The area (in sq. ,units) of the region,given | vedclass

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(A) The region is bounded by the curves $y = 2x^2$ and $y = 4-2x$ for $x \geq 0$.To find the intersection points,set $2x^2 = 4-2x$,which simplifies to

Vedclass · Accepted Answer

$\frac{7}{3}$

Vedclass · Answer

(A) The region is bounded by the curves $y = 2x^2$ and $y = 4-2x$ for $x \geq 0$.To find the intersection points,set $2x^2 = 4-2x$,which simplifies to $x^2 + x - 2 = 0$.Factoring gives $(x+2)(x-1) = 0$,so $x = 1$ or $x = -2$.Since $x \geq 0$,we consider the interval $[0, 1]$.The area is given by the integral $\int_{0}^{1} ((4-2x) - 2x^2) dx$.$= \int_{0}^{1} (4 - 2x - 2x^2) dx$$= [4x - x^2 - \frac{2x^3}{3}]_{0}^{1}$$= (4(1) - (1)^2 - \frac{2(1)^3}{3}) - (0)$$= 4 - 1 - \frac{2}{3} = 3 - \frac{2}{3} = \frac{7}{3} \, sq. \, units$.

The area (in $sq. \,units$) of the region,given by the set $\{(x, y) \in R \times R \mid x \geq 0, 2x^2 \leq y \leq 4-2x\}$ is:

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