Math Core

You know that $\int_a^b f(x)\,dx$ gives the area under a positive curve. The next step is the area of a region trapped between two curves. It is one of the most common setups on the AP exam, and the same "top minus bottom" thinking is the foundation for the volume lessons later in this unit.

Top minus bottom

Suppose $f(x) \ge g(x)$ for every $x$ in $[a, b]$ . Slice the region into thin vertical rectangles of width $\Delta x$ . A rectangle at position $x$ reaches from the lower curve up to the upper curve, so its height is $f(x) - g(x)$ and its area is about $\big(f(x) - g(x)\big)\Delta x$ . Adding the rectangles and letting $\Delta x \to 0$ gives an integral.

Area between two curves

If $f(x) \ge g(x)$ on $[a, b]$ , the area of the region between them is

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

This works even when part of the region is below the $x$ -axis. The height of a slice is always top minus bottom, whatever the signs of the two functions.

The four-step method

Sketch the curves (or picture them on a calculator) so you know which is on top.
Find the intersections by solving $f(x) = g(x)$ . If the problem doesn't give $a$ and $b$ , these are your limits.
Set up $\int_a^b (\text{top} - \text{bottom})\,dx$ .
Evaluate. The answer must be positive; a negative area means you subtracted in the wrong order.

The region between y = 2x + 3 (top) and y = x² (bottom). A vertical slice has height (2x + 3) − x².Open in grapher →

When the curves cross

If the curves trade places inside the interval, "top" changes. Split the integral at each crossing point and use the correct order on each piece. Equivalently, the area is

A = \int_a^b |f(x) - g(x)|\,dx,

which is exactly what you type into a calculator. Without a calculator, you must split by hand.

Common mistake

Don't integrate $f - g$ straight across a crossing point. The positive and negative pieces cancel and you get a number smaller than the true area, sometimes even 0. Always check for intersections inside $[a, b]$ , not only at the ends.

Worked examples

Worked example: Limits from the intersections

Find the area of the region bounded by $y = x^2$ and $y = 2x + 3$ .

Solution. Intersections: $x^2 = 2x + 3$ gives $x^2 - 2x - 3 = 0$ , so $(x - 3)(x + 1) = 0$ and $x = -1$ or $x = 3$ . At $x = 0$ the line is at 3 and the parabola at 0, so the line is on top.

\begin{aligned} A &= \int_{-1}^{3} \big(2x + 3 - x^2\big)\,dx = \left[x^2 + 3x - \frac{x^3}{3}\right]_{-1}^{3} \\ &= (9 + 9 - 9) - \left(1 - 3 + \frac{1}{3}\right) = 9 + \frac{5}{3} = \frac{32}{3}. \end{aligned}

Worked example: A parabola and a line through the origin

Find the area of the region between $y = 4x - x^2$ and $y = x$ .

Solution. $4x - x^2 = x$ gives $x(3 - x) = 0$ , so $x = 0$ and $x = 3$ . Testing $x = 1$ : the parabola is at 3 and the line at 1, so the parabola is on top.

A = \int_0^3 \big(4x - x^2 - x\big)\,dx = \int_0^3 (3x - x^2)\,dx = \left[\frac{3x^2}{2} - \frac{x^3}{3}\right]_0^3 = \frac{27}{2} - 9 = \frac{9}{2}.

Worked example: Curves that switch

Find the area between $y = \sin x$ and $y = \cos x$ on $\left[0, \dfrac{\pi}{2}\right]$ .

Solution. They cross where $\sin x = \cos x$ , at $x = \dfrac{\pi}{4}$ . Cosine is on top before that and sine is on top after.

\begin{aligned} A &= \int_0^{\pi/4} (\cos x - \sin x)\,dx + \int_{\pi/4}^{\pi/2} (\sin x - \cos x)\,dx \\ &= \Big[\sin x + \cos x\Big]_0^{\pi/4} + \Big[-\cos x - \sin x\Big]_{\pi/4}^{\pi/2} \\ &= (\sqrt{2} - 1) + (-1 + \sqrt{2}) = 2\sqrt{2} - 2. \end{aligned}

If you had integrated $\cos x - \sin x$ over the whole interval, you would have gotten 0.

Between 0 and π/2 the curves cross at π/4, so the region is two pieces with different tops.Open in grapher →

Tip

To decide which curve is on top, plug one easy test value between the intersection points into both functions. It is faster and more reliable than guessing from memory of the graphs.

Calculator questions

On calculator-active AP questions, intersections are often not nice numbers. Store each intersection in your calculator at full precision, then integrate. Round only the final answer, to three decimal places, unless told otherwise.