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

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 \[ \boxed{\frac{94}{3}}. \]