The area bounded by the curve y = begin{cases | vedclass

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(C) The curve is defined as $y = x^2$ for $x < 0$ and $y = x$ for $x \geq 0$. The line is $y = 4$.For $x < 0$,the curve is $y = x^2$,which implies $x

Vedclass · Accepted Answer

$\frac{40}{3}$

Vedclass · Answer

(C) The curve is defined as $y = x^2$ for $x For $x $A_1 = \int_{0}^{4} |-\sqrt{y}| dy = \int_{0}^{4} y^{1/2} dy = \left[ \frac{2}{3} y^{3/2} \right]_{0}^{4} = \frac{2}{3}(8) = \frac{16}{3}$.For $x \geq 0$,the curve is $y = x$. The intersection with $y=4$ is at $x = 4$. The area $A_2$ in the first quadrant is the area of the triangle formed by the vertices $(0,0)$,$(4,0)$,and $(4,4)$,or the integral of $x$ with respect to $y$ from $y=0$ to $y=4$:$A_2 = \int_{0}^{4} x dy = \int_{0}^{4} y dy = \left[ \frac{y^2}{2} \right]_{0}^{4} = \frac{16}{2} = 8$.The total area is $A_1 + A_2 = \frac{16}{3} + 8 = \frac{16 + 24}{3} = \frac{40}{3}$ square units.

The area bounded by the curve $y = \begin{cases} x^2, & x < 0 \\ x, & x \geq 0 \end{cases}$ and the line $y = 4$ is

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