How do you find the area enclosed between two graphs?

The basic rule: upper curve minus lower curve

When determining the area enclosed between two graphs, the key idea is subtraction. Instead of calculating areas relative to the horizontal axis separately, you subtract the lower function from the upper function inside one definite integral.

The boundaries of this region are the points where the two curves meet. Integrating the difference between the curves gives the exact geometric area enclosed between them, even if part or all of the region dips below the x-axis.

Three steps to set up and evaluate the integral

First, find the points of intersection. Set f(x) equal to g(x) and solve for x. For polynomials, this usually means moving all terms to one side to solve a quadratic or cubic equation. These x-values become your lower limit, a, and upper limit, b.

Second, decide which curve lies on top between a and b. You can plug any test value strictly between a and b into both equations. Whichever gives the higher y-value is the upper function.

Third, integrate (upper minus lower) from a to b. To integrate a polynomial term, raise the exponent by one and divide by the new exponent. Substitute the upper limit, substitute the lower limit, and subtract.

A worked example: y = x + 2 and y = x^2

Find where the line and the parabola meet: set x + 2 = x^2. Rearranging gives x^2 - x - 2 = 0, which factors into (x - 2)(x + 1) = 0. The intersection points are x = -1 and x = 2.

Test a point between -1 and 2, such as x = 0. The line gives y = 2 and the parabola gives y = 0. Because 2 is greater than 0, the line lies above the parabola.

Set up the integral from -1 to 2 of (x + 2 - x^2) dx. The antiderivative is (x^2 / 2) + 2x - (x^3 / 3). Evaluating at 2 yields 10/3, and at -1 yields -7/6. Subtracting (10/3 - (-7/6)) gives 27/6, or 4.5 square units.

Common pitfalls to avoid

Subtracting in the wrong order leads to a negative area. Area is always positive; if your final answer is negative, verify which curve is on top rather than merely dropping the minus sign.

If the curves cross each other within the interval, the upper and lower functions switch places. In that case, split the calculation into separate integrals at each crossing point so areas do not cancel out.