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(D) The integral $\int_{0}^{11} [x] dx$ can be split into intervals of unit length where $[x]$ is constant:$\int_{0}^{11} [x] dx = \sum_{k=0}^{10} \in

Vedclass · Accepted Answer

$55$

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(D) The integral $\int_{0}^{11} [x] dx$ can be split into intervals of unit length where $[x]$ is constant:$\int_{0}^{11} [x] dx = \sum_{k=0}^{10} \int_{k}^{k+1} [x] dx$Since for $x \in [k, k+1)$,$[x] = k$,we have:$\int_{k}^{k+1} [x] dx = \int_{k}^{k+1} k dx = k(k+1 - k) = k$Therefore,the sum becomes:$\int_{0}^{11} [x] dx = \sum_{k=0}^{10} k = 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10$Using the sum formula for the first $n$ natural numbers,$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$:$\sum_{k=1}^{10} k = \frac{10 	imes 11}{2} = 55$Thus,the value of the integral is $55$.

If $[x]$ is the greatest integer function not greater than $x$,then $\int_{0}^{11} [x] dx$ is equal to

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