Using a double integral, find the area of the region between y = 4x - x^2 and the line y = x.

Step-by-step solution

1. 1

   Find where the two curves meet. Setting $4x-x^{2}=x$ gives $3x-x^{2}=0$, i.e. $x(3-x)=0$, so the curves cross at $x=0$ and $x=3$. Since $y=x$ there, the intersection points are $(0,0)$ and $(3,3)$. These two values become the outer limits of the integral.

2. 2

   Decide which curve is on top. Test an interior point: at $x=1$ the parabola gives $4(1)-1^{2}=3$ while the line gives $1$. So the parabola is above the line throughout $0<x<3$, and the region is $R=\left\{(x,y):0\le x\le 3,\ x\le y\le 4x-x^{2}\right\}.$

3. 3

   Set up the area as a double integral. Area is the double integral of the constant $1$ over the region, so with the vertical-strip description above, $A=\iint_R 1\,dA=\int_{0}^{3}\int_{x}^{4x-x^{2}}1\,dy\,dx .$

4. 4

   Do the inner integration. Integrating $1$ with respect to $y$ just measures the strip height, $\int_{x}^{4x-x^{2}}1\,dy=(4x-x^{2})-x=3x-x^{2},$ which is the familiar top-minus-bottom integrand of a single-variable area problem.

5. 5

   Do the outer integration. $A=\int_{0}^{3}\left(3x-x^{2}\right)dx=\left[\frac{3x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{3}=\frac{27}{2}-9=\frac{9}{2}.$ The integrand $3x-x^{2}$ is non-negative on $[0,3]$, so the answer is positive as an area must be.