Let [] denote the greatest integer function. Then the value of is: [2026]

Question

Let [ $\cdot$ ] denote the greatest integer function. Then the value of $\int_{0}^{3} (\frac{e^{x} + e^{- x}}{[x]!}) d x$ is: [2026]

Smart Achievers · Accepted Answer

(2)

$\int_{0}^{3} (\frac{e^{x} + e^{- x}}{[x]!}) d x = {\frac{e^{x} + e^{- x}}{1}}_{0}^{1} + {\frac{e^{x} + e^{- x}}{1}}_{1}^{2} + {\frac{e^{x} + e^{- x}}{2}}_{2}^{3}$

$= (e - \frac{1}{e}) + (e^{2} - \frac{1}{e^{2}}) - (e - \frac{1}{e}) + (\frac{e^{3} - \frac{1}{e^{3}}}{2}) - (\frac{e^{2} - \frac{1}{e^{2}}}{2})$

$= \frac{1}{2} {(e^{2} - \frac{1}{e^{2}}) + (e^{3} - \frac{1}{e^{3}})}$

Solution

Q.
Let [ $\cdot$ ] denote the greatest integer function. Then the value of $\int_{0}^{3} (\frac{e^{x} + e^{- x}}{[x]!}) d x$ is: [2026]

1 $e^{2} + e^{3} - \frac{1}{e^{2}} - \frac{1}{e^{3}}$

2 $\frac{1}{2} (e^{2} + e^{3} - \frac{1}{e^{2}} - \frac{1}{e^{3}})$

3 $e^{2} + e^{3} - \frac{1}{2 e^{2}} - \frac{1}{2 e^{3}}$

4 $\frac{1}{2} (e^{2} + e^{3}) - \frac{1}{e^{2}} - \frac{1}{e^{3}}$

Want Us to Email you About Special Offers & Updates?

Solution

Q. Let [·] denote the greatest integer function. Then the value of ∫03(ex+e–x[x]!)dx is: [2026]

1 e2+e3–1e2–1e3

2 12(e2+e3–1e2–1e3)

3 e2+e3–12e2–12e3

4 12(e2+e3)–1e2–1e3