Given: f(x) = begin{cases} x, & 0 leq x

Question

(B) The region is bounded by $x = \frac{1}{2}$ and $x = \frac{\sqrt{3}}{2}$. In this interval,$f(x) = 1-x$ and $g(x) = (x-\frac{1}{2})^2$.The area $A$

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{4}-\frac{1}{3}$

Vedclass · Answer

(B) The region is bounded by $x = \frac{1}{2}$ and $x = \frac{\sqrt{3}}{2}$. In this interval,$f(x) = 1-x$ and $g(x) = (x-\frac{1}{2})^2$.The area $A$ is given by the integral:$A = \int_{1/2}^{\sqrt{3}/2} (f(x) - g(x)) dx$$A = \int_{1/2}^{\sqrt{3}/2} ((1-x) - (x-\frac{1}{2})^2) dx$Let $u = x - \frac{1}{2}$,then $du = dx$. When $x = 1/2, u = 0$. When $x = \sqrt{3}/2, u = \frac{\sqrt{3}-1}{2}$.$A = \int_{0}^{(\sqrt{3}-1)/2} (1 - (u + 1/2) - u^2) du = \int_{0}^{(\sqrt{3}-1)/2} (1/2 - u - u^2) du$$A = [\frac{1}{2}u - \frac{u^2}{2} - \frac{u^3}{3}]_{0}^{(\sqrt{3}-1)/2}$Substituting the upper limit:$A = \frac{1}{2}(\frac{\sqrt{3}-1}{2}) - \frac{1}{2}(\frac{3+1-2\sqrt{3}}{4}) - \frac{1}{3}(\frac{3\sqrt{3}-9+3\sqrt{3}-1}{8})$$A = \frac{\sqrt{3}-1}{4} - \frac{4-2\sqrt{3}}{8} - \frac{6\sqrt{3}-10}{24} = \frac{6\sqrt{3}-6 - 12+6\sqrt{3} - 6\sqrt{3}+10}{24} = \frac{6\sqrt{3}-8}{24} = \frac{\sqrt{3}}{4} - \frac{1}{3}$.

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