Area under a curve

Signed area versus actual area

• The definite integral gives signed area:
  • Above the $x$-axis — positive
  • Below the $x$-axis — negative
• "Area" always means a positive quantity. If part of the curve dips below the axis, the integral under-reports the area, or even cancels it to zero.

The method that always works

• Sketch the curve over the interval — this is the step that prevents every error on this page.

• Find the $x$-intercepts inside the interval by solving $f(x) = 0$. These are where the sign changes.

• Split the integral at each intercept.

• Evaluate each piece separately.

• Take the absolute value of any negative piece.

• Add the magnitudes.

• Worked through — the area between $y = x^2 - 4$ and the $x$-axis from $x = 0$ to $x = 3$:

  • Intercepts: $x = \pm 2$, so $x = 2$ is inside the interval
  • From 0 to 2 the curve is below the axis; from 2 to 3 it is above
  • $$\displaystyle\int_0^2\left(x^2-4\right)dx = \left[\frac{x^3}{3}-4x\right]_0^2 = \frac{8}{3}-8 = -\frac{16}{3}$$
  • $$\displaystyle\int_2^3\left(x^2-4\right)dx = \left(9-12\right)-\left(\frac{8}{3}-8\right) = -3 + \frac{16}{3} = \frac{7}{3}$$
  • Area $= \left|-\frac{16}{3}\right| + \frac{7}{3} = \frac{23}{3}$ square units
  • The integral over the whole interval would have been $-3$ — a completely different number.

When splitting is not needed

• If the curve stays entirely above the axis on the interval, the integral is the area and no splitting is required.
• If it stays entirely below, the area is the magnitude of the integral.
• Check by sketching, not by assuming.

Area between a curve and the $y$-axis

• Some questions ask for the area between a curve and the vertical axis. Then you integrate with respect to $y$:

$$A = \int_c^d x\,dy$$

• Rearrange the equation to make $x$ the subject, and use $y$-limits.
• For $y = x^2$ with $x \geq 0$, the area between the curve and the $y$-axis from $y = 0$ to $y = 4$ is $\displaystyle\int_0^4 \sqrt{y}\,dy = \left[\frac{2}{3}y^{3/2}\right]_0^4 = \frac{16}{3}$.

Units

• Area from an integral is in square units of whatever the axes measure.
• If $x$ is in seconds and $y$ is in metres per second, the "area" is a distance in metres — the units multiply.
• Always interpret the units in context when the question is applied rather than purely geometric.

Curves with an unknown boundary

• A common Merit/Excellence turn: "the region enclosed between the curve and the axis has area 36; find $a$".
  • Set up the integral with $a$ as a limit.
  • Evaluate in terms of $a$.
  • Set equal to the given area and solve.
  • Check the solution makes sense — a boundary outside the region is invalid.