Question

Math

Posted about 2 months ago

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

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Answer from Sia

Posted about 2 months ago

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.

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