int_0^2 |2x - 3| , dx = | vedclass

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(B) Let $I = \int_0^2 |2x - 3| \, dx$. Since $|2x - 3| = 3 - 2x$ for $x < \frac{3}{2}$ and $|2x - 3| = 2x - 3$ for $x \ge \frac{3}{2}$,we split the in

Vedclass · Accepted Answer

$\frac{5}{2}$

Vedclass · Answer

(B) Let $I = \int_0^2 |2x - 3| \, dx$. Since $|2x - 3| = 3 - 2x$ for $x $I = \int_0^{3/2} (3 - 2x) \, dx + \int_{3/2}^2 (2x - 3) \, dx$. Evaluating the first part: $\int_0^{3/2} (3 - 2x) \, dx = [3x - x^2]_0^{3/2} = (3(\frac{3}{2}) - (\frac{3}{2})^2) - 0 = \frac{9}{2} - \frac{9}{4} = \frac{9}{4}$. Evaluating the second part: $\int_{3/2}^2 (2x - 3) \, dx = [x^2 - 3x]_{3/2}^2 = (2^2 - 3(2)) - ((\frac{3}{2})^2 - 3(\frac{3}{2})) = (4 - 6) - (\frac{9}{4} - \frac{9}{2}) = -2 - (-\frac{9}{4}) = -2 + \frac{9}{4} = \frac{1}{4}$. Adding both parts: $I = \frac{9}{4} + \frac{1}{4} = \frac{10}{4} = \frac{5}{2}$.

$\int_0^2 |2x - 3| \, dx = $

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