Area Between Two Curves

In calculus, the area between two curves refers to the region enclosed by two functions on a given interval.
It is one of the most fundamental applications of integrals, especially in AP Calculus BC.

The key idea:

Area is the integral of “top function minus bottom function.”

Area Between Curves (Vertical Slicing)

Most commonly, we compute area using vertical slices (integration with respect to x).

Suppose:

• y=f(x) is the upper function

• y=g(x) is the lower function

• They are continuous on [a,b]

Then the area between them is:

$$\int_{a}^{b}[f(x)-g(x)]dx$$

Geometry of Vertical Slices

Imagine slicing the area into thin rectangles:

• Height = f(x)−g(x)

• Width = dx

Summing the rectangles gives the integral:

$$\int(top-bottom)dx$$

Example (Vertical Slicing)

Find the area between:

$$y=x^{2}$$ and $$y=2x$$ 

Step 1: Find intersection points

Solve:

$$x^{2}=2x \longrightarrow x^{2}-2x=0$$ and factoring $$x(x-2)=0 \longrightarrow x=0 , x=2$$

Step 2: Identify top and bottom

For $$0\leq{x}\leq2$$

• 2x is the top

• $$x^{2}$$ is the bottom

Step 3: Set up the integral

A=$$\int_{0}^{2}(2x-x^{2})dx$$

Compute:

A=$$\left[x^{2}-\frac{1}{3}x^{3}\right]=(4-\frac{8}{3})-0=\frac{4}{3}$$​

✔️ Final Answer:

$$A=\frac{4}{3}$$​

Horizontal Slicing

Sometimes vertical slices do not work well, such as when:

• The region is bounded horizontally

• A function fails the vertical line test

• Equations given explicitly as x=f(y)

Then you use horizontal slices:

Area=$$\int_{a}^{b}[right function – left function]dy$$

Where:

• a , b = intersection points in terms of y

Example (Horizontal Slicing)

Find area between:

$$x=y^{2}$$ and $$x=4-y^{2}$$

These curves open left/right, so horizontal slices are easier.

Step 1: Find intersections

Solve:

$$y^{2}=4-y^{2} \longrightarrow 2y^{2}-4=0$$ then $$y^{2}-2=0 \longrightarrow (y+\sqrt{2})(y-\sqrt{2})=0$$

$$y=\pm\sqrt{2}$$​

Step 2: Identify left/right curves

For this region:

• Right : $$x=4-y^{2}$$

• Left : $$x=y^{2}$$

Step 3: Set up integral

A=$$\int_{-\sqrt{2}}^{\sqrt{2}}[(4-y^{2})-y^{2}]dx$$

Simplify:

A=$$\int_{-\sqrt{2}}^{\sqrt{2}}(4-2y^{2})dx$$

Compute:

A=$$\left[4y-\frac{2y^{3}}{3}\right]_{-\sqrt{2}}^{\sqrt{2}}$$​​

Answer:

A=$$\frac{16\sqrt{2}}{3}$$​​

Piecewise Area

If the top/bottom function switches, the area must be split at that point.

Example:

$$y=x$$ and $$y=|x|$$

The functions change relationship at $x=0$, so break integral:

A=$$\int_{-1}^{0}[|x|-x]dx + \int_{0}^{1}[x-|x|]dx$$