Question

Find the area of the shaped region.

Short answer

The area of the shaded region is \(\frac{{32}}{3}\)

Step-by-step solution

Given information

\(\begin{aligned}{l}y = 5x - {x^2}\\y = x\end{aligned}\)

The graph

Finding the area of shaped region

The area of the blue region given by these functions must be determined.

\(\begin{aligned}{l}y = 5x - {x^2}\\y = x\end{aligned}\)

First, we must determine where these two functions cross. We can see which points are intersection on the graph, but let's calculate:

\(y = y\)

\(5x - {x^2} = x\)

\(5x - {x^2} - x = 0\)

\(\begin{aligned}{l} - {x^2} + 4x = 0 \Rightarrow x\left( { - x + 4} \right) = 0 \Rightarrow \left\{ {x = 0} \right.\\ - x + 4 = 0 \Rightarrow x = 4\end{aligned}\)

We've reached the end of our integration capabilities.

To get the area, we must first compute the integral

\(\int\limits_0^4 {\left( {\left( {5x - {x^2}} \right) - x} \right)} dx\)

The intersection points are the integration limits.

\(\int\limits_0^4 {\left( {4x - {x^2}} \right)} dx = \int\limits_0^4 {4xdx - } \int\limits_0^4 {{x^2}dx} \)

\(\left( {use:\int {\alpha f(x)dx} = \alpha \int {f(x)dx} } \right)\)

\( = 4\int\limits_0^4 {xdx - } \int\limits_0^4 {{x^2}dx} \)

\( = \left. {4.\frac{{{x^2}}}{2}} \right|_0^4 - \left. {\frac{{{x^3}}}{3}} \right|_0^4\)

\(\begin{aligned}{l} = 2.\left( {{4^2} - 0} \right) - \frac{1}{3}.\left( {{4^3} - 0} \right)\\ = 32 - \frac{{64}}{3}\end{aligned}\)

The area of the shaded region is \(\frac{{32}}{3}\)