Area Under a Curve

Video lesson · from 26:592 worked examples

One lesson video covers all of Integration, so it opens at 26:59 for this topic — not from the beginning.

In general, we can find the area enclosed by a curve and the $x$ -axis between $x=a$ and $x=b$ by using:

\int_a^b f(x) \,dx

Find the limits if you are not given them: "the area between the curve and the $x$ -axis" means you must solve $y = 0$ to get the roots first.
An area is never negative: if the integral comes out negative the region is below the axis. Take the positive value and say why.
Split where the curve crosses the axis: integrating straight through a root lets positive and negative parts cancel, which understates the area.
Use brackets when substituting: particularly with a negative limit, where a missing bracket changes every sign.
State the units: areas are in square units.

Example 1

The graph of $y=x^2-4x$ is shown below.
Calculate the shaded area.

The area is between $x=4$ and $x=5$ .

\begin{aligned} A &= \int_4^5 (x^2 - 4x) \,dx \\ &= \left[ \frac{x^3}{3} - 2x^2 \right]_4^5 \\ &= \left( \frac{125}{3} - 2(25) \right) - \left( \frac{64}{3} - 2(16) \right) \\ &= \left( \frac{125}{3} - 50 \right) - \left( \frac{64}{3} - 32 \right) \\ &= \left( -\frac{25}{3} \right) - \left( -\frac{32}{3} \right) \\ &= \frac{7}{3} \text{ square units} \end{aligned}

Example 2

Find the area enclosed by the graph of $y = -x^2 + 5x - 6$ and the $x$ -axis.

First find the roots (where $y=0$ ):

\begin{aligned} -x^2 + 5x - 6 &= 0 \\ x^2 - 5x + 6 &= 0 \\ (x-2)(x-3) &= 0 \end{aligned}

The limits are $x=2$ to $x=3$ .

\begin{aligned} A &= \int_2^3 (-x^2 + 5x - 6) \,dx \\ &= \left[ -\frac{x^3}{3} + \frac{5x^2}{2} - 6x \right]_2^3 \\ &= \left( -9 + \frac{45}{2} - 18 \right) - \left( -\frac{8}{3} + 10 - 12 \right) \\ &= \left( -\frac{9}{2} \right) - \left( -\frac{14}{3} \right) \\ &= -\frac{27}{6} + \frac{28}{6} = \frac{1}{6} \text{ square units} \end{aligned}