Question

What are the values of a such that the average value of f(x) = 1 2x − x 2 on [0, a] is equal to 1?

Answer 1

The average value of the given function is $f_{avg} = \frac{-2a^{2} + 6a + 6 }{6}$.

According to the statement

we have given that the function f(x) and we have to find the average value of that function.

So, For this purpose, we know that the

The given function f(x) is

$f(x) = -x^{2} + 2x +1$

And now integrate this function with the limit 0 to a then

$f_{avg} = \frac{1}{b - a} \int\limits^a_0 {f(x)} \, dx = -x^{2} + 2x +1$

Now integrate this then

$f_{avg} = \frac{1}{a} \int\limits^a_0 {-x^{2} + 2x +1} \, dx$

Then the value becomes according to the integration rules is:

$f_{avg} = \frac{1}{a} \int\limits^a_0 {-\frac{x^{3} }{3} + \frac{2x^{2} }{2} +x} \,$

Now put the limits then answer will become as output is:

$f_{avg} = \frac{1}{a} [ {-\frac{a^{3} }{3} + \frac{2a^{2} }{2} +a} \,]$

Now solve this equation then

$f_{avg} = [ {-\frac{a^{2} }{3} + \frac{2a }{2} +1} \,]$

Now

$f_{avg} = \frac{-2a^{2} + 6a + 6 }{6}$

This is the value which represent the average of the given function in the statement.

So, The average value of the given function is $f_{avg} = \frac{-2a^{2} + 6a + 6 }{6}$.

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