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JEE Mains · Maths · STD 12 - 7.2 definite integral

Let \([\cdot]\) denote the greatest integer function. Then the value of \(\displaystyle\int_0^3 \left(\dfrac{e^x + e^{-x}}{[x]!}\right) dx\) is :

  1. A \(e^2 + e^3 - \dfrac{1}{e^2} - \dfrac{1}{e^3}\)
  2. B \(\dfrac{1}{2}\left(e^2 + e^3 - \dfrac{1}{e^2} - \dfrac{1}{e^3}\right)\)
  3. C \(e^2 + e^3 - \dfrac{1}{2e^2} - \dfrac{1}{2e^3}\)
  4. D \(\dfrac{1}{2}(e^2 + e^3) - \dfrac{1}{e^2} - \dfrac{1}{e^3}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\dfrac{1}{2}\left(e^2 + e^3 - \dfrac{1}{e^2} - \dfrac{1}{e^3}\right)\)

Step-by-step Solution

Detailed explanation

The given integral can be split at integer values of \(x\) because of the greatest integer function \([x]\). \(\displaystyle I = \int_0^3 \left(\dfrac{e^x + e^{-x}}{[x]!}\right) dx\)…