To find the average value $h_{\text{ave}}$ of the function $h(x) = 7 \cos^4(x) \sin(x)$ over the interval $[0, \pi]$, we use the formula for the average value of a function:

$h_{\text{ave}} = \frac{1}{b-a} \int_a^b h(x) \, dx$

Step 1: Define the integral and interval

Here:

• $h(x) = 7 \cos^4(x) \sin(x)$
• Interval: $[0, \pi]$
• $a = 0$, $b = \pi$

$h_{\text{ave}} = \frac{1}{\pi - 0} \int_0^\pi 7 \cos^4(x) \sin(x) \, dx$

Step 2: Simplify the integral

Factor out the constant $7$ and simplify: $h_{\text{ave}} = \frac{7}{\pi} \int_0^\pi \cos^4(x) \sin(x) \, dx$

Step 3: Use substitution

Let $u = \cos(x)$, so that $du = -\sin(x) dx$. The limits of integration change as follows:

• When $x = 0$, $u = \cos(0) = 1$
• When $x = \pi$, $u = \cos(\pi) = -1$

The integral becomes: $\int_0^\pi \cos^4(x) \sin(x) \, dx = -\int_1^{-1} u^4 \, du$

Reversing the limits of integration to simplify: $\int_1^{-1} u^4 \, du = \int_{-1}^1 u^4 \, du$

Step 4: Evaluate the integral

The integral of $u^4$ is: $\int u^4 \, du = \frac{u^5}{5} + C$

Apply the limits: $\int_{-1}^1 u^4 \, du = \left[ \frac{u^5}{5} \right]_{-1}^1 = \frac{(1)^5}{5} - \frac{(-1)^5}{5}$

Simplify: $\int_{-1}^1 u^4 \, du = \frac{1}{5} - \left(-\frac{1}{5}\right) = \frac{2}{5}$

Step 5: Calculate $h_{\text{ave}}$

Substitute the result of the integral back into the formula for $h_{\text{ave}}$: $h_{\text{ave}} = \frac{7}{\pi} \cdot \frac{2}{5} = \frac{14}{5\pi}$

Final Answer:

$h_{\text{ave}} = \frac{14}{5\pi}$

Would you like further details or explanation?

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Related Questions:

1. How is the substitution method applied to integrals involving trigonometric functions?
2. What are common techniques to handle even powers of trigonometric functions?
3. Can this problem be solved without substitution? If so, how?
4. How does the average value of a function relate to its integral over an interval?
5. How would the solution change if the interval were $[0, \frac{\pi}{2}]$?

Tip:

For definite integrals involving even powers of trigonometric functions, substitution often simplifies the computation significantly.