Question - Solving Absolute Value Integral
Solution:
To solve the integral $$\int_{-1}^{2} |x - x^2|\, dx$$, we need to consider when the expression inside the absolute value, $$x - x^2$$, is positive or negative.First, let's find the points where $$x - x^2 = 0$$:$$x - x^2 = x(1 - x) = 0$$This gives us two points $$x = 0$$ and $$x = 1$$.Now we divide the interval $$[-1, 2]$$ based on these points and analyze the sign of the expression $$x - x^2$$ in each subinterval:1. For $$x \in [-1, 0]$$, the expression is always negative since both $$x$$ and $$x^2$$ are non-positive and $$x^2$$ is greater in magnitude than $$x$$.2. For $$x \in [0, 1]$$, the expression is positive since $$x$$ is greater than or equal to $$x^2$$.3. For $$x \in [1, 2]$$, the expression is negative since $$x^2$$ is greater than $$x$$.Thus, we need to change the sign of the expression in the intervals where it is negative, which gives us:$$\int_{-1}^{0} -(x - x^2)\, dx + \int_{0}^{1} (x - x^2)\, dx + \int_{1}^{2} -(x - x^2)\, dx$$Now we can integrate each term separately:1. $$\int_{-1}^{0} -(x - x^2)\, dx = -\frac{1}{2} x^2 + \frac{1}{3}x^3\Big|_{-1}^{0}$$2. $$\int_{0}^{1} (x - x^2)\, dx = \frac{1}{2} x^2 - \frac{1}{3}x^3\Big|_{0}^{1}$$3. $$\int_{1}^{2} -(x - x^2)\, dx = -\frac{1}{2} x^2 + \frac{1}{3}x^3\Big|_{1}^{2}$$Now, evaluate each integral:1. $$-\frac{1}{2} (0)^2 + \frac{1}{3}(0)^3 - (-\frac{1}{2} (-1)^2 + \frac{1}{3}(-1)^3) = 0 - (-\frac{1}{2} + \frac{1}{3}) = \frac{1}{3} - \frac{1}{2} = -\frac{1}{6}$$2. $$\frac{1}{2} (1)^2 - \frac{1}{3}(1)^3 - (\frac{1}{2} (0)^2 - \frac{1}{3}(0)^3) = \frac{1}{2} - \frac{1}{3} - 0 = \frac{1}{6}$$3. $$-\frac{1}{2} (2)^2 + \frac{1}{3}(2)^3 - (-\frac{1}{2} (1)^2 + \frac{1}{3}(1)^3) = -2 + \frac{8}{3} - (-\frac{1}{2} + \frac{1}{3}) = -2 + \frac{8}{3} + \frac{1}{6} = -2 + \frac{8}{3} + \frac{1}{6}$$To add $$-2 + \frac{8}{3} + \frac{1}{6}$$, first find a common denominator, which is 6:$$-2 + \frac{8 \cdot 2}{3 \cdot 2} + \frac{1}{6} = -2 + \frac{16}{6} + \frac{1}{6} = -2 + \frac{17}{6}$$$$-2$$ is equivalent to $$- \frac{12}{6}$$, so we have:$$-\frac{12}{6} + \frac{17}{6} = \frac{5}{6}$$Adding the results of all the terms:$$-\frac{1}{6} + \frac{1}{6} + \frac{5}{6} = \frac{5}{6}$$So, the value of the integral $$\int_{-1}^{2} |x - x^2|\, dx$$ is $$\frac{5}{6}$$.
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