Practicing Success
Practicing Success
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Value of $\int\limits_0^n[x] d x$ (where n ∈ N) is |
$\frac{n(n+1)}{2}$ $\frac{n(n-1)}{2}$ $n(n-1)$ None of these |
$\frac{n(n-1)}{2}$ |
$\int\limits_0^n[x] d x=\sum\limits_{i=1}^n \int\limits_{i-1}^i[x] d x$ $\sum\limits_{i=1}^n(i-1) d x=\frac{n(n-1)}{2}$ Hence (2) is the correct answer. |
