^2(4x-x^2-x^2)dx =\int\limits_0^2(4x-2x^2)dx$

$=(2x^2-\frac{2}3x^3)|_0^2 =8-\frac{16}3=\frac{8}3.$