The value of $I = \int_{0}^{3} ([x] + [x + \frac{1}{3}] + [x + \frac{2}{3}]) d x$ , where $[\cdot]$ denotes the greatest integer function, is equal to

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Solution

The correct option is C
12

Let $I = \int_{0}^{\frac{1}{3}} 0 d x + \int_{\frac{1}{3}}^{\frac{2}{3}} 1 d x + \int_{\frac{2}{3}}^{1} 2 d x + \int_{1}^{\frac{4}{3}} 3 d x + \int_{\frac{4}{3}}^{\frac{5}{3}} 4 d x + \int_{\frac{5}{3}}^{\frac{6}{3}} 5 d x + \int_{\frac{6}{3}}^{\frac{7}{3}} 6 d x + \int_{\frac{7}{3}}^{\frac{8}{3}} 7 d x + \int_{\frac{8}{3}}^{\frac{9}{3}} 8 d x = \frac{1}{3} (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) = 12$

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The value of I=∫30([x]+[x+13]+[x+23])dx, where [⋅] denotes the greatest integer function, is equal to

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The value of $I = \int_{0}^{3} ([x] + [x + \frac{1}{3}] + [x + \frac{2}{3}]) d x$ , where $[\cdot]$ denotes the greatest integer function, is equal to