Find the area of the region {(x, y): 0 leq y | vedclass

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(A) The region is defined by the intersection of $y \leq x^2 + 1$,$y \leq x + 1$,and $0 \leq x \leq 2$. First,find the intersection points of the curv

Vedclass · Accepted Answer

$\frac{23}{6}$

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(A) The region is defined by the intersection of $y \leq x^2 + 1$,$y \leq x + 1$,and $0 \leq x \leq 2$. First,find the intersection points of the curves $y = x^2 + 1$ and $y = x + 1$: $x^2 + 1 = x + 1 \implies x^2 - x = 0 \implies x(x - 1) = 0$. So,the curves intersect at $x = 0$ and $x = 1$. For $0 \leq x \leq 1$,the region is bounded by $y \leq x + 1$ (since $x+1 \geq x^2+1$ in this interval). For $1 \leq x \leq 2$,the region is bounded by $y \leq x^2 + 1$ (since $x^2+1 \geq x+1$ in this interval). Wait,looking at the graph provided,the region is bounded by the lower of the two curves at any point $x$. For $0 \leq x \leq 1$,the lower curve is $y = x^2 + 1$. For $1 \leq x \leq 2$,the lower curve is $y = x + 1$. Area $= \int_{0}^{1} (x^2 + 1) dx + \int_{1}^{2} (x + 1) dx$. $= [\frac{x^3}{3} + x]_{0}^{1} + [\frac{x^2}{2} + x]_{1}^{2}$. $= (\frac{1}{3} + 1) - 0 + ((\frac{4}{2} + 2) - (\frac{1}{2} + 1))$. $= \frac{4}{3} + (4 - \frac{3}{2}) = \frac{4}{3} + \frac{5}{2} = \frac{8 + 15}{6} = \frac{23}{6}$.

Find the area of the region $\{(x, y): 0 \leq y \leq x^{2}+1, 0 \leq y \leq x+1, 0 \leq x \leq 2\}$.

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