$$ 1 \leftarrow $$ Find the average value of the following function over the given interval. Dri $$ f(x)=x(x-1) ;[4,8] $$

Question

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

$$
A = \frac{1}{b-a}\int_a^b f(x)\,dx.
$$

For the function

$$
f(x) = x(x-1) = x^2 - x
$$

and the interval $$[4, 8]$$, we have $$ a = 4 $$ and $$ b = 8 $$.

1. **Setup the Integral**

$$
   A = \frac{1}{8-4}\int_4^8 (x^2-x)\,dx = \frac{1}{4}\int_4^8 (x^2-x)\,dx.
   $$

2. **Compute the Antiderivative**

Find the antiderivative of $$ f(x) $$:

$$
   \int (x^2-x)\,dx = \frac{x^3}{3}-\frac{x^2}{2}+C.
   $$

3. **Evaluate the Definite Integral**

Evaluate the antiderivative from $$ x=4 $$ to $$ x=8 $$:

$$
   \left[\frac{x^3}{3}-\frac{x^2}{2}\right]_4^8 = \left(\frac{8^3}{3}-\frac{8^2}{2}\right) - \left(\frac{4^3}{3}-\frac{4^2}{2}\right).
   $$

Compute each part:

- For $$ x=8 $$:

$$
     \frac{8^3}{3}-\frac{8^2}{2} = \frac{512}{3} - \frac{64}{2} = \frac{512}{3} - 32.
     $$

- For $$ x=4 $$:

$$
     \frac{4^3}{3}-\frac{4^2}{2} = \frac{64}{3} - \frac{16}{2} = \frac{64}{3} - 8.
     $$

Now, subtract:

$$
   \left(\frac{512}{3}-32\right) - \left(\frac{64}{3}-8\right) = \frac{512-64}{3} - (32-8) = \frac{448}{3} - 24.
   $$

Express $$ 24 $$ as a fraction with denominator 3:

$$
   24 = \frac{72}{3},
   $$
   
   so,

$$
   \frac{448}{3} - \frac{72}{3} = \frac{376}{3}.
   $$

4. **Determine the Average Value**

Substitute the result of the integral into the formula for $$ A $$:

$$
   A = \frac{1}{4} \times \frac{376}{3} = \frac{376}{12}.
   $$

Simplify the fraction by dividing both numerator and denominator by 4:

$$
   \frac{376 \div 4}{12 \div 4} = \frac{94}{3}.
   $$

The average value of the function $$ f(x) = x(x-1) $$ on the interval $$[4, 8]$$ is

$$
\boxed{\frac{94}{3}}.
$$
The average value of the function $$ f(x) = x(x-1) $$ over the interval $$[4, 8]$$ is $$ \frac{94}{3} $$.

Find the average value of the following function over the given interval. Dri