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

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(D) To evaluate the integral $\int_{0}^{2} f(x) \, dx$,we split the integral at the point of discontinuity $x = 1$: $\int_{0}^{2} f(x) \, dx = \int_{0

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$\frac{55}{42}$

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(D) To evaluate the integral $\int_{0}^{2} f(x) \, dx$,we split the integral at the point of discontinuity $x = 1$: $\int_{0}^{2} f(x) \, dx = \int_{0}^{1} \sqrt{1 - x} \, dx + \int_{1}^{2} (7x - 6)^{-1/3} \, dx$ For the first part: $\int_{0}^{1} (1 - x)^{1/2} \, dx = \left[ -\frac{2}{3} (1 - x)^{3/2} \right]_{0}^{1} = -\frac{2}{3} (0 - 1) = \frac{2}{3}$ For the second part: $\int_{1}^{2} (7x - 6)^{-1/3} \, dx = \left[ \frac{1}{7} \cdot \frac{(7x - 6)^{2/3}}{2/3} \right]_{1}^{2} = \left[ \frac{3}{14} (7x - 6)^{2/3} \right]_{1}^{2}$ $= \frac{3}{14} ((14 - 6)^{2/3} - (7 - 6)^{2/3}) = \frac{3}{14} (8^{2/3} - 1^{2/3}) = \frac{3}{14} (4 - 1) = \frac{3}{14} \cdot 3 = \frac{9}{14}$ Adding the two parts: $\frac{2}{3} + \frac{9}{14} = \frac{28 + 27}{42} = \frac{55}{42}$ Thus,the correct option is $D$.

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

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