Area Between Curves

Main idea

Slice the region between two curves into thin vertical strips: each strip's height is the top curve minus the bottom one, so the area is the integral of top minus bottom. The places where the curves cross tell you where to start and stop.

A = \int_a^b\big[{\htmlData{sym=0}{f(x)}} - {\htmlData{sym=1}{g(x)}}\big]\,dx \quad (f \ge g \text{ on } [a, b])

What the symbols mean

$f(x)$
Top curve
The upper boundary.
$g(x)$
Bottom curve
The lower boundary. The difference is each slice's height.

Slicing horizontally

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

Explanation

The area between curves is simply the amount of space trapped between two graphs. All we're really doing is what we did for the area under one curve, except the bottom of each rectangle is now another curve instead of the x-axis. The formula has three pieces: $f(x)$ (the top curve), $g(x)$ (the bottom curve) and $f(x) - g(x)$ (the gap between them, which is the height of each thin strip). Integrate the gap from where the region starts to where it ends. That's it.

You've actually done this before without calling it that. The area under $f$ is the area between $f$ and the curve $y = 0$ , and $f(x) - 0$ is just $f(x)$ . Now the bottom is allowed to be any curve.

Each thin vertical slice is a rectangle of height $f(x) - g(x)$ and width $dx$ . Integrating adds them up. Because it's a difference, it doesn't matter whether the curves are above or below the axis: the gap between them is the same either way.

A common mistake. It's tempting to integrate one curve minus the other across the whole interval even when they cross in the middle, because it's one tidy integral. It doesn't work: past the crossing, top and bottom swap, the difference turns negative, and that piece of area subtracts instead of adding. Split at every crossing.

Steps:

Sketch the region and decide which curve is on top.
Solve $f(x) = g(x)$ for the crossing points.
If the curves swap places, split the integral at each crossing.

Switch to $y$ when the region is bounded by curves like $x = y^2$ , or when vertical slices would need several formulas. Then slices are horizontal: right minus left.

See it

This is a demand curve, the price buyers would pay for each extra unit, with a horizontal line at the market price. The area between them is the consumer surplus: what buyers would have been willing to pay, minus what they actually pay, added up over every unit sold. Notice that lowering the price both deepens the region and widens it, since more people buy, so the surplus grows quickly: from about 11 at a price of 45 to about 239 at a price of 10.

Worked examples

Example 1

Find the area between $y = x^2$ and $y = x + 2$ .

1
Find where they cross: $x^2 = x + 2$ means $(x - 2)(x + 1) = 0$ , so $x = -1$ and $x = 2$ . At $x = 0$ , between them, the line is at $2$ and the parabola at $0$ , so the line is on top.

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Find the area between $y = x$ and $y = x^2$ on $[0, 1]$ .

Problem 2+5 XP

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

Problem 3+5 XP

Find the area between $y = x^2$ and $y = 2x$ .

Problem 4+5 XP

Find the area between $y = \cos x$ and $y = \sin x$ from $0$ to $\frac\pi4$ .

Problem 5+5 XP

Find the area between $x = y^2$ and $x = y + 2$ (integrate in $y$ ).

Problem 6+5 XP

Two apps earn revenue at the rates $f(t) = 5 + 0.4t$ and $g(t) = 3 + 0.8t$ million dollars per year. The first one leads until the rates cross. How much more does it earn up to that point?

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Find the total area enclosed between $y = x^3 - 3x$ and $y = x$ .

Where it's used

Consumer and producer surplus in economics are areas between a demand or supply curve and a price line. The gap between two growth curves over time is total excess.

Consumer SurplusCalc II · Economics & Finance

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