8.4 Finding the Area Between Curves Expressed as Functions of \(x\)

8.4 Finding the Area Between Curves Expressed as Functions of (x)

Topic 8.3 used a definite integral to add up a changing rate. The same adding-up works in geometry: cut a region of the plane into thin vertical strips, and the integral of the strip heights is the region's area. This page starts with the area under a single curve, then finds the area enclosed between two curves written as functions of $x$.

Definite integrals as geometric area

When $f(x)\ge 0$ on $[a,b]$, the definite integral

$$ \int_a^b f(x),dx $$

equals the geometric area between the graph of $f$, the $x$-axis, and the vertical lines $x=a$ and $x=b$. If $f(x)$ is below the $x$-axis, the integral is negative because it represents signed area.

CHA-5: Definite integrals allow us to solve problems involving the accumulation of change in area or volume over an interval.

For Topic 8.4, the learning objective is CHA-5.A: Calculate areas in the plane using the definite integral, supported by CHA-5.A.1: Areas of regions in the plane can be calculated with definite integrals. The associated communication skill is 4.C: Use appropriate mathematical symbols and notation. A correct answer requires more than a final decimal: the bounds, integrand, subtraction order, and differential must communicate the region precisely.

Area between curves expressed as functions of $x$

For two curves written as $y=f(x)$ and $y=g(x)$, a vertical slice has height

$$ \text{top}-\text{bottom}. $$

Thus, if $f(x)$ is above $g(x)$ on $[a,b]$,

$$ \boxed{ \text{Area}=\int_a^b\bigl(f(x)-g(x)\bigr),dx }. $$

The bounds $a$ and $b$ usually come from the intersection points, found by solving

$$ f(x)=g(x). $$

Worked example: a region bounded by two curves

Let

$$ f(x)=x+2 \qquad\text{and}\qquad g(x)=x^2. $$

The intersection points satisfy

$$ x+2=x^2, $$

so

$$ x^2-x-2=(x-2)(x+1)=0. $$

Therefore, the region extends from $x=-1$ to $x=2$. On $x=0$, $f(0)=2$ and $g(0)=0$, so $f$ is above $g$. The area is

$$ \int_{-1}^{2}\left[(x+2)-x^2\right]dx. $$

Compute:

$$ \begin{aligned} \text{Area} &=\left[\frac{x^2}{2}+2x-\frac{x^3}{3}\right]_{-1}^{2}\ &=\frac{10}{3}-\left(-\frac{7}{6}\right)\ &=\frac{9}{2}. \end{aligned} $$

The region has area $\frac{9}{2}$ square units.

Choosing the direction of the slices

Integrating with respect to $x$ uses vertical slices, so compare which curve is above the other. Integrating with respect to $y$ uses horizontal slices, so compare which curve is to the right:

$$ \text{horizontal width}=\text{right}-\text{left}. $$

For curves expressed as functions of $y$, the corresponding setup is

$$ \int_c^d\bigl(x_{\text{right}}(y)-x_{\text{left}}(y)\bigr),dy. $$

This is the same geometric idea—adding thin slices—but the next topic develops that $y$-based method in detail.

Misconception check — “Subtracting the curves automatically gives area.” Not always. If the curves cross inside the interval, $f(x)-g(x)$ changes sign and the integral may cancel positive and negative pieces. Find all intersections first, test the ordering on each subinterval, and split the integral whenever the top curve changes.

Retrieval check

A region is bounded by $y=3x$, $y=x^2$, and the $y$-axis. The curves intersect when $3x=x^2$, giving $x=0$ and $x=3$. Since $3x$ is above $x^2$ on $0<x<3$, write the area integral—not its evaluation:

$$ \boxed{\int_0^3(3x-x^2),dx}. $$

The essential questions are always: What does a slice measure? Where are the bounds? Which curve is on top—or, for horizontal slices, which is on the right?