Question

Sketch the region enclosed by the given curves and find its area.

Sketch the region enclosed by the given curves.

\(12. y = {x^2},\quad y = 4x - {x^2}\)

Short answer

The resulting area is \(\frac{8}{3}\)

Step-by-step solution

Let us define the following functions:

\(y = {x^2},\quad y = 4x - {x^2}\)

Let us sketch the graph presenting those functions:

Find the intersections by setting the functions equal to each other

\(4x - {x^2} = {x^2}\)

\(0 = 2{x^2} - 4x\)

\(0 = 2x(x - 2)\)

\(x = 0,2\)

These will be the integral limits. The top function is\(y = 4x - {x^2}\) the bottom is \(y = {x^2}\)

Final Proof

Subtract the bottom function from the top one.

\(A = \int_0^2 {\left( {\left( {4x - {x^2}} \right) - {x^2}} \right)} dx\)

\( = \int_0^2 {\left( {4x - 2{x^2}} \right)} dx\)

\( = \left( {2{x^2} - 2 \times \frac{1}{3}{x^3}} \right)_0^2\)

\( = 2{(2)^2} - \frac{2}{3}{(2)^3} - (0 - 0)\)

\( = 8 - \frac{{16}}{3} = \frac{{24 - 16}}{3} = \frac{8}{3}\)

The resulting area is \(\frac{8}{3}\)