Area Between Curves

A single definite integral measures the area between a curve and the x-axis. Replace that axis with a second curve and you get the area trapped between the two. If, on [a, b], the curve y = f(x) sits on top and y = g(x) underneath (so f(x) \ge g(x) throughout), the region between them has area

A = \int_a^{b} \big(f(x) - g(x)\big)\,dx.

Read it as "top minus bottom, integrated across the strip". The lower curve is free to dip below the axis — the subtraction still measures the genuine vertical gap.

Why "top minus bottom"? A strip-by-strip derivation

The formula is not an extra rule to memorise — it falls straight out of the Riemann-sum picture.

Step 1 — slice the region into thin vertical strips. Chop [a, b] into n pieces of width \Delta x. Over the i-th piece, at a sample point x_i, approximate the region by a rectangle that reaches up to the top curve and down to the bottom curve.

Step 2 — find one strip's area. That rectangle has width \Delta x and height equal to the vertical gap between the curves, f(x_i) - g(x_i). So its area is

\Delta A_i = \big(f(x_i) - g(x_i)\big)\,\Delta x.

Step 3 — add up all the strips. The whole region is approximately the sum of these rectangles — a Riemann sum:

A \approx \sum_{i=1}^{n} \big(f(x_i) - g(x_i)\big)\,\Delta x.

Step 4 — refine to a limit. Let the strips get infinitely thin (n \to \infty, \Delta x \to 0). The Riemann sum converges to the definite integral, and the approximation becomes exact:

A = \lim_{n \to \infty} \sum_{i=1}^{n} \big(f(x_i) - g(x_i)\big)\,\Delta x = \int_a^{b} \big(f(x) - g(x)\big)\,dx.

Worked example: between y = x and y = x^2

Find the area enclosed between the line y = x and the parabola y = x^2.

Step 1 — find where they meet (the limits of integration). Set the curves equal:

x = x^2 \;\Longrightarrow\; x^2 - x = 0 \;\Longrightarrow\; x(x - 1) = 0 \;\Longrightarrow\; x = 0,\ 1.

Step 2 — decide which curve is on top. Test a point inside (0, 1), say x = \tfrac12: x = 0.5 while x^2 = 0.25. The line is higher, so f(x) = x is the top and g(x) = x^2 the bottom.

Step 3 — set up the integral.

A = \int_0^{1} \big(x - x^2\big)\,dx.

Step 4 — antidifferentiate and evaluate.

A = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^{1} = \left(\frac{1}{2} - \frac{1}{3}\right) - 0 = \frac{3 - 2}{6} = \frac{1}{6}.

The little sliver between line and parabola has area exactly \tfrac16.

Let f and g be continuous on [a, b] with f(x) \ge g(x) there. The area of the region bounded above by y = f(x) and below by y = g(x) is

A = \int_a^{b} \big(f(x) - g(x)\big)\,dx.

Vertical strips are natural when one curve is cleanly above the other across the whole span. But for a region better described as "right curve minus left curve" — bounded on the sides rather than the top and bottom — slice it horizontally and integrate in y instead:

A = \int_c^{d} \big(x_{\text{right}}(y) - x_{\text{left}}(y)\big)\,dy.

Take the region between x = y^2 and x = y + 2. As a function of x the boundary is a multi-valued mess (the sideways parabola fails the vertical-line test). As a function of y it is two honest curves: they meet where y^2 = y + 2, i.e. y = -1, 2, and the line x = y + 2 is to the right throughout. One clean horizontal integral replaces two awkward vertical ones. The lesson: slice in whichever direction the boundary becomes single-valued.

See the region, change the strip

Here are the line y = x and the parabola y = x^2, with the region between them shaded. Drag the limits a and b: the shaded area updates live, and the readout reports \int_a^{b}(x - x^2)\,dx exactly. At a = 0,\ b = 1 you recover the \tfrac16 above.