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If y=[x]2+2{x}[x]+100∑r=1{x+r}2100, then ∫y⋅dx=(where [.] and {.} are greatest integer function and fractional part function respectively and c is integration constant) |
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Answer» If y=[x]2+2{x}[x]+100∑r=1{x+r}2100, then ∫y⋅dx= |
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