Question

Let \([\,]\) denote the greatest integer function. Then the value of} \[ \int_{0}^{3}\left(\frac{e^x+e^{-x}}{[x]!}\right)dx \] is:

• A. \(e^2+e^3-\frac{1}{e^2}-\frac{1}{e^3}\)
• B. \(\frac{1}{2}\left(e^2+e^3-\frac{1}{e^2}-\frac{1}{e^3}\right)\)
• C. \(e^2+e^3-\frac{1}{2e^2}-\frac{1}{2e^3}\)
• D. \(\frac{1}{2}(e^2+e^3)-\frac{1}{e^2}-\frac{1}{e^3}\)

Answer 1

The Correct Option is B

Concept: The greatest integer function \([x]\) changes value at integers. Hence the integral must be split over intervals: \[ [0,1),\;[1,2),\;[2,3) \] Also, \[ \int (e^x+e^{-x})dx=e^x-e^{-x} \]
Step 1:Split the integral.} For \(0\le x<1\): \( [x]=0 \Rightarrow [x]!=0!=1 \) \[ I_1=\int_0^1 (e^x+e^{-x})dx \] For \(1\le x<2\): \( [x]=1 \Rightarrow 1!=1 \) \[ I_2=\int_1^2 (e^x+e^{-x})dx \] For \(2\le x\le3\): \( [x]=2 \Rightarrow 2!=2 \) \[ I_3=\frac12\int_2^3 (e^x+e^{-x})dx \]
Step 2:Evaluate each integral.} \[ I_1=(e-e^{-1}) \] \[ I_2=(e^2-e)-(e^{-2}-e^{-1}) \] \[ I_3=\frac12[(e^3-e^2)-(e^{-3}-e^{-2})] \]
Step 3:Add the values.} \[ I=I_1+I_2+I_3 \] After simplification, \[ I=\frac12\left(e^2+e^3-\frac1{e^2}-\frac1{e^3}\right) \]