Question

Draw the graph of the function \(f(x)=x(x-2)(x-4)=x^{3}-6 x^{2}+8 x\), and indicate the region \(\mathrm{P}^{+}\)defined by the inequalities \(0 \leq \mathrm{x} \leq 3\) and \(0 \leq y \leq f(x)\), and the region \(P^{-}\)defined by \(0 \leq x \leq 3\) and \(\mathrm{f}(\mathrm{x}) \leq \mathrm{y} \leq 0 .\) Let \(\mathrm{P}=\mathrm{P}^{+} \cup \mathrm{P}^{-}\), and suppose that \(\int_{0}^{2} \mathrm{f}(\mathrm{x}) \mathrm{dx}=4\) and \(\int_{0}^{3} \mathrm{f}(\mathrm{x}) \mathrm{dx}=2 \frac{1}{4} .\) Find area \(\left(\mathrm{P}^{+}\right)\), area \(\left(\mathrm{P}^{-}\right)\), and area \((\mathrm{P}) .\)

Short answer

Answer: The areas of the regions are: Area\(\left(P^{+}\right) = 4\), Area\(\left(P^{-}\right) = \frac{7}{4}\) , and Area\((P) = \frac{23}{4}\).

Step-by-step solution

Draw the graph of the function \(f(x)=x^3-6x^2+8x\)

The given function is a cubic function. The function can be written as \(f(x)=x(x-2)(x-4)\), which indicates that the function has roots at \(x=0, 2, 4\). This will help us in sketching the graph.

Find the intercepts and turning points

We know the x-intercepts are 0, 2, and 4, which are the points where the graph touches the x-axis. Now take the first derivative: \(f'(x) = 3x^2-12x+8\) To find the turning points, we have to solve for \(x\) when \(f'(x) = 0\). \(3x^2-12x+8=0\) This is a quadratic equation which can be solved using the quadratic formula or by factoring: \((3x - 4)(x - 2) = 0\) So the critical points are x=2 and x=4/3. We know that \(0 \leq x \leq 3\), so the only turning point that matters is at \(x=4/3\). Calculate \(f(4/3)= 4(4/3-2)(4/3-4)= -16/9\).

Find the areas of regions \(P^{+}\) and \(P^{-}\)

We are given that \(\int_0^2 f(x)dx = 4\). This indicates the area of region \(P^{+}\). We are also given that \(\int_0^3 f(x)dx = 2\frac{1}{4}\). To find the area of region \(P^{-}\), we first have to find the area of the region under the curve for \(x\) values between 2 and 3. To do this, we can calculate \(\int_2^3 f(x)dx\). It is important to remember that the area under the curve is equal to the absolute value of the integral. Let \(A\) be this value. So, \(4 + A = 2\frac{1}{4}\) This gives: \(A= -\frac{7}{4}\) Since the area cannot be negative, we take the absolute value of A. Therefore, the area of the region \(P^{-}\) is \(\frac{7}{4}\).

Find the area of the region \(P\)

To find the area of \(P=P^{+}\cup P^{-}\), we simply add the areas of regions \(P^{+}\) and \(P^{-}\). Area of \(P=P^{+}+P^{-}= 4+\frac{7}{4}=4\frac{7}{4}=\frac{23}{4}\). Therefore, the area of the region \(P\) is \(\frac{23}{4}\). So in conclusion: Area\(\left(P^{+}\right) = 4\), Area\(\left(P^{-}\right) = \frac{7}{4}\) , and Area\((P) = \frac{23}{4}\).