If f(x) = begin{cases} sqrt{1-x} & 0 leqslant | vedclass

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(A) We need to evaluate the integral $\int_{0}^{2} f(x) dx$. Since the function is defined piecewise,we split the integral at $x=1$: $\int_{0}^{2} f(x

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$\frac{55}{42}$

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(A) We need to evaluate the integral $\int_{0}^{2} f(x) dx$. Since the function is defined piecewise,we split the integral at $x=1$: $\int_{0}^{2} f(x) dx = \int_{0}^{1} \sqrt{1-x} dx + \int_{1}^{2} (7x-6)^{-1/3} dx$. For the first part,let $u = 1-x$,then $du = -dx$. When $x=0, u=1$; when $x=1, u=0$: $\int_{0}^{1} (1-x)^{1/2} dx = -\int_{1}^{0} u^{1/2} du = \int_{0}^{1} u^{1/2} du = [\frac{2}{3} u^{3/2}]_{0}^{1} = \frac{2}{3}$. For the second part,let $v = 7x-6$,then $dv = 7 dx$,so $dx = \frac{1}{7} dv$. When $x=1, v=1$; when $x=2, v=8$: $\int_{1}^{2} (7x-6)^{-1/3} dx = \frac{1}{7} \int_{1}^{8} v^{-1/3} dv = \frac{1}{7} [\frac{3}{2} v^{2/3}]_{1}^{8} = \frac{1}{7} \cdot \frac{3}{2} (8^{2/3} - 1^{2/3}) = \frac{3}{14} (4 - 1) = \frac{3}{14} \cdot 3 = \frac{9}{14}$. Adding the two parts: $\frac{2}{3} + \frac{9}{14} = \frac{28 + 27}{42} = \frac{55}{42}$.

If $f(x) = \begin{cases} \sqrt{1-x} & 0 \leqslant x \leqslant 1 \\ (7x-6)^{-1/3} & 1 < x \leqslant 2 \end{cases}$,then $\int_{0}^{2} f(x) dx$ is equal to

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