If f(x) = begin{cases} 2x^2 + 1, & x leq 1 4 | vedclass

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(A) Given, $f(x) = \begin{cases} 2x^2 + 1, & x \leq 1 \ 4x^3 - 1, & x > 1 \end{cases}$.We need to evaluate $\int_{0}^{2} f(x) dx$.Since the function

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$47/3$

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(A) Given, $f(x) = \begin{cases} 2x^2 + 1, & x \leq 1 \ 4x^3 - 1, & x > 1 \end{cases}$.We need to evaluate $\int_{0}^{2} f(x) dx$.Since the function definition changes at $x = 1$, we split the integral:$\int_{0}^{2} f(x) dx = \int_{0}^{1} (2x^2 + 1) dx + \int_{1}^{2} (4x^3 - 1) dx$.Evaluating the first part:$\int_{0}^{1} (2x^2 + 1) dx = \left[ \frac{2x^3}{3} + x ight]_{0}^{1} = \left( \frac{2(1)^3}{3} + 1 ight) - (0) = \frac{2}{3} + 1 = \frac{5}{3}$.Evaluating the second part:$\int_{1}^{2} (4x^3 - 1) dx = \left[ x^4 - x ight]_{1}^{2} = (2^4 - 2) - (1^4 - 1) = (16 - 2) - (0) = 14$.Adding both parts:$\int_{0}^{2} f(x) dx = \frac{5}{3} + 14 = \frac{5 + 42}{3} = \frac{47}{3}$.

If $f(x) = \begin{cases} 2x^2 + 1, & x \leq 1 \\ 4x^3 - 1, & x > 1 \end{cases}$, then $\int_{0}^{2} f(x) dx$ is

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