To compute the average value of a function $f(x) = x \sin(4x)$ on the interval $[0, \frac{\pi}{2}]$, we use the formula for the average value of a continuous function over an interval $[a, b]$:

$\text{Average value of } f(x) = \frac{1}{b - a} \int_a^b f(x) \, dx$

Here, $a = 0$ and $b = \frac{\pi}{2}$, so the formula becomes:

$\text{Average value of } f(x) = \frac{1}{\frac{\pi}{2} - 0} \int_0^{\frac{\pi}{2}} x \sin(4x) \, dx$

This simplifies to:

$\text{Average value of } f(x) = \frac{2}{\pi} \int_0^{\frac{\pi}{2}} x \sin(4x) \, dx$

Step 1: Compute the integral $\int_0^{\frac{\pi}{2}} x \sin(4x) \, dx$

We can solve this integral using integration by parts. Let:

• $u = x$, so $du = dx$,
• $dv = \sin(4x) \, dx$, so $v = -\frac{1}{4} \cos(4x)$.

Using the integration by parts formula $\int u \, dv = uv - \int v \, du$, we get:

$\int x \sin(4x) \, dx = -\frac{x}{4} \cos(4x) + \frac{1}{4} \int \cos(4x) \, dx$

Now, compute the second integral:

$\int \cos(4x) \, dx = \frac{1}{4} \sin(4x)$

Thus:

$\int x \sin(4x) \, dx = -\frac{x}{4} \cos(4x) + \frac{1}{16} \sin(4x)$

Step 2: Evaluate the definite integral from 0 to $\frac{\pi}{2}$

Now, we evaluate the integral:

$\int_0^{\frac{\pi}{2}} x \sin(4x) \, dx = \left[ -\frac{x}{4} \cos(4x) + \frac{1}{16} \sin(4x) \right]_0^{\frac{\pi}{2}}$

At $x = \frac{\pi}{2}$:

• $\cos(4 \times \frac{\pi}{2}) = \cos(2\pi) = 1$,
• $\sin(4 \times \frac{\pi}{2}) = \sin(2\pi) = 0$.

So, we get:

$-\frac{\frac{\pi}{2}}{4} \cdot 1 + \frac{1}{16} \cdot 0 = -\frac{\pi}{8}$

At $x = 0$:

• $\cos(4 \times 0) = \cos(0) = 1$,
• $\sin(4 \times 0) = \sin(0) = 0$.

Thus:

$-\frac{0}{4} \cdot 1 + \frac{1}{16} \cdot 0 = 0$

So, the definite integral evaluates to:

$\int_0^{\frac{\pi}{2}} x \sin(4x) \, dx = -\frac{\pi}{8} - 0 = -\frac{\pi}{8}$

Step 3: Compute the average value

Now, multiply by the constant factor $\frac{2}{\pi}$:

$\text{Average value of } f(x) = \frac{2}{\pi} \times \left( -\frac{\pi}{8} \right) = -\frac{1}{4}$

Thus, the average value of $f(x) = x \sin(4x)$ on the interval $[0, \frac{\pi}{2}]$ is:

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

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Would you like further details or explanations on any part of this solution?

Here are some related questions for further exploration:

1. How do you apply integration by parts for different types of integrals?
2. What is the physical interpretation of the average value of a function?
3. Can you solve integrals involving products of trigonometric and polynomial functions using other methods?
4. How does changing the limits of integration affect the average value?
5. What are some common applications of the average value of a function in physics or engineering?

Tip: Remember to always check the behavior of trigonometric functions at the boundaries of your integration interval—they often simplify nicely at multiples of $\pi$ or $2\pi$.