Finding areas

Finding areas when the curve is below the x-axis or crosses it by splitting at the x-intercepts, and finding areas between a curve and the y-axis.

Integral or area?

An integral counts area below the axis as negative, but an area is always positive. So when part of the region is below the axis, you must split the integral.

Positive and negative pieces

The blue piece counts as negative in the integral. At b = 2 the integral is 2/3, but the total shaded area is 2.

Try this

Find the value of $b$ that makes the integral zero.

Example 1: below the axis●

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

The curve cuts the axis at $x = \pm 2$ and is below it in between.

$\displaystyle\int_{-2}^{2} (x^2 - 4)\,dx = \Big[\tfrac{x^3}{3} - 4x\Big]_{-2}^{2} = -\tfrac{16}{3} - \tfrac{16}{3} = -\tfrac{32}{3}$.

Area $= \tfrac{32}{3}$ square units.

Example 2: crossing the axis●●

Find the area between $y = x^2 - 1$ and the $x$-axis from $x = 0$ to $x = 2$.

The curve crosses at $x = 1$, so split there.

$\displaystyle\int_0^1 (x^2 - 1)\,dx = \tfrac13 - 1 = -\tfrac23$ and $\displaystyle\int_1^2 (x^2 - 1)\,dx = \tfrac23 - \left(-\tfrac23\right) = \tfrac43$.

Area $= \tfrac23 + \tfrac43 = 2$ square units.

Areas against the $y$-axis

For a region between a curve and the $y$-axis, write $x$ in terms of $y$ and integrate with respect to $y$: $$\text{Area} = \int_c^d x\,dy.$$

Example 3●●

Find the area between $y = x^2$ (with $x \ge 0$), the $y$-axis and the line $y = 4$.

$x = \sqrt{y}$, and $y$ runs from $0$ to $4$.

$\displaystyle\int_0^4 y^{\frac12}\,dy = \Big[\tfrac23 y^{\frac32}\Big]_0^4 = \tfrac23 \times 8 = \tfrac{16}{3}$ square units.

Flashcards

Area below the $x$-axis

The absolute value of the integral.

A curve that crosses the axis

Split at the $x$-intercepts and add the absolute values.

Area against the $y$-axis

$\displaystyle\int_c^d x\,dy$

Practice

Find the area between $y = x^2 - 4$ and the $x$-axis.
Evaluate $\displaystyle\int_0^2 (x^2 - 1)\,dx$.
Find the area between $y = x^2 - 1$ and the $x$-axis from $x = 0$ to $x = 2$.
Find the area between $y = x^2 - 3x$ and the $x$-axis.
Find the total area between $y = x^3$ and the $x$-axis from $x = -1$ to $x = 1$.
Find the area between $y = x^2$ ($x \ge 0$), the $y$-axis and $y = 4$.
Find the area between $y = x^2$ ($x \ge 0$), the $y$-axis, $y = 1$ and $y = 9$.
A curve crosses the $x$-axis inside the interval. The first step to find the area is:
The curve is below the axis on $[a, b]$ and $\displaystyle\int_a^b f(x)\,dx = -6$. The area is: