The area enclosed by the curve y^2 + x^4 = x^ | vedclass

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(B) The given curve is $y^2 = x^2 - x^4 = x^2(1 - x^2)$.Taking the square root,we get $y = \pm x \sqrt{1 - x^2}$.The curve is symmetric about both the

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$\frac{4}{3}$

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(B) The given curve is $y^2 = x^2 - x^4 = x^2(1 - x^2)$.Taking the square root,we get $y = \pm x \sqrt{1 - x^2}$.The curve is symmetric about both the $x$-axis and the $y$-axis.Therefore,the total area $A$ is $4$ times the area in the first quadrant.$A = 4 \int_{0}^{1} x \sqrt{1 - x^2} dx$.Let $u = 1 - x^2$,then $du = -2x dx$,or $x dx = -\frac{1}{2} du$.When $x = 0$,$u = 1$. When $x = 1$,$u = 0$.$A = 4 \int_{1}^{0} \sqrt{u} \left(-\frac{1}{2}ight) du = 2 \int_{0}^{1} u^{1/2} du$.$A = 2 \left[ \frac{u^{3/2}}{3/2} ight]_{0}^{1} = 2 	imes \frac{2}{3} [1 - 0] = \frac{4}{3}$ square units.

The area enclosed by the curve $y^2 + x^4 = x^2$ is :

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