91Ó°ÊÓ

To find the points of intersection, we need to set these two functions equal to each other: \(x = x^3\) To solve this equation, set the left side to 0 by subtracting the right side: \(0 = x^3 - x\) Next, factor out a common term, which is x: \(x(x^2 - 1) = 0\) Now we have a quadratic equation in the parentheses. We can further factor this equation: \(x(x - 1)(x + 1) = 0\) Now, we have found that the points of intersection occur when x = -1, x = 0, and x = 1. These correspond to the points (-1, -1), (0, 0), and (1, 1) respectively.

To find the area between the curves, we will integrate the difference of the functions between the points of intersection found in step 1. Since the functions intersect between x = -1 and x = 1, we will use these values as limits of integration. The difference of the functions is given by: \[y = x - x^3\] Now we can set up the integral to find the area: \[A = \int_{-1}^{1} (x - x^3) dx\] To integrate this function, we will find the antiderivative: \[A = \frac{1}{2}x^2 - \frac{1}{4}x^4\] Now we will evaluate this antiderivative at the limits of integration: \[A = \frac{1}{2}(1)^2 - \frac{1}{4}(1)^4 - \left(\frac{1}{2}(-1)^2 - \frac{1}{4}(-1)^4\right)\] \[A = \frac{1}{2} - \frac{1}{4} - \frac{1}{2} + \frac{1}{4}\] \[A = 0\] However, since the area is enclosed, there is a possibility that one curve is above the other in one side of the interval and vice versa. To have an accurate result, we must split the range into two intervals and find the area for each interval prior to adding them up. We will consider the intervals from -1 to 0 and 0 to 1. For the interval from -1 to 0, the curve of \(y = x^3\) is above \(y = x\). Therefore, the area for this interval is given by: \[A_1 = \int_{-1}^{0} (x^3 - x) dx\] \[A_1 = \frac{1}{4}x^4 - \frac{1}{2}x^2\] \[A_1 = \frac{1}{4}(0)^4 - \frac{1}{2}(0)^2 - \left(\frac{1}{4}(-1)^4 - \frac{1}{2}(-1)^2\right)\] \[A_1 = \frac{1}{4}\] For the interval from 0 to 1, the curve of \(y = x\) is above \(y = x^3\). Therefore, the area for this interval is given by: \[A_2 = \int_{0}^{1} (x - x^3) dx\] \[A_2 = \frac{1}{2}x^2 - \frac{1}{4}x^4\] \[A_2 = \frac{1}{2}(1)^2 - \frac{1}{4}(1)^4 - \left(\frac{1}{2}(0)^2 - \frac{1}{4}(0)^4\right)\] \[A_2 = \frac{1}{4}\] Now we can add up the areas of both intervals: \[A = A_1 + A_2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\] So, the area enclosed by the functions \(y = x\) and \(y = x^3\) is \(\frac{1}{2}\).