The value of _0^3e^x+e^-x[x]!,dx, where [] is the greatest integer function, is equal to:

Question

The value of $\displaystyle\int_0^3\frac{e^x+e^{-x}}{[x]!}\,dx$, where $[\cdot]$ is the greatest integer function, is equal to:

Accepted Answer

For $x\in[0,2)$, $[x]!\in\{0!,1!\}=1$; for $x\in[2,3)$, $[x]!=2!=2$.

$$I = \int_0^2 (e^x+e^{-x})\,dx + \frac12\int_2^3(e^x+e^{-x})\,dx$$

$$= \left(e^2-e^{-2}\right) + \frac12\left[(e^3-e^{-3})-(e^2-e^{-2})\right] = \frac12(e^2+e^3) - \frac12\left(\frac{1}{e^2}+\frac{1}{e^3}\right)$$

$$I = \frac12\left(e^2+e^3-\frac{1}{e^2}-\frac{1}{e^3}\right)$$

Solutions · 1