Find the area bounded by the curves x = sqrt{ | vedclass

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(C) Given curves are $x = \sqrt{y - 1}$ and $y = x + 1$.From the first equation,$x^2 = y - 1$,which implies $y = x^2 + 1$.To find the intersection poi

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$1/6$

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(C) Given curves are $x = \sqrt{y - 1}$ and $y = x + 1$.From the first equation,$x^2 = y - 1$,which implies $y = x^2 + 1$.To find the intersection points,set $x^2 + 1 = x + 1$.$x^2 - x = 0 \implies x(x - 1) = 0$.So,the intersection points are $x = 0$ and $x = 1$.The area $A$ is given by $\int_{0}^{1} |(x + 1) - (x^2 + 1)| dx$.$A = \int_{0}^{1} (x - x^2) dx$.$A = [\frac{x^2}{2} - \frac{x^3}{3}]_{0}^{1}$.$A = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$ square units.

Find the area bounded by the curves $x = \sqrt{y - 1}$ and $y = x + 1$.

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