Find the average value of the following function over the given interval. Draw a graph of the function and indicate the average value. \[ f(x)=x(x-1) ;[3,5] \] The average value of the function is \( f=\square \).
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To find the average value of \( f(x) = x(x-1) \) over the interval \([3, 5]\), we first compute the definite integral of the function over the interval and then divide by the length of the interval. 1. Calculate the definite integral: \[ \int_{3}^{5} f(x) \, dx = \int_{3}^{5} x(x-1) \, dx = \int_{3}^{5} (x^2 - x) \, dx \] The antiderivative is: \[ \frac{x^3}{3} - \frac{x^2}{2} \] 2. Evaluate from 3 to 5: \[ \left[\frac{5^3}{3} - \frac{5^2}{2}\right] - \left[\frac{3^3}{3} - \frac{3^2}{2}\right] \] Calculating: \[ =\left[\frac{125}{3} - \frac{25}{2}\right] - \left[\frac{27}{3} - \frac{9}{2}\right] \] Finding a common denominator (6) for both parts: \[ =\left[\frac{250}{6} - \frac{75}{6}\right] - \left[\frac{54}{6} - \frac{27}{6}\right] \] \[ =\frac{175}{6} - \frac{27}{6} = \frac{148}{6} = \frac{74}{3} \] 3. Calculate the average value over the interval length, which is \(5 - 3 = 2\): \[ \text{Average value} = \frac{1}{2} \int_{3}^{5} f(x) \, dx = \frac{1}{2} \cdot \frac{74}{3} = \frac{74}{6} = \frac{37}{3} \] Therefore, the average value of the function \( f = \frac{37}{3} \) is approximately \(12.33\). Now for the graph, you would draw the parabolic curve of \( f(x) = x^2 - x \) which opens upwards. The points \( (3, f(3)) \) and \( (5, f(5)) \) would be plotted, and the average value \( y = \frac{37}{3} \) would be indicated as a horizontal line across the graph between these two points.
