f(x) = begin{cases} x^2, & 0 leq x

Question

(A) To evaluate the integral $\int_0^2 f(x) \, dx$,we split the interval $[0, 2]$ at $x = 1$ based on the definition of $f(x)$.$\int_0^2 f(x) \, dx =

Vedclass · Accepted Answer

$\frac{4 \sqrt{2}-1}{3}$

Vedclass · Answer

(A) To evaluate the integral $\int_0^2 f(x) \, dx$,we split the interval $[0, 2]$ at $x = 1$ based on the definition of $f(x)$.$\int_0^2 f(x) \, dx = \int_0^1 x^2 \, dx + \int_1^2 \sqrt{x} \, dx$Evaluating the first part: $\int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} - 0 = \frac{1}{3}$Evaluating the second part: $\int_1^2 x^{1/2} \, dx = \left[ \frac{x^{3/2}}{3/2} \right]_1^2 = \left[ \frac{2}{3} x^{3/2} \right]_1^2 = \frac{2}{3} (2^{3/2} - 1^{3/2}) = \frac{2}{3} (2\sqrt{2} - 1)$Adding both parts: $\frac{1}{3} + \frac{2}{3}(2\sqrt{2} - 1) = \frac{1 + 4\sqrt{2} - 2}{3} = \frac{4\sqrt{2} - 1}{3}$

$f(x) = \begin{cases} x^2, & 0 \leq x < 1 \\ \sqrt{x}, & 1 \leq x \leq 2 \end{cases} \implies \int_0^2 f(x) \, dx = ?$

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