Area under a curve
Net area and total area
\int_0^{2\pi}\sin x\,dx=0. Yet the graph of \sin x clearly encloses some area between 0 and 2\pi. So what exactly does the integral measure, and how would you find the area you can see?
Drag the endpoints of the interval below. Ochre regions lie above the axis and indigo regions below. Compare the integral, which is the net area, with the total area you would paint.
Drag the ochre endpoints along the axis.
Net area and total area
The definite integral adds up f(x)\,\Delta x, and where f(x)<0 those terms are negative. So the integral measures net area:
The total area counts every region as positive. It is the integral of |f(x)|, and you compute it in three steps:
- 1. Find the zeros
Solve f(x)=0 to find where the graph crosses the axis inside [a,b].
- 2. Split the integral
Integrate separately over each piece between consecutive zeros.
- 3. Add the sizes
Add the absolute values of the pieces.
For \sin x on [0,2\pi], the pieces are \int_0^{\pi}\sin x\,dx=2 and \int_{\pi}^{2\pi}\sin x\,dx=-2. The net area is 0 and the total area is 4.
When geometry is faster
If the region is a familiar shape, use its area formula. The graph of \sqrt{4-x^2} is the top half of a circle of radius 2, so \int_{-2}^{2}\sqrt{4-x^2}\,dx=\tfrac12\pi(2)^2=2\pi, with no antiderivative needed. Which of the functions above are made of triangles?
Solve it step by step
Find the net area and the total area. Answers can use pi, as in 2pi. The score counts steps you answer correctly on the first try.
Common mistakes
- Reporting the integral as the area
If the graph dips below the axis, the integral is smaller than the area you can see, and can even be 0 or negative.
- Taking the absolute value at the end
\left|\int_a^bf(x)\,dx\right| is not the total area. The cancellation has already happened inside. Split at the zeros first.
- Missing a zero
Check every zero inside the interval, not just the obvious one. x^3-4x on [-2,2] crosses at 0 as well as at the endpoints.