# `y = x^2, y = 4x - x^2` Sketch the region enclosed by the given curves and find its area.

You need to determine first the points of intersection between curves `y = x^2 ` and `y = 4x - x^2` , by solving the equation, such that:

`x^2 = 4x - x^2 => 2x^2 - 4x = 0`

Factoring out 2x yields:

`2x(x - 2) = 0 => 2x...

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You need to determine first the points of intersection between curves `y = x^2 ` and `y = 4x - x^2` , by solving the equation, such that:

`x^2 = 4x - x^2 => 2x^2 - 4x = 0`

Factoring out 2x yields:

`2x(x - 2) = 0 => 2x = 0 or x - 2 = 0`

Hence, the endpoints of integral are x = 0 and x = 2.

You need to decide what curve is greater than the other on the interval [0,2]. You need to notice that `x^2 < 4x - x^2` on the interval [0,2], hence, you may evaluate the area of the region enclosed by the given curves, such that:

`int_a^b (f(x) - g(x))dx` , where `f(x) > g(x)` for `x in [a,b]`

`int_0^2 (4x - x^2 - x^2)dx = int_0^2 (4x - 2x^2)dx`

`int_0^2 (4x - 2x^2)dx = int_0^2 (4x)dx - int_0^2 (2x^2)dx`

`int_0^2 (4x - 2x^2)dx = 4x^2/2|_0^2 - 2x^3/3|_0^2`

`int_0^2 (4x - 2x^2)dx = 2x^2|_0^2 - 2x^3/3|_0^2`

`int_0^2 (4x - 2x^2)dx = 2(2^2 - 0^2) - (2/3)(2^3 - 0^3)`

`int_0^2 (4x - 2x^2)dx = 8 - 16/3`

`int_0^2 (4x - 2x^2)dx = (24-16)/3`

`int_0^2 (4x - 2x^2)dx = 8/3`

Hence, evaluating the area of the region enclosed by the given curves, yields `int_0^2 (4x - 2x^2)dx = 8/3.`

The area of the region enclosed by the given curves is found between the red and orange curves, for `x in [0,2]` .

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