Area Between Curves and Average Values

From slices to a total

A vertical strip between an upper curve $f$ and a lower curve $g$ has approximate area $(f(x)-g(x))\Delta x$ . Adding strips and taking a limit gives $A=\int_a^b(f-g)dx$ when $f\ge g$ throughout the interval. If the curves cross, find the intersections and split the integral so each integrand is nonnegative. Equivalently, integrate $|f-g|$ .

Horizontal strips can be simpler: integrate right boundary minus left boundary with respect to $y$ . Draw the region before selecting a variable; the best choice may avoid several separate pieces.

Visual guide

VISUAL GUIDEReplace varying height by an equal-area rectangle

x²
Average-height rectangle

The area under x² on [0, 2] is 8/3. Dividing by interval length 2 gives average height 4/3. The dashed rectangle has exactly the same area as the shaded curved region.

Worked example: a bounded region

The curves $y=x$ and $y=x^2$ meet at $x=0,1$ . Between them, $x\ge x^2$ , so

A=\int_0^1(x-x^2)dx=\frac12-\frac13=\frac16.

Checking a midpoint establishes which curve is higher. Integrating the difference in the opposite order would produce signed area $-1/6$ , not the geometric area.

Worked example: an average reading

The average value of an integrable function on $[a,b]$ , with $a<b$ , is

f_{\mathrm{avg}}=\frac1{b-a}\int_a^bf(x)dx.

For temperature $T(t)=20+3t^2$ degrees Celsius over $0\le t\le2$ hours, the accumulated temperature-time value is $48$ degree-hours. Dividing by two hours gives $24$ degrees Celsius. Averaging just the endpoint temperatures would give $26$ , which is incorrect for this curved history.

For continuous $f$ , the integral mean value theorem guarantees at least one $c\in[a,b]$ with $f(c)=f_{\mathrm{avg}}$ . Here $20+3c^2=24$ gives $c=2/\sqrt3$ . This is an existence statement; it does not say the midpoint always works.

Practice

Find the area between $y=x$ and $y=-x$ on $[-1,1]$ .
Find the average of $\sin x$ on $[0,\pi]$ .
The region has $0\le y\le1$ and $y^2\le x\le y$ . Write a horizontal-strip integral.

Show worked solutions

Split at zero: $\int_{-1}^0(-2x)dx+\int_0^12x dx=2$ .
$\pi^{-1}[-\cos x]_0^\pi=2/\pi$ .
$A=\int_0^1(y-y^2)dy=1/6$ . Right minus left is the width.