Find the area of the region bounded by $y = x^2 - 2x$ and $y = -x$.

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∻Solution by Steps∻ ‖ step 1 ⋮ To find the area between the curves $ y = x^2 - 2x $ and $ y = -x $, we need to set up the integral of the top function minus the bottom function from the leftmost to the rightmost points of intersection. ‖ ‖ step 2 ⋮ First, find the points of intersection by setting $ x^2 - 2x = -x $. Solving this gives $ x^2 - x = 0 $, which has solutions $ x = 0 $ and $ x = 1 $. ‖ ‖ step 3 ⋮ The integral setup is $ \int_{0}^{1} ((x^2 - 2x) - (-x)) dx $, which simplifies to $ \int_{0}^{1} (x^2 - x) dx $. ‖ ‖ step 4 ⋮ Compute the integral $ \int_{0}^{1} (x^2 - x) dx $ to find the area. ‖ ‖ step 5 ⋮ The antiderivative of $ x^2 - x $ is $ \frac{x^3}{3} - \frac{x^2}{2} $. ‖ ‖ step 6 ⋮ Evaluate the antiderivative from 0 to 1: $ \left[\frac{x^3}{3} - \frac{x^2}{2}\right]_{0}^{1} = \left(\frac{1^3}{3} - \frac{1^2}{2}\right) - \left(\frac{0^3}{3} - \frac{0^2}{2}\right) $. ‖ ‖ step 7 ⋮ Simplify to get the area: $ \frac{1}{3} - \frac{1}{2} = \frac{2}{6} - \frac{3}{6} = -\frac{1}{6} $. Since area cannot be negative, we take the absolute value. ‖ ∻Answer∻ ⚹ $\frac{1}{6}$ ⚹ ∻Key Concept∻ ⚹Area between two curves⚹ ∻Explanation∻ ⚹The area between two curves is found by integrating the difference of the functions (top function minus bottom function) over the interval defined by their points of intersection.⚹◊What are the x-coordinates of the intersection points of the two functions $y = x^2 - 2x$ and $y = -x$?⍭ Generate me a similar question◊

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