The area enclosed by the curves y=x|x|,x=-1 a | vedclass

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Question

(B) The given curve is $y = x|x|$. We can define this as a piecewise function:$y = \begin{cases} x^2 & 	ext{if } x \geq 0 \ -x^2 & 	ext{if } x < 0

Vedclass · Accepted Answer

$\frac{2}{3}$

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(B) The given curve is $y = x|x|$. We can define this as a piecewise function:$y = \begin{cases} x^2 & 	ext{if } x \geq 0 \ -x^2 & 	ext{if } x We need to find the area enclosed by this curve between $x = -1$ and $x = 1$.The total area $A$ is given by the integral of the absolute value of $y$ from $-1$ to $1$:$A = \int_{-1}^{1} |y| dx = \int_{-1}^{0} |-x^2| dx + \int_{0}^{1} |x^2| dx$$A = \int_{-1}^{0} x^2 dx + \int_{0}^{1} x^2 dx$Calculating the integrals:$A = \left[ \frac{x^3}{3} ight]_{-1}^{0} + \left[ \frac{x^3}{3} ight]_{0}^{1}$$A = (0 - (-\frac{1}{3})) + (\frac{1}{3} - 0)$$A = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$Thus,the area is $\frac{2}{3}$ sq units.

The area enclosed by the curves $y=x|x|$,$x=-1$ and $x=1$ is.......... sq units.

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