If the area of the region {(x, y): 1+x^2 leq | vedclass

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(A) The region is bounded by the parabola $y = 1+x^2$ and the lines $y = x+7$ and $y = 11-3x$. First,find the intersection points: $1+x^2 = x+7 \Right

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(A) The region is bounded by the parabola $y = 1+x^2$ and the lines $y = x+7$ and $y = 11-3x$. First,find the intersection points: $1+x^2 = x+7 \Rightarrow x^2-x-6 = 0 \Rightarrow (x-3)(x+2) = 0$. Since the region is in the interval $[-2, 2]$,we take $x = -2$. $1+x^2 = 11-3x \Rightarrow x^2+3x-10 = 0 \Rightarrow (x+5)(x-2) = 0$. We take $x = 2$. $x+7 = 11-3x \Rightarrow 4x = 4 \Rightarrow x = 1$. The area $A$ is given by: $A = \int_{-2}^{1} ((x+7) - (1+x^2)) dx + \int_{1}^{2} ((11-3x) - (1+x^2)) dx$ $A = \int_{-2}^{1} (6+x-x^2) dx + \int_{1}^{2} (10-3x-x^2) dx$ $A = [6x + \frac{x^2}{2} - \frac{x^3}{3}]_{-2}^{1} + [10x - \frac{3x^2}{2} - \frac{x^3}{3}]_{1}^{2}$ $A = (6 + \frac{1}{2} - \frac{1}{3}) - (-12 + 2 + \frac{8}{3}) + (20 - 6 - \frac{8}{3}) - (10 - \frac{3}{2} - \frac{1}{3})$ $A = (\frac{36+3-2}{6}) - (\frac{-36+6+8}{3}) + (\frac{14}{1} - \frac{8}{3}) - (\frac{60-9-2}{6})$ $A = \frac{37}{6} - (-\frac{22}{3}) + \frac{34}{3} - \frac{49}{6} = \frac{37+44+68-49}{6} = \frac{100}{6} = \frac{50}{3}$. Therefore,$3A = 3 	imes \frac{50}{3} = 50$.

If the area of the region $\{(x, y): 1+x^2 \leq y \leq \min \{x+7, 11-3x\}\}$ is $A$,then $3A$ is equal to

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