To solve this problem, we need to:

1. Find the average value of the function $f(x) = 1 - \frac{x^2}{a^2}$ on the interval $[0, a]$.
2. Determine the $x$-values where $f(x)$ equals this average value.

Step 1: Calculate the Average Value of $f(x)$

The formula for the average value of a function $f(x)$ over the interval $[0, a]$ is:

$\text{Average value} = \frac{1}{a - 0} \int_0^a f(x) \, dx = \frac{1}{a} \int_0^a \left( 1 - \frac{x^2}{a^2} \right) \, dx$

Expanding this integral:

$\frac{1}{a} \int_0^a \left( 1 - \frac{x^2}{a^2} \right) \, dx = \frac{1}{a} \left( \int_0^a 1 \, dx - \int_0^a \frac{x^2}{a^2} \, dx \right)$

Calculating each part:

• Integral of 1: $\int_0^a 1 \, dx = x \Big|_0^a = a$
• Integral of $\frac{x^2}{a^2}$:

$\int_0^a \frac{x^2}{a^2} \, dx = \frac{1}{a^2} \int_0^a x^2 \, dx = \frac{1}{a^2} \cdot \frac{x^3}{3} \Big|_0^a = \frac{1}{a^2} \cdot \frac{a^3}{3} = \frac{a}{3}$

Putting it all together:

$\text{Average value} = \frac{1}{a} \left( a - \frac{a}{3} \right) = \frac{1}{a} \cdot \frac{2a}{3} = \frac{2}{3}$

So, the average value of $f(x)$ on $[0, a]$ is $\frac{2}{3}$.

Step 2: Set $f(x) = \frac{2}{3}$ and Solve for $x$

We want to find $x$-values such that:

$1 - \frac{x^2}{a^2} = \frac{2}{3}$

Rearrange to isolate $x^2$:

$\frac{x^2}{a^2} = 1 - \frac{2}{3} = \frac{1}{3}$

Multiply both sides by $a^2$:

$x^2 = \frac{a^2}{3}$

Take the square root of both sides:

$x = \pm \frac{a}{\sqrt{3}} = \pm \frac{a \sqrt{3}}{3}$

Final Answer

The points at which $f(x)$ equals its average value on the interval $[0, a]$ are:

$x = \frac{a \sqrt{3}}{3} \text{ and } x = -\frac{a \sqrt{3}}{3}$

Since we are considering only the interval $[0, a]$, the relevant solution is:

$x = \frac{a \sqrt{3}}{3}$

Let me know if you want further details or have questions about each step!

Related Questions to Explore

1. How do you compute the average value of a more complex function over a given interval?
2. What changes if we consider a different interval, like $[-a, a]$, for the same function?
3. How does changing the limits of integration impact the average value calculation?
4. What happens if we set different values for $a$ in the function?
5. Can the average value of a function ever exceed the maximum value of the function itself?

Quick Tip

Remember that the average value of a function over an interval gives the "height" at which the area under the function would be constant across that interval.