int_0^1 |5x - 3| , dx = | vedclass

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(B) To evaluate the integral $I = \int_0^1 |5x - 3| \, dx$,we first find the point where the expression inside the modulus changes sign. $5x - 3 = 0 \

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$\frac{13}{10}$

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(B) To evaluate the integral $I = \int_0^1 |5x - 3| \, dx$,we first find the point where the expression inside the modulus changes sign. $5x - 3 = 0 \implies x = \frac{3}{5}$. Since $\frac{3}{5} \in [0, 1]$,we split the integral at $x = \frac{3}{5}$: $I = \int_0^{3/5} -(5x - 3) \, dx + \int_{3/5}^1 (5x - 3) \, dx$ $I = \int_0^{3/5} (3 - 5x) \, dx + \int_{3/5}^1 (5x - 3) \, dx$ Evaluating the first part: $\left[ 3x - \frac{5x^2}{2} \right]_0^{3/5} = (3(\frac{3}{5}) - \frac{5}{2}(\frac{9}{25})) - 0 = \frac{9}{5} - \frac{9}{10} = \frac{18-9}{10} = \frac{9}{10}$. Evaluating the second part: $\left[ \frac{5x^2}{2} - 3x \right]_{3/5}^1 = (\frac{5}{2} - 3) - (\frac{5}{2}(\frac{9}{25}) - 3(\frac{3}{5})) = (-\frac{1}{2}) - (\frac{9}{10} - \frac{9}{5}) = -\frac{1}{2} - (-\frac{9}{10}) = -\frac{5}{10} + \frac{9}{10} = \frac{4}{10} = \frac{2}{5}$. Adding both parts: $I = \frac{9}{10} + \frac{4}{10} = \frac{13}{10}$.

$\int_0^1 |5x - 3| \, dx = $

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