The area of the region A = { (x,y) : 0 le y l | vedclass

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(C) The given region is bounded by $y = x|x| + 1$ and the $x$-axis for $-1 \le x \le 1$.We can define $y$ as:$y = \begin{cases} -x^2 + 1, & 	ext{if }

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$2$

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(C) The given region is bounded by $y = x|x| + 1$ and the $x$-axis for $-1 \le x \le 1$.We can define $y$ as:$y = \begin{cases} -x^2 + 1, & 	ext{if } -1 \le x The area is given by the integral:$	ext{Area} = \int_{-1}^{0} (-x^2 + 1) dx + \int_{0}^{1} (x^2 + 1) dx$Evaluating the first integral:$\int_{-1}^{0} (-x^2 + 1) dx = \left[ -\frac{x^3}{3} + x ight]_{-1}^{0} = 0 - (\frac{1}{3} - 1) = \frac{2}{3}$Evaluating the second integral:$\int_{0}^{1} (x^2 + 1) dx = \left[ \frac{x^3}{3} + x ight]_{0}^{1} = (\frac{1}{3} + 1) - 0 = \frac{4}{3}$Total Area $= \frac{2}{3} + \frac{4}{3} = \frac{6}{3} = 2$ square units.

The area of the region $A = \{ (x,y) : 0 \le y \le x|x| + 1, -1 \le x \le 1 \}$ in square units is:

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