The area of the region enclosed between the graphs of two functions is obtained by integrating their difference (in absolute value): A=abf(x)g(x)dx.A = \int_a^b|f(x)-g(x)|\,dx. If fgf\ge g on the whole of [a,b][a,b], the absolute value can be dropped: A=ab(fg)dxA = \int_a^b(f-g)\,dx. The endpoints a,ba,b are usually the abscissae of the intersection points of the two curves.

Example — Area between f(x)=x2+4xf(x) = -x^2+4x and g(x)=1+xg(x) = 1+x

The intersections are found from x2+4x=1+x-x^2+4x = 1+x, that is x23x+1=0x^2-3x+1=0: xA0,38x_A\approx 0{,}38 and xB2,62x_B\approx 2{,}62. Over the interval the parabola lies above the line, so: A=0,382,62[(x2+4x)(1+x)]dx=0,382,62(x2+3x1)dx.A = \int_{0{,}38}^{2{,}62}\bigl[(-x^2+4x)-(1+x)\bigr]\,dx = \int_{0{,}38}^{2{,}62}(-x^2+3x-1)\,dx.

The green region between the parabola ff (blue) and the line gg (red): the area is the integral of the difference fgf-g between the two intersections.

Topics: Integrale
Concepts: Area tra due curve · Integrale definito
Skills: Geometria analitica · Integrare · Interpretare grafico