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 [x] dx
now solve it
Evaluate 0òx [x] dx .
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 [x] dx
now solve it