The area bounded by the curve y=x|x|,the x-ax | vedclass

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Question

(A) The function is defined as $y = x|x| = \begin{cases} x^2, & x \ge 0 \ -x^2, & x < 0 \end{cases}$.
The required area is the sum of the

Vedclass · Accepted Answer

$2/3$

Vedclass · Answer

(A) The function is defined as $y = x|x| = \begin{cases} x^2, & x \ge 0 \ -x^2, & x < 0 \end{cases}$.
The required area is the sum of the absolute values of the integrals over the intervals $[-1, 0]$ and $[0, 1]$.
Area $= \int_{-1}^{1} |y| dx = \int_{-1}^{0} |x^2| dx + \int_{0}^{1} |x^2| dx$.
Since the area must be positive,we calculate the integral of the absolute value of the function.
Area $= \int_{-1}^{0} |-x^2| dx + \int_{0}^{1} |x^2| dx = \int_{-1}^{0} x^2 dx + \int_{0}^{1} x^2 dx$.
Area $= \left[ \frac{x^3}{3} ight]_{-1}^{0} + \left[ \frac{x^3}{3} ight]_{0}^{1}$.
Area $= \left( 0 - \left( \frac{(-1)^3}{3} ight) ight) + \left( \frac{1^3}{3} - 0 ight) = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$ sq. units.
Thus,the correct answer is $A$.

The area bounded by the curve $y=x|x|$,the $x$-axis,and the ordinates $x=-1$ and $x=1$ is given by [Hint: $y=x^{2}$ if $x>0$ and $y=-x^{2}$ if $x < 0$].

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