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Find:\( \int_{1}^{2}[2 x] d x \) |
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Answer» \(\int\limits_1^2[2x]dx\) = \(\int\limits_1^{3/2}[2x]dx\) = \(\int\limits_{3/2}^2[2x]dx\) = \(\int\limits_1^{3/2}2dx+\int\limits_{3/2}^2 3dx\) = \(2[x]_1^{3/2}+3[x]^2_{3/2}\) = 2(32/ - 1) + 3(2 - 3/2) = 2 x 1/2 + 3 x 1/2 = 5/2 |
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