Problem 125
Medium Special Techniques

Find the average value of the function on the given

Find the average value of the function on the given interval: $$ y = \sin^4 x, \quad [0, \pi] $$

Explanation of Method

The average value of a function $f$ on an interval $[a,b]$ is computed from the Mean Value Theorem for integrals formula, $\bar f=\frac{1}{b-a}\int_a^b f(x)\,dx$ (interpreted as a limit when the interval is unbounded), reducing the problem to evaluating a definite or improper integral.

Definition of average value

The average value of a function $f(x)$ on the interval $[a, b]$ is:

$$\bar{f} = \frac{1}{b-a} \int_a^b f(x) \, dx$$

Step 1: Set up the formula

For $f(x) = \sin^4 x$ on $[0, \pi]$:

$$\bar{y} = \frac{1}{\pi - 0} \int_0^\pi \sin^4 x \, dx = \frac{1}{\pi} \int_0^\pi \sin^4 x \, dx$$

Step 2: Simplify $\sin^4 x$ using trigonometric identities

We use the power-reduction formula. First, recall:

$$\sin^2 x = \frac{1 - \cos 2x}{2}$$

Therefore:

$$\sin^4 x = (\sin^2 x)^2 = \left(\frac{1 - \cos 2x}{2}\right)^2 = \frac{(1 - \cos 2x)^2}{4}$$

Expanding:

$$(1 - \cos 2x)^2 = 1 - 2\cos 2x + \cos^2 2x$$

Now use the power-reduction formula again for $\cos^2 2x$:

$$\cos^2 2x = \frac{1 + \cos 4x}{2}$$

Therefore:

$$\sin^4 x = \frac{1 - 2\cos 2x + \frac{1 + \cos 4x}{2}}{4}$$

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

$$= \frac{1}{4}\left(\frac{3}{2} - 2\cos 2x + \frac{\cos 4x}{2}\right)$$

$$= \frac{3}{8} - \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x$$

Step 3: Integrate term by term

$$\int_0^\pi \sin^4 x \, dx = \int_0^\pi \left(\frac{3}{8} - \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x\right) dx$$

$$= \left[\frac{3x}{8} - \frac{\sin 2x}{4} + \frac{\sin 4x}{32}\right]_0^\pi$$

At $x = \pi$:

$$\frac{3\pi}{8} - \frac{\sin 2\pi}{4} + \frac{\sin 4\pi}{32} = \frac{3\pi}{8} - 0 + 0 = \frac{3\pi}{8}$$

At $x = 0$:

$$\frac{0}{8} - \frac{\sin 0}{4} + \frac{\sin 0}{32} = 0 - 0 + 0 = 0$$

Therefore:

$$\int_0^\pi \sin^4 x \, dx = \frac{3\pi}{8} - 0 = \frac{3\pi}{8}$$

Step 4: Calculate the average value

$$\bar{y} = \frac{1}{\pi} \int_0^\pi \sin^4 x \, dx = \frac{1}{\pi} \cdot \frac{3\pi}{8} = \frac{3}{8}$$

$$\boxed{\bar{y} = \frac{3}{8}}$$

Step 5: Numerical value

$$\bar{y} = \frac{3}{8} = 0.375$$

$$\boxed{\bar{y} = 0.375}$$

Verification using symmetry

Note that $\sin^4 x$ has the following properties on $[0, \pi]$:

The average of $\frac{3}{8} = 0.375$ is reasonable as it lies between the minimum (0) and maximum (1).

Alternative: Wallis integral formula

The integral $\int_0^\pi \sin^n x \, dx$ can be computed using Wallis formulas.

For even $n = 2m$:

$$\int_0^\pi \sin^{2m} x \, dx = \frac{(2m-1)!!}{(2m)!!} \cdot \pi$$

For $n = 4$ (so $m = 2$):

$$\int_0^\pi \sin^4 x \, dx = \frac{3!!}{4!!} \cdot \pi = \frac{3 \cdot 1}{4 \cdot 2} \cdot \pi = \frac{3}{8}\pi$$

This confirms our result. ✓

Summary

Average value of $y = \sin^4 x$ on $[0, \pi]$:

$$\boxed{\bar{y} = \frac{3}{8} = 0.375}$$

Key steps:

  1. Use formula: $\bar{f} = \frac{1}{b-a} \int_a^b f(x) \, dx$
  2. Apply power-reduction: $\sin^4 x = \frac{3}{8} - \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x$
  3. Integrate: $\int_0^\pi \sin^4 x \, dx = \frac{3\pi}{8}$
  4. Calculate average: $\bar{y} = \frac{1}{\pi} \cdot \frac{3\pi}{8} = \frac{3}{8}$

Key insights:


Result

$$\boxed{\bar{y} = \frac{3}{8}}$$

$$\boxed{\bar{y} = 0.375}$$

Average value of $y = \sin^4 x$ on $[0, \pi]$:

$$\boxed{\bar{y} = \frac{3}{8} = 0.375}$$