Let [cdot] denote the greatest integer functi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The integral is $\int_0^3 \frac{e^x + e^{-x}}{[x]!} dx$. We split the interval $[0, 3)$ into $[0, 1), [1, 2), [2, 3)$.For $x \in [0, 1)$, $[x] = 0

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The integral is $\int_0^3 \frac{e^x + e^{-x}}{[x]!} dx$. We split the interval $[0, 3)$ into $[0, 1), [1, 2), [2, 3)$.For $x \in [0, 1)$, $[x] = 0$, so $[x]! = 0! = 1$. The integral is $\int_0^1 (e^x + e^{-x}) dx = [e^x - e^{-x}]_0^1 = (e - e^{-1}) - (1 - 1) = e - e^{-1}$.For $x \in [1, 2)$, $[x] = 1$, so $[x]! = 1! = 1$. The integral is $\int_1^2 (e^x + e^{-x}) dx = [e^x - e^{-x}]_1^2 = (e^2 - e^{-2}) - (e - e^{-1}) = e^2 - e^{-2} - e + e^{-1}$.For $x \in [2, 3)$, $[x] = 2$, so $[x]! = 2! = 2$. The integral is $\int_2^3 \frac{e^x + e^{-x}}{2} dx = \frac{1}{2} [e^x - e^{-x}]_2^3 = \frac{1}{2} ((e^3 - e^{-3}) - (e^2 - e^{-2})) = \frac{1}{2}e^3 - \frac{1}{2}e^{-3} - \frac{1}{2}e^2 + \frac{1}{2}e^{-2}$.Summing these: $(e - e^{-1}) + (e^2 - e^{-2} - e + e^{-1}) + (\frac{1}{2}e^3 - \frac{1}{2}e^{-3} - \frac{1}{2}e^2 + \frac{1}{2}e^{-2}) = \frac{1}{2}e^3 + \frac{1}{2}e^2 - \frac{1}{2}e^{-2} - \frac{1}{2}e^{-3} = \frac{1}{2}(e^3 + e^2 - e^{-2} - e^{-3})$.

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

Similar Questions

Let $f(x) = \max \{x+|x|, x-[x]\}$, where $[x]$ denotes the greatest integer $\leq x$. Then, the value of $\int_{-3}^{3} f(x) dx$ is

Let $f(x) = |x - 2|$ and $g(x) = f(f(x))$,$x \in [0, 4]$. Then $\int_{0}^{3} (g(x) - f(x)) \, dx$ is equal to

$\int_{0}^{1} \frac{d}{dx} \left[ \sin^{-1} \left( \frac{2x}{1+x^2} \right) \right] dx$ is equal to

$\int\limits_{\frac{\pi }{6}}^{\frac{5\pi }{6}} {\left( {\frac{1}{2}{{(3\sin \theta )}^2} - \frac{1}{2}{{(1 + \sin \theta )}^2}} \right)\,d\theta } $

$\int_0^{\pi / 4} \frac{x^2}{(x \sin x+\cos x)^2} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module