To solve for the value(s) of $a$ such that the average value of the function $f(x) = x^2 - 2ax + 1$ over the interval $[0, a]$ is 0, we need to follow these steps:

Step 1: Determine the formula for the average value of a function

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

$\text{Average value} = \frac{1}{c-b} \int_b^c f(x) \, dx$

Step 2: Apply the formula to the given function

Here, $f(x) = x^2 - 2ax + 1$, and the interval is $[0, a]$. So, the average value of $f(x)$ over $[0, a]$ is:

$\frac{1}{a-0} \int_0^a (x^2 - 2ax + 1) \, dx = \frac{1}{a} \int_0^a (x^2 - 2ax + 1) \, dx$

Step 3: Compute the integral

We need to compute the integral $\int_0^a (x^2 - 2ax + 1) \, dx$. This can be done by integrating each term separately:

$\int_0^a x^2 \, dx = \left[\frac{x^3}{3}\right]_0^a = \frac{a^3}{3}$

$\int_0^a -2ax \, dx = -2a\left[\frac{x^2}{2}\right]_0^a = -2a\left(\frac{a^2}{2}\right) = -a^3$

$\int_0^a 1 \, dx = \left[x\right]_0^a = a$

So, the integral becomes:

$\int_0^a (x^2 - 2ax + 1) \, dx = \frac{a^3}{3} - a^3 + a = \frac{a^3}{3} - \frac{3a^3}{3} + a = -\frac{2a^3}{3} + a$

Step 4: Find the average value and set it equal to 0

The average value is:

$\frac{1}{a} \left(-\frac{2a^3}{3} + a \right) = -\frac{2a^2}{3} + 1$

We set this equal to 0 to find $a$:

$-\frac{2a^2}{3} + 1 = 0$

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

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

$a = \sqrt{\frac{3}{2}} = \frac{\sqrt{6}}{2}$

Final Answer:

The value of $a$ that satisfies the condition is $a = \frac{\sqrt{6}}{2}$.

Would you like more details, or do you have any questions?

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Here are 5 related questions to expand your understanding:

1. How do you calculate the average value of a different quadratic function on an interval?
2. What is the geometric interpretation of finding the average value of a function on an interval?
3. Can the method used be applied to functions that are not quadratic? If so, how?
4. What happens if the function is shifted vertically? How would the calculation change?
5. What if $a$ were a negative constant? How would that affect the solution?

Tip: When solving problems involving the average value of a function, ensure that the interval and function are correctly set up to simplify calculations.