Question

Evaluate 0òx [x] dx .

Joshi sir comment

 

 

integer nearst to x and less than x will be [x]

so 0òx [x] dx = 0ò1 0 dx + 1ò2 1 dx + 2ò3 2 dx + ..................... + [x]-1[x] [x]-1 dx + [x][x] dx

now solve it

 

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