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Question

Evaluate: 0[[x+1]e5x+2]dx

where, [x] denotes the greator integer less than or equal to 'x'

A
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Solution

The correct option is A 0
[[x+1]e5x+2]dxletf(x)=[x+1]ex+2valueoff(0)=1e2(0,1)f(0)=2e3(0,1)f(0)=3e4(0,1)andsoonforallvalueofxf(x)(0,1)=[f(x)]=0[x]G.1.F=0[f(x)]dx=00dx=0

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