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Question
Math
Posted about 2 months ago
Compute the area between the curves y=x3xy = x^3 - x and y=xy = 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=x3xy = x^3 - x and y=xy = x, we first need to determine the points of intersection by solving x3x=xx^3 - x = x
step 2
Solving x3x=xx^3 - x = x gives x32x=0x^3 - 2x = 0. Factoring out xx, we get x(x22)=0x(x^2 - 2) = 0
step 3
The solutions to x(x22)=0x(x^2 - 2) = 0 are x=0x = 0 and x=±2x = \pm \sqrt{2}
step 4
The area between the curves from x=2x = -\sqrt{2} to x=2x = \sqrt{2} is given by the integral 22(x(x3x))dx\int_{-\sqrt{2}}^{\sqrt{2}} (x - (x^3 - x)) \, dx
step 5
Simplifying the integrand, we get 22(xx3+x)dx=22(2xx3)dx\int_{-\sqrt{2}}^{\sqrt{2}} (x - x^3 + x) \, dx = \int_{-\sqrt{2}}^{\sqrt{2}} (2x - x^3) \, dx
step 6
We now integrate 22(2xx3)dx\int_{-\sqrt{2}}^{\sqrt{2}} (2x - x^3) \, dx
step 7
The integral of 2x2x is x2x^2 and the integral of x3-x^3 is x44-\frac{x^4}{4}. Thus, (2xx3)dx=x2x44+C\int (2x - x^3) \, dx = x^2 - \frac{x^4}{4} + C
step 8
Evaluating the definite integral from 2-\sqrt{2} to 2\sqrt{2}, we get [x2x44]22\left[ x^2 - \frac{x^4}{4} \right]_{-\sqrt{2}}^{\sqrt{2}}
step 9
Substituting the limits, we get ((2)2(2)44)((2)2(2)44)\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 (244)(244)=(21)(21)=11=0\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=x3xy = x^3 - x and y=xy = x is 00
Answer
The area between the curves y=x3xy = x^3 - x and y=xy = x is 00.
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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