91Ó°ÊÓ

To find the x-intercepts of the function, set \(y = 0\) and solve for x. Therefore, we have: \( 0 = x^3 - 9x^2 + 23x - 15 \) We need to find the values of x where this equation holds.

In order to find the x-intercepts, we can factor this polynomial. Factoring the polynomial helps us break it into simpler terms from which we can find the roots. \( x^3 - 9x^2 + 23x - 15 = (x - 1)(x - 3)(x - 5) \) Now, we can easily find the x-intercepts of the function: \( x - 1 = 0 \Rightarrow x = 1 \) \( x - 3 = 0 \Rightarrow x = 3 \) \( x - 5 = 0 \Rightarrow x = 5 \) The x-intercepts of the function are: x = 1, x = 3, and x = 5.

Since the function is over the \(\mathrm{x}\)-axis between x=1 and x=3, and between x=3 and x=5, we need to calculate the area under the curve in these intervals. We can achieve this by integrating the given function in each interval and finding the difference: \( \int_{1}^{3} (x^3 - 9x^2 + 23x - 15)dx + \int_{3}^{5} (x^3 - 9x^2 + 23x - 15)dx\) Now, we will integrate the function concerning x: \( \int (x^3 - 9x^2 + 23x - 15)dx = \frac{1}{4}x^4 - 3x^3 + \frac{23}{2}x^2 - 15x + C \) Now, calculate each integral: \( \left[\frac{1}{4}(3)^4 - 3(3)^3 + \frac{23}{2}(3)^2 - 15(3)\right] - \left[\frac{1}{4}(1)^4 - 3(1)^3 + \frac{23}{2}(1)^2 - 15(1)\right] + \\ \left[\frac{1}{4}(5)^4 - 3(5)^3 + \frac{23}{2}(5)^2 - 15(5)\right] - \left[\frac{1}{4}(3)^4 - 3(3)^3 + \frac{23}{2}(3)^2 - 15(3)\right]\) Plugging the given x values into the integrated function and evaluating the expression gives: \[ \left[-72\right] - \left[-34\right] + \left[-225\right] - \left[-72\right] = 38 + 153 = 191 \]

Now that we have calculated the value of the integral, we can interpret its meaning. The total area under the curve and above the \(\mathrm{x}\)-axis for the intervals we have considered (x = 1 to x = 3 and x = 3 to x = 5) is 191 square units.