Areas of compound regions

Finding the area between two curves by integrating the upper curve minus the lower curve, finding points of intersection, and splitting compound regions into simpler pieces.

The area between two curves

Upper minus lower

$$\text{Area} = \int_a^b \big(\text{upper} - \text{lower}\big)\,dx$$

This works wherever the region is, even below the $x$-axis, because the difference of the heights is always the height of the strip.

The region between y = x² and y = 2x

Each vertical strip has height 2x − x². The shaded area is 4/3.

Example 1●

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

$x^2 = 2x$ gives $x = 0$ or $x = 2$.

Intersections

The line is on top: $\displaystyle\int_0^2 (2x - x^2)\,dx = 4 - \tfrac83 = \tfrac43$ square units.

Integrate

Example 2●●

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

$4 - x^2 = x + 2$ gives $x^2 + x - 2 = 0$, so $x = -2$ or $x = 1$.

Intersections

$(4 - x^2) - (x + 2) = 2 - x - x^2$.

Upper minus lower

$\Big[2x - \tfrac{x^2}{2} - \tfrac{x^3}{3}\Big]_{-2}^{1} = \tfrac76 - \left(-\tfrac{10}{3}\right) = \tfrac92$ square units.

Compound regions

When the upper or lower boundary changes part way across, split the region where it changes.

Example 3●●

Find the area bounded by $y = x^2$, the line $y = 2 - x$ and the $x$-axis.

The curve and line meet at $x = 1$, and the line meets the $x$-axis at $x = 2$. The top boundary is $y = x^2$ from $0$ to $1$, then the line from $1$ to $2$.

$\displaystyle\int_0^1 x^2\,dx = \tfrac13$, and the piece under the line is a triangle of area $\tfrac12 \times 1 \times 1 = \tfrac12$.

Area $= \tfrac13 + \tfrac12 = \tfrac56$ square units.

Flashcards

Area between two curves

$\displaystyle\int_a^b (\text{upper} - \text{lower})\,dx$

Finding the limits

Solve the two equations simultaneously.

The boundary changes part way

Split the region where it changes.

Practice

Find the area enclosed between $y = x^2$ and $y = 2x$.
Find the area enclosed between $y = x^2$ and $y = x$.
Find the area enclosed between $y = x^2$ and $y = 4$.
Find the area enclosed between $y = x^2 - 2x$ and $y = x$.
Find the area enclosed between $y = 4 - x^2$ and $y = x + 2$.
Find the area between $y = x$ and $y = x^3$ from $x = 0$ to $x = 1$.
Find the area bounded by $y = x^2$, $y = 2 - x$ and the $x$-axis.
The area between two curves is:
Part of the region lies below the $x$-axis. Upper minus lower: