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 :
- A \(e^2 + e^3 - \dfrac{1}{e^2} - \dfrac{1}{e^3}\)
- B \(\dfrac{1}{2}\left(e^2 + e^3 - \dfrac{1}{e^2} - \dfrac{1}{e^3}\right)\)
- C \(e^2 + e^3 - \dfrac{1}{2e^2} - \dfrac{1}{2e^3}\)
- D \(\dfrac{1}{2}(e^2 + e^3) - \dfrac{1}{e^2} - \dfrac{1}{e^3}\)
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 \([x]\). \(\displaystyle I = \int_0^3 \left(\dfrac{e^x + e^{-x}}{[x]!}\right) dx\)…
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