Area Between Curves (Functions of x) — AP® Calculus AB Notes

From area under a curve to area between curves

You already know that $\int_a^b f(x)\,dx$ adds up thin rectangles that stand on the $x$ -axis and reach up to the graph of $f$ . Now the region you care about has a curve on top and a different curve on the bottom. The $x$ -axis is not part of the story at all.

The idea is the same as before. Cut the region into thin vertical rectangles. Each rectangle stands at some $x$ , has a small width $\Delta x$ , and reaches from the bottom curve up to the top curve. If $f$ is the top and $g$ is the bottom, the rectangle's height is

\text{height} = f(x) - g(x).

So one rectangle has area $\left(f(x) - g(x)\right)\Delta x$ . Adding all of them is a Riemann sum for the function $f - g$ , and letting the rectangles get thinner turns the sum into a definite integral.

The region between the parabola y = 6 − x² (on top) and the line y = x (on the bottom), shaded from x = −3 to x = 2, where they meet at (−3, −3) and (2, 2). One thin vertical rectangle of width Δx has height (6 − x²) − x. A side panel shows the limits from 6 − x² = x and the area, the integral from −3 to 2 of [(6 − x²) − x] dx = 125/6. — **Figure 1.** One thin rectangle between $y = 6 - x^2$ and $y = x$ . Its height is top minus bottom, and the region runs between the two crossings. LevlPrep original diagram.

The area between curves. If $f(x) \ge g(x)$ for every $x$ in $[a, b]$ , the area of the region between their graphs from $x = a$ to $x = b$ is

A = \int_a^b \left(f(x) - g(x)\right)\,dx.

Read it as "top minus bottom, from left to right." The formula does not care whether the curves are above or below the $x$ -axis. Move both graphs down 10 units and every height $f(x) - g(x)$ stays the same, so the area stays the same.

The three steps

Find the limits. When a region is "bounded by" two curves, it starts and ends where they meet. Solve $f(x) = g(x)$ . The solutions are $x$ -values, which is what an integral in $dx$ needs.
Decide which curve is on top. Pick any $x$ between the limits and evaluate both functions. The larger value is the top.
Set up and integrate top minus bottom. Put the bottom function in parentheses before subtracting, so every one of its terms changes sign.

Worked example. Find the area of the region bounded by $y = 6 - x^2$ and $y = x$ (the figure above).

Limits. Set the functions equal: $6 - x^2 = x$ , so $x^2 + x - 6 = 0$ , which factors as $(x + 3)(x - 2) = 0$ . The curves meet at $x = -3$ and $x = 2$ .

Top. Test $x = 0$ : the parabola gives 6 and the line gives 0. The parabola is on top.

Integral. Top minus bottom is $(6 - x^2) - x$ :

A = \int_{-3}^{2} \left(6 - x^2 - x\right)\,dx = \left[6x - \frac{x^3}{3} - \frac{x^2}{2}\right]_{-3}^{2}.

At $x = 2$ : $12 - \frac{8}{3} - 2 = \frac{22}{3}$ . At $x = -3$ : $-18 + 9 - \frac{9}{2} = -\frac{27}{2}$ . So $A$ $= \frac{22}{3}$ $+ \frac{27}{2}$ $= \frac{44 + 81}{6}$ $= \frac{125}{6}$ .

The region dips below the $x$ -axis on the left, but nothing special happens there: the height $(6 - x^2) - x$ is still positive.

Parentheses and other places points go missing

The most common algebra slip is dropping parentheses. With top $y = x + 1$ and bottom $y = x^2 - 1$ , the height is

(x + 1) - (x^2 - 1) = x + 1 - x^2 + 1 = x + 2 - x^2.

Writing $x + 1 - x^2 - 1$ forgets that subtracting $-1$ adds 1, and every number after that is wrong.

Worked example. Find the area of the region bounded by $y = e^x$ , $y = 1$ and $x = \ln 3$ .

Limits. This time one boundary is a vertical line, so one limit is given: $x = \ln 3$ . The other is where $e^x = 1$ , which is $x = 0$ .

Top. For $x > 0$ , $e^x > 1$ , so $e^x$ is on top and the line $y = 1$ is the bottom (not the $x$ -axis).

Integral.

A = \int_0^{\ln 3} \left(e^x - 1\right)\,dx = \Big[e^x - x\Big]_0^{\ln 3} = (3 - \ln 3) - (1 - 0) = 2 - \ln 3.

How to check an answer

What you see	What it usually means	Fix
A negative area	You subtracted bottom minus top	Test a point between the limits to find the top curve
An answer that matches the area under the top curve	The bottom curve was left out	Subtract the bottom function, even when it is a line like $y = 1$
Limits like the $y$ -values where the curves meet	Limits in the wrong variable	With $dx$ , use $x$ -values
Limits where one curve meets the $x$ -axis	The $x$ -axis is not a boundary here	Solve $f(x) = g(x)$

A quick size check helps too. The region is never taller than the largest gap between the curves and never wider than $b - a$ , so the area is less than that height times that width.

With a calculator

When $f(x) = g(x)$ cannot be solved by algebra, as with $2\cos x = x^2 - 1$ , use the calculator to find the crossings (intersect or a solver, in radian mode). Store each crossing in the calculator instead of retyping a rounded value. Then write the setup, for example $\int_{-a}^{a} \left(2\cos x - (x^2 - 1)\right)\,dx$ , and let the calculator evaluate it. Report the answer correct to three decimal places, and round only at the end.