The area (in sq. units) of the region {(x, y) | vedclass

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(B) The region is defined by $0 \leq y \leq \min(2|x|+1, x^2+1)$ for $x \in [-3, 3]$.Due to symmetry about the $y$-axis,the total area is $2 	imes \i

Vedclass · Accepted Answer

$\frac{64}{3}$

Vedclass · Answer

(B) The region is defined by $0 \leq y \leq \min(2|x|+1, x^2+1)$ for $x \in [-3, 3]$.Due to symmetry about the $y$-axis,the total area is $2 	imes \int_0^3 \min(2x+1, x^2+1) dx$.First,find the intersection of $y = 2x+1$ and $y = x^2+1$ for $x > 0$:$x^2+1 = 2x+1 \Rightarrow x^2 - 2x = 0 \Rightarrow x(x-2) = 0$.So,the curves intersect at $x = 0$ and $x = 2$.For $x \in [0, 2]$,$x^2+1 \leq 2x+1$,so $\min(2x+1, x^2+1) = x^2+1$.For $x \in [2, 3]$,$2x+1 \leq x^2+1$,so $\min(2x+1, x^2+1) = 2x+1$.Thus,the area is $2 \left[ \int_0^2 (x^2+1) dx + \int_2^3 (2x+1) dx ight]$.$= 2 \left[ \left( \frac{x^3}{3} + x ight)_0^2 + \left( x^2 + x ight)_2^3 ight]$$= 2 \left[ \left( \frac{8}{3} + 2 ight) + ((9+3) - (4+2)) ight]$$= 2 \left[ \frac{14}{3} + (12 - 6) ight] = 2 \left[ \frac{14}{3} + 6 ight] = 2 \left[ \frac{14+18}{3} ight] = 2 \left( \frac{32}{3} ight) = \frac{64}{3}$.

The area (in sq. units) of the region $\{(x, y): 0 \leq y \leq 2|x|+1, 0 \leq y \leq x^2+1, |x| \leq 3\}$ is

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