Let [·] denote the greatest integer function. Then the value of ∫_0^3 ( e^x + e^ -x [x]! ) dx is :

Question

Quantrex Academy · Accepted Answer

Correct answer: B. (1)/(2)(e^2 + e^3 - (1)/(e^2) - (1)/(e^3))

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

\displaystyle I = \int_0^1 \dfrac{e^x + e^{-x}}{0!} dx + \int_1^2 \dfrac{e^x + e^{-x}}{1!} dx + \int_2^3 \dfrac{e^x + e^{-x}}{2!} dx

Since

0! = 1

,

1! = 1

, and

2! = 2

, we get:

\displaystyle I = \int_0^2 (e^x + e^{-x}) dx + \dfrac{1}{2} \int_2^3 (e^x + e^{-x}) dx

Integrating the terms:

\displaystyle I = \left[ e^x - e^{-x} \right]_0^2 + \dfrac{1}{2} \left[ e^x - e^{-x} \right]_2^3

\displaystyle I = (e^2 - e^{-2} - (1 - 1)) + \dfrac{1}{2} (e^3 - e^{-3} - (e^2 - e^{-2}))

\displaystyle I = e^2 - e^{-2} + \dfrac{1}{2} e^3 - \dfrac{1}{2} e^{-3} - \dfrac{1}{2} e^2 + \dfrac{1}{2} e^{-2}

\displaystyle I = \dfrac{1}{2} e^2 + \dfrac{1}{2} e^3 - \dfrac{1}{2} e^{-2} - \dfrac{1}{2} e^{-3}

\displaystyle I = \dfrac{1}{2} \left( e^2 + e^3 - \dfrac{1}{e^2} - \dfrac{1}{e^3} \right)

Answer: $\dfrac{1}{2}\left(e^2 + e^3 - \dfrac{1}{e^2} - \dfrac{1}{e^3}\right)

Let [·] denote the greatest integer function. Then the value of...