91Ó°ÊÓ

To find the antiderivative of the piecewise function, f(x), we need to derive the antiderivatives of the individual functions in the given interval: 1. For \(0 \leq x \leq 1\): \(F_1(x) = \int x^2 dx = \frac{1}{3}x^3 + C_1\) 2. For \(1 < x \leq 2\): \(F_2(x) = \int (2-x) dx = 2x - \frac{1}{2}x^2 + C_2\) Now, we have to adjust the constants of integration, C1 and C2, to make the antiderivative continuous on the whole interval [0, 2]. Since \(f(x)\) is defined on the interval [0, 1] and f(1) = 1, we can find the value of C1 by plugging x = 1 into \(F_1(x)\): \(F_1(1) = \frac{1}{3} + C_1 = 1 \Rightarrow C_1 = \frac{2}{3}\) Now, let's plug x = 1 into \(F_2(x)\): \(F_2(1) = 2 - \frac{1}{2} + C_2 = F_1(1) \Rightarrow C_2 = -\frac{1}{2}\) So, the antiderivative of f(x) on the whole interval [0, 2] is: $F(x) = \left\\{\begin{array}{ll}\frac{1}{3}x^3 + \frac{2}{3} & \text { for } 0 \leq x \leq 1 \\\ 2x - \frac{1}{2}x^2 -\frac{1}{2} & \text { for } 1<x \leq 2\end{array}\right.$ Now, we can calculate the integral \(\int_{0}^{2} f(x) dx = F(2) - F(0)\): \(= \left(2(2)-\frac{1}{2}(2)^2-\frac{1}{2}\right) - \left(\frac{1}{3}(0)^3 + \frac{2}{3}\right) = 2-\frac{1}{2}=-\frac{1}{2}\) So, the integral of f(x) is equal to -1/2.

In this approach, we will calculate the integral over each subinterval separately and then add them together: 1. For the interval [0, 1]: \(\int_{0}^{1} x^2 dx = \left[\frac{1}{3}x^3\right]_0^1 = \frac{1}{3}(1)^3 - \frac{1}{3}(0)^3 = \frac{1}{3}\) 2. For the interval [1, 2]: \(\int_{1}^{2} (2-x) dx = \left[2x-\frac{1}{2}x^2\right]_1^2 = \left(2(2) -\frac{1}{2}(2)^2\right) - \left(2(1)-\frac{1}{2}(1)^2\right) = - (\frac{1}{2})\) Now we add the integrals: \(\int_{0}^{2} f(x) dx = \int_{0}^{1} x^2 dx + \int_{1}^{2} (2-x) dx = \frac{1}{3} - \frac{1}{2} = -\frac{1}{6}\) So, the integral of f(x) is equal to -1/6. Note: The correct answer is -1/6. There was an error in the first approach. Please disregard the result obtained in the first approach.