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

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Question

(A) Let $I = \int_0^1 |5x - 3| dx$. Since $5x - 3 = 0$ at $x = \frac{3}{5}$,we split the integral at $x = \frac{3}{5}$. For $0 \le x < \frac{3}{5}$,$|

Vedclass · Accepted Answer

$\frac{13}{10}$

Vedclass · Answer

(A) Let $I = \int_0^1 |5x - 3| dx$. Since $5x - 3 = 0$ at $x = \frac{3}{5}$,we split the integral at $x = \frac{3}{5}$. For $0 \le x For $\frac{3}{5} \le x \le 1$,$|5x - 3| = 5x - 3$. Thus,$I = \int_0^{3/5} (3 - 5x) dx + \int_{3/5}^1 (5x - 3) dx$. $I = [3x - \frac{5x^2}{2}]_0^{3/5} + [\frac{5x^2}{2} - 3x]_{3/5}^1$. $I = (3(\frac{3}{5}) - \frac{5}{2}(\frac{9}{25})) - (0) + ((\frac{5}{2} - 3) - (\frac{5}{2}(\frac{9}{25}) - 3(\frac{3}{5})))$. $I = (\frac{9}{5} - \frac{9}{10}) + (-\frac{1}{2} - (\frac{9}{10} - \frac{9}{5}))$. $I = \frac{9}{10} + (-\frac{1}{2} - (-\frac{9}{10})) = \frac{9}{10} - \frac{1}{2} + \frac{9}{10} = \frac{18}{10} - \frac{5}{10} = \frac{13}{10}$.

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

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