Compute the area between the curves $y = x^3 - x$ and $y = x$.

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∻Solution by Steps∻ ‖ step 1 ⋮ To find the area between the curves $y = x^3 - x$ and $y = x$, we first need to determine the points of intersection by solving $x^3 - x = x$. ‖ ‖ step 2 ⋮ Solving $x^3 - x = x$ gives $x^3 - 2x = 0$. Factoring out $x$, we get $x(x^2 - 2) = 0$. ‖ ‖ step 3 ⋮ The solutions to $x(x^2 - 2) = 0$ are $x = 0$ and $x = \pm \sqrt{2}$. ‖ ‖ step 4 ⋮ The area between the curves from $x = -\sqrt{2}$ to $x = \sqrt{2}$ is given by the integral $\int_{-\sqrt{2}}^{\sqrt{2}} (x - (x^3 - x)) \, dx$. ‖ ‖ step 5 ⋮ Simplifying the integrand, we get $\int_{-\sqrt{2}}^{\sqrt{2}} (x - x^3 + x) \, dx = \int_{-\sqrt{2}}^{\sqrt{2}} (2x - x^3) \, dx$. ‖ ‖ step 6 ⋮ We now integrate $\int_{-\sqrt{2}}^{\sqrt{2}} (2x - x^3) \, dx$. ‖ ‖ step 7 ⋮ The integral of $2x$ is $x^2$ and the integral of $-x^3$ is $-\frac{x^4}{4}$. Thus, $\int (2x - x^3) \, dx = x^2 - \frac{x^4}{4} + C$. ‖ ‖ step 8 ⋮ Evaluating the definite integral from $-\sqrt{2}$ to $\sqrt{2}$, we get $\left[ x^2 - \frac{x^4}{4} \right]_{-\sqrt{2}}^{\sqrt{2}}$. ‖ ‖ step 9 ⋮ Substituting the limits, we get $\left( (\sqrt{2})^2 - \frac{(\sqrt{2})^4}{4} \right) - \left( (-\sqrt{2})^2 - \frac{(-\sqrt{2})^4}{4} \right)$. ‖ ‖ step 10 ⋮ Simplifying, we get $\left( 2 - \frac{4}{4} \right) - \left( 2 - \frac{4}{4} \right) = (2 - 1) - (2 - 1) = 1 - 1 = 0$. ‖ ‖ step 11 ⋮ Therefore, the area between the curves $y = x^3 - x$ and $y = x$ is $0$. ‖ ∻Answer∻ ⚹ The area between the curves $y = x^3 - x$ and $y = x$ is $0$. ⚹ ∻Key Concept∻ ⚹Area between curves⚹ ∻Explanation∻ ⚹To find the area between two curves, we integrate the difference of the functions over the interval defined by their points of intersection.⚹◊What is the intersection point(s) of the curves $y = x^3 - x$ and $y = x$?⍭ Generate me a similar question◊

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