Areas of Compound Regions (HSC SSCE Mathematics Advanced): Revision Notes

Worked Example: Two parabolas meeting at a point

Question: The curves $y = x^2$ and $y = (x - 2)^2$ intersect at $(1, 1)$. Find the area of the region bounded by these curves and the $x$-axis.

Solution:

First, sketch both curves on the same axes. You can verify that $(1, 1)$ lies on both curves by substitution.

The complete region sits above the $x$-axis, but we need to find the areas to the left and right of $x = 1$ separately.

For the left section (bounded by $y = x^2$):

$\int_0^1 x^2 \, dx = \left[\frac{x^3}{3}\right]_0^1$

$= \frac{1}{3}$

For the right section (bounded by $y = (x-2)^2$):

$\int_1^2 (x - 2)^2 \, dx = \left[\frac{(x - 2)^3}{3}\right]_1^2$

$= 0 - \left(-\frac{1}{3}\right)$

$= \frac{1}{3}$

Combining both sections:

Total area $= \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$ square units

Worked Example: Parabola and straight line

Question: Find the area between the curve $y = (x - 2)^2$ and the line $y = x$.

Part a: Find intersection points

Set the equations equal:

$(x - 2)^2 = x$

$x^2 - 4x + 4 = x$

$x^2 - 5x + 4 = 0$

$(x - 1)(x - 4) = 0$

Therefore $x = 1$ or $x = 4$

The curves intersect at (1, 1) and (4, 4).

Part b: Calculate the area

In the shaded region, the straight line sits above the parabola.

$\text{Area} = \int_1^4 (x - (x - 2)^2) \, dx$

$= \int_1^4 (x - (x^2 - 4x + 4)) \, dx$

$= \int_1^4 (-x^2 + 5x - 4) \, dx$

$= \left[-\frac{x^3}{3} + \frac{5x^2}{2} - 4x\right]_1^4$

$= \left(-21\frac{1}{3} + 40 - 16\right) - \left(-\frac{1}{3} + 2\frac{1}{2} - 4\right)$

$= 2\frac{2}{3} + 1\frac{5}{6}$

$= 4\frac{1}{2} \text{ square units}$

Worked Example: Region crossing the x-axis

Question: Find the area between the curves $y = x^2 - 4x + 2$ and $y = x - 2$.

Part a: Find intersection points

Set equations equal:

$x^2 - 4x + 2 = x - 2$

$x^2 - 5x + 4 = 0$

$(x - 1)(x - 4) = 0$

Therefore $x = 1$ or $x = 4$

The curves intersect at (1, -1) and (4, 2).

Part b: Calculate the area

The line remains above the parabola throughout the interval.

$\text{Area} = \int_1^4 ((x - 2) - (x^2 - 4x + 2)) \, dx$

$= \int_1^4 (-x^2 + 5x - 4) \, dx$

$= \left[-\frac{x^3}{3} + \frac{5x^2}{2} - 4x\right]_1^4$

$= \left(-21\frac{1}{3} + 40 - 16\right) - \left(-\frac{1}{3} + 2\frac{1}{2} - 4\right)$

$= 4\frac{1}{2} \text{ square units}$

Worked Example: Two parabolas that cross

Question: The curves $y = -x^2 + 4x - 4$ and $y = x^2 - 8x + 12$ meet at $(2, 0)$ and $(4, -4)$. Find the shaded area between them.

Solution:

In the left-hand region (from $x = 0$ to $x = 2$), the second curve sits above the first.

Left area:

$= \int_0^2 ((x^2 - 8x + 12) - (-x^2 + 4x - 4)) \, dx$

$= \int_0^2 (2x^2 - 12x + 16) \, dx$

$= \left[\frac{2x^3}{3} - 6x^2 + 16x\right]_0^2$

$= 5\frac{1}{3} - 24 + 32$

$= 13\frac{1}{3} \text{ square units}$

In the right-hand region (from $x = 2$ to $x = 4$), the first curve sits above the second.

Right area:

$= \int_2^4 ((-x^2 + 4x - 4) - (x^2 - 8x + 12)) \, dx$

$= \int_2^4 (-2x^2 + 12x - 16) \, dx$

$= \left[-\frac{2x^3}{3} + 6x^2 - 16x\right]_2^4$

$= \left(-42\frac{2}{3} + 96 - 64\right) - \left(-5\frac{1}{3} + 24 - 32\right)$

$= -10\frac{2}{3} + 13\frac{1}{3}$

$= 2\frac{2}{3} \text{ square units}$

Total area $= 13\frac{1}{3} + 2\frac{2}{3} = 16$ square units