If f(x) = begin{cases} 4x + 3, & 1 le x le 2 | vedclass

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(D) To evaluate the integral $\int_1^4 f(x) \, dx$,we split the interval $[1, 4]$ at $x = 2$ based on the definition of $f(x)$:$\int_1^4 f(x) \, dx =

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$37$

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(D) To evaluate the integral $\int_1^4 f(x) \, dx$,we split the interval $[1, 4]$ at $x = 2$ based on the definition of $f(x)$:$\int_1^4 f(x) \, dx = \int_1^2 (4x + 3) \, dx + \int_2^4 (3x + 5) \, dx$First part: $\int_1^2 (4x + 3) \, dx = [2x^2 + 3x]_1^2 = (2(2)^2 + 3(2)) - (2(1)^2 + 3(1)) = (8 + 6) - (2 + 3) = 14 - 5 = 9$Second part: $\int_2^4 (3x + 5) \, dx = [\frac{3x^2}{2} + 5x]_2^4 = (\frac{3(16)}{2} + 5(4)) - (\frac{3(4)}{2} + 5(2)) = (24 + 20) - (6 + 10) = 44 - 16 = 28$Total integral: $9 + 28 = 37$.

If $f(x) = \begin{cases} 4x + 3, & 1 \le x \le 2 \\ 3x + 5, & 2 < x \le 4 \end{cases}$,then $\int_1^4 f(x) \, dx = $

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