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07/02/2022 11:50 am
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Let [t] denote the greatest integer ≤ t. Then the value of 8. \(\int_{-\frac{1}{2}}^1\)([2x]+ |x|) dx is .......
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07/02/2022 11:58 am
I = \(\int_{-\frac{1}{2}}^1\)([2x]+ |x|) dx
= \(\int_{-\frac{1}{2}}^1\)[2x]dx + \(\int_{-\frac{1}{2}}^1\)|x|dx
= 0 + \(\int_{-\frac{1}{2}}^1\)(-x)dx + \(\int_0^1\)x dx
= \(\Big(-\frac{x^2}{2}\Big)^0_{\frac{1}{2}}\) + \(\Big(\frac{x^2}{2}\Big)^1_0\)
= \(\Big(0 + \frac{1}{8}\Big) + \frac{1}{2}\)
= \(\frac{5}{8}\)
8I = 5