Applications of Integration (HSC SSCE Mathematics Advanced): Revision Notes

Worked Example: Area under $y = \cos\frac{1}{2}x$

Let's find the area between $y = \cos\frac{1}{2}x$ and the $x$-axis for $0 \leq x \leq 4\pi$.

Step 1: Sketch the curve

First, we identify the properties of the function:

• Amplitude = $1$
• Period = $\frac{2\pi}{\frac{1}{2}} = 4\pi$
• The x-intercepts in the interval occur at $x = \pi$ and $x = 3\pi$

Step 2: Identify the intervals

The curve crosses the $x$-axis at $x = \pi$ and $x = 3\pi$, dividing our region into three parts:

• $[0, \pi]$ - curve is above the x-axis
• $[\pi, 3\pi]$ - curve is below the $x$-axis
• $[3\pi, 4\pi]$ - curve is above the x-axis

Step 3: Integrate over each interval

For the first interval $[0, \pi]$:

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

$= 2\sin\frac{\pi}{2} - 2\sin 0$

$= 2 - 0$

$= 2$

This is positive because the curve is above the $x$-axis.

For the second interval $[\pi, 3\pi]$:

$\int_{\pi}^{3\pi} \cos\frac{1}{2}x \, dx = \left[2\sin\frac{1}{2}x\right]_{\pi}^{3\pi}$

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

$= -2 - 2$

$= -4$

This is negative because the curve is below the $x$-axis.

For the third interval $[3\pi, 4\pi]$:

$\int_{3\pi}^{4\pi} \cos\frac{1}{2}x \, dx = \left[2\sin\frac{1}{2}x\right]_{3\pi}^{4\pi}$

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

$= 0 - (-2)$

$= 2$

This is positive because the curve is above the $x$-axis.

Step 4: Calculate total area

Total area = $2 + 4 + 2 = 8$ square units

Worked Example: Area between $y = \sin x$ and $y = \sin 2x$

Let's find the area between these curves in the interval $\left[0, \frac{\pi}{3}\right]$.

Step 1: Verify the intersection point

The curves intersect at $x = \frac{\pi}{3}$ because:

$\sin\frac{\pi}{3} = \sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2}$

Step 2: Sketch the curves

In the interval $0 \leq x \leq \frac{\pi}{3}$, the curve $y = \sin 2x$ is always above $y = \sin x$.

Step 3: Set up and evaluate the integral

$\text{Area} = \int_0^{\frac{\pi}{3}} (\sin 2x - \sin x) \, dx$

$= \left[-\frac{1}{2}\cos 2x + \cos x\right]_0^{\frac{\pi}{3}}$

$= \left(-\frac{1}{2}\cos\frac{2\pi}{3} + \cos\frac{\pi}{3}\right) - \left(-\frac{1}{2}\cos 0 + \cos 0\right)$

Using $\cos 0 = 1$, $\cos\frac{\pi}{3} = \frac{1}{2}$, and $\cos\frac{2\pi}{3} = -\frac{1}{2}$:

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

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

Worked Example: Area between $y = \sin x$ and $y = \cos x$

Let's find the area between these curves in the interval $[0, 2\pi]$.

Step 1: Find intersection points

Setting $\sin x = \cos x$:

$\tan x = 1$

$x = \frac{\pi}{4} \text{ or } \frac{5\pi}{4}$

The curves intersect at $\left(\frac{\pi}{4}, \frac{\sqrt{2}}{2}\right)$ and $\left(\frac{5\pi}{4}, -\frac{\sqrt{2}}{2}\right)$.

Step 2: Sketch the curves

Between $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$, the curve $y = \sin x$ is above $y = \cos x$.

Step 3: Calculate the area

$\text{Area} = \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} (\sin x - \cos x) \, dx$

$= \left[-\cos x - \sin x\right]_{\frac{\pi}{4}}^{\frac{5\pi}{4}}$

$= -\cos\frac{5\pi}{4} - \sin\frac{5\pi}{4} + \cos\frac{\pi}{4} + \sin\frac{\pi}{4}$

$= -\left(-\frac{\sqrt{2}}{2}\right) - \left(-\frac{\sqrt{2}}{2}\right) + \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}$

$= 2\sqrt{2} \text{ square units}$