a) \(E = \int_{0}^{5} |x-2| dx\)
Ta có ; \(x-2 < 0\) khi \(x < 2\)
\(x-2 \ge 0\) khi\( x \ge 2\)
\(E = \int_{0}^{2} (2-x) dx + \int_{2}^{5} (x-2) dx\)
\(\int_{0}^{2} (2-x) dx = [2x - \frac{x^2}{2}]_{0}^{2} = (4 - 2) - (0 - 0) = 2\)\(\int_{2}^{5} (x-2) dx = [\frac{x^2}{2} - 2x]_{2}^{5} = (\frac{25}{2} - 10) - (2 - 4) = \frac{5}{2} + 2 = \frac{9}{2} \)
Vậy, \(E = 2 + \frac{9}{2} = \frac{13}{2}\)
b) \(E = \int_{-1}^{2} |2x-1| dx\)
Ta có 2x-1 < 0 khi \(x < \frac{1}{2}\)
\(2x-1 \ge 0 \)khi \(x \ge \frac{1}{2}\)
\(E = \int_{-1}^{\frac{1}{2}} (1-2x) dx + \int_{\frac{1}{2}}^{2} (2x-1) dx \)
\(\int_{-1}^{\frac{1}{2}} (1-2x) dx = [x - x^2]_{-1}^{\frac{1}{2}} = (\frac{1}{2} - \frac{1}{4}) - (-1 - 1) = \frac{1}{4} + 2 = \frac{9}{4}\)
\(\int_{\frac{1}{2}}^{2} (2x-1) dx = [x^2 - x]_{\frac{1}{2}}^{2} = (4 - 2) - (\frac{1}{4} - \frac{1}{2}) = 2 + \frac{1}{4} = \frac{9}{4}\)
\(Vậy, E = \frac{9}{4} + \frac{9}{4} = \frac{18}{4} = \frac{9}{2}\)
