Finding Areas by Integration (HSC SSCE Mathematics Advanced): Revision Notes

Worked Example: Parabola Above the $x$-axis

Find the area of the region bounded by the curve $y = 4 - x^2$ and the $x$-axis.

Solution:

First, we find where the curve meets the $x$-axis by setting $y = 0$:

$4 - x^2 = 0$

$x^2 = 4$

$x = \pm 2$

So the $x$-intercepts are at $(2, 0)$ and $(-2, 0)$.

Since the region lies entirely above the $x$-axis, we evaluate:

$\int_{-2}^{2} (4 - x^2) \, dx = \left[ 4x - \frac{x^3}{3} \right]_{-2}^{2}$

$= \left( 8 - \frac{8}{3} \right) - \left( -8 + \frac{8}{3} \right)$

$= \frac{16}{3} - \left( -\frac{16}{3} \right)$

$= \frac{32}{3}$

$= 10\frac{2}{3}$

This integral is positive because the region lies entirely above the $x$-axis.

Therefore, the required area is $10\frac{2}{3}$ square units.

Worked Example: Parabola Below the $x$-axis

Find the area of the region bounded by the curve $y = x^2 - 1$ and the $x$-axis.

Solution:

The curve meets the $x$-axis when $y = 0$:

$x^2 - 1 = 0$

$x^2 = 1$

$x = \pm 1$

So the $x$-intercepts are at $(1, 0)$ and $(-1, 0)$.

The curve is a parabola opening upward with vertex at $(0, -1)$, so the region between $x = -1$ and $x = 1$ lies entirely below the $x$-axis.

We evaluate the integral:

$\int_{-1}^{1} (x^2 - 1) \, dx = \left[ \frac{x^3}{3} - x \right]_{-1}^{1}$

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

$= -\frac{2}{3} - \frac{2}{3}$

$= -\frac{4}{3}$

$= -1\frac{1}{3}$

This integral is negative because the region lies entirely below the $x$-axis.

Therefore, the required area is $1\frac{1}{3}$ square units.

Worked Example: Cubic Curve Crossing the $x$-axis

a) Sketch the cubic curve $y = x(x + 1)(x - 2)$, showing the $x$-intercepts.

b) Shade the region enclosed between the $x$-axis and the curve, and find its area.

c) Find $\int_{-1}^{2} x(x + 1)(x - 2) \, dx$ and explain why this integral does not represent the area of the region described in part b.

Solution:

a) The curve has $x$-intercepts at $x = -1$, $x = 0$, and $x = 2$.

The cubic curve crosses the $x$-axis at these three points, creating regions both above and below the axis.

b) To find the area, we first expand the cubic:

$y = x(x + 1)(x - 2)$

$= x(x^2 - x - 2)$

$= x^3 - x^2 - 2x$

For the region above the $x$-axis (between $x = -1$ and $x = 0$):

$\int_{-1}^{0} (x^3 - x^2 - 2x) \, dx = \left[ \frac{x^4}{4} - \frac{x^3}{3} - x^2 \right]_{-1}^{0}$

$= (0 - 0 - 0) - \left( \frac{1}{4} + \frac{1}{3} - 1 \right)$

$= 0 - \left( -\frac{5}{12} \right)$

$= \frac{5}{12}$

So the area above the $x$-axis is $\frac{5}{12}$ square units.

For the region below the $x$-axis (between $x = 0$ and $x = 2$):

$\int_{0}^{2} (x^3 - x^2 - 2x) \, dx = \left[ \frac{x^4}{4} - \frac{x^3}{3} - x^2 \right]_{0}^{2}$

$= \left( 4 - \frac{8}{3} - 4 \right) - (0 - 0 - 0)$

$= -\frac{8}{3}$

$= -2\frac{2}{3}$

So the area below the $x$-axis is $2\frac{2}{3}$ square units.

Adding these areas:

Total area $= \frac{5}{12} + 2\frac{2}{3} = \frac{5}{12} + \frac{32}{12} = \frac{37}{12} = 3\frac{1}{12}$ square units.

c) Evaluating the integral from $-1$ to $2$:

$\int_{-1}^{2} x(x + 1)(x - 2) \, dx = \left[ \frac{x^4}{4} - \frac{x^3}{3} - x^2 \right]_{-1}^{2}$

$= \left( 4 - \frac{8}{3} - 4 \right) - \left( \frac{1}{4} + \frac{1}{3} - 1 \right)$

$= -\frac{8}{3} + \frac{5}{12}$

$= -2\frac{2}{3} + \frac{5}{12}$

$= -2\frac{1}{4}$

This integral represents the difference between the two areas: $2\frac{2}{3} - \frac{5}{12} = 2\frac{1}{4}$. The result is negative because the area below the $x$-axis is larger than the area above. This is not the same as the total area we found in part b.

Worked Example: Odd Function

a) Show that $y = x^3 - x$ is an odd function.

b) Using part a, find the area between the curve $y = x^3 - x$ and the $x$-axis.

Solution:

a) Let $f(x) = x^3 - x$.

Then $f(-x) = (-x)^3 - (-x)$

$= -x^3 + x$

$= -(x^3 - x)$

$= -f(x)$

Since $f(-x) = -f(x)$, the function is odd.

b) First, we factorise to find the $x$-intercepts:

$y = x(x^2 - 1)$

$= x(x - 1)(x + 1)$

So the $x$-intercepts are $x = -1$, $x = 0$, and $x = 1$.

Because the function is odd, the two shaded regions have equal areas. This means we only need to calculate one area and double it.

We calculate the area of the region below the $x$-axis (between $x = 0$ and $x = 1$):

$\int_{0}^{1} (x^3 - x) \, dx = \left[ \frac{x^4}{4} - \frac{x^2}{2} \right]_{0}^{1}$

$= \left( \frac{1}{4} - \frac{1}{2} \right) - (0 - 0)$

$= -\frac{1}{4}$

So the area below the $x$-axis is $\frac{1}{4}$ square units.

Doubling this for both regions:

Total area $= 2 \times \frac{1}{4} =$ $\frac{1}{2}$ square units.

Worked Example: Triangular Region

a) Sketch the lines $y = x + 1$ and $y = 5$, and shade the region between these lines to the right of the $y$-axis.

b) Write the equation of the line so that $x$ is a function of $y$.

c) Use integration with respect to $y$ to find the area of this region.

d) Confirm the result using the triangle area formula.

Solution:

a) The line $y = x + 1$ has $y$-intercept at $(0, 1)$ and the horizontal line $y = 5$ meets it at $(4, 5)$.

b) Starting with $y = x + 1$, we rearrange to make $x$ the subject:

$x = y - 1$

c) The region extends from $y = 1$ to $y = 5$, and we integrate $x = y - 1$ with respect to $y$:

$\int_{1}^{5} (y - 1) \, dy = \left[ \frac{y^2}{2} - y \right]_{1}^{5}$

$= \left( \frac{25}{2} - 5 \right) - \left( \frac{1}{2} - 1 \right)$

$= \frac{15}{2} - \left( -\frac{1}{2} \right)$

$= \frac{15}{2} + \frac{1}{2}$

$= 8$

This integral is positive because the region lies to the right of the $y$-axis.

Therefore, the required area is 8 square units.

d) Using the triangle area formula:

$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$

$= \frac{1}{2} \times 4 \times 4$

$= 8 \text{ square units}$

This confirms our result.

Worked Example: Cubic Region with respect to $y$

The curve shown is the cubic $y = x^3$.

a) Write the equation of the cubic so that $x$ is a function of $y$.

b) Use integration with respect to $y$ to find the areas of the shaded regions to the right and left of the $y$-axis.

c) Find the total area of the two shaded regions.

Solution:

a) Starting with $y = x^3$, we rearrange:

$x^3 = y$

$x = y^{\frac{1}{3}} \text{ (taking the cube root)}$

b) For the region to the right of the $y$-axis (from $y = 0$ to $y = 8$):

$\int_{0}^{8} y^{\frac{1}{3}} \, dy = \left[ \frac{3}{4} y^{\frac{4}{3}} \right]_{0}^{8}$

$= \frac{3}{4} \times (8^{\frac{4}{3}} - 0)$

Now, $8^{\frac{4}{3}} = (8^{\frac{1}{3}})^4 = 2^4 = 16$

$= \frac{3}{4} \times 16$

$= 12$

So the area to the right is 12 square units.

For the region to the left of the $y$-axis (from $y = -1$ to $y = 0$):

$\int_{-1}^{0} y^{\frac{1}{3}} \, dy = \left[ \frac{3}{4} y^{\frac{4}{3}} \right]_{-1}^{0}$

$= \frac{3}{4} \times (0 - (-1)^{\frac{4}{3}})$

Now, $(-1)^{\frac{4}{3}} = ((-1)^{\frac{1}{3}})^4 = (-1)^4 = 1$

$= \frac{3}{4} \times (0 - 1)$

$= -\frac{3}{4}$

So the area to the left is $\frac{3}{4}$ square units.

c) Adding the two areas:

Total area $= 12 + \frac{3}{4} =$ $12\frac{3}{4}$ square units.