The value of sum_{n=1}^{100} int_{n-1}^{n} e^ | vedclass

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

Question

(A) Let $f(x) = e^{x-[x]} = e^{\{x\}},$ where $\{x\}$ is the fractional part of $x.$Since the function $f(x) = e^{\{x\}}$ is periodic with period $1,$

Vedclass · Accepted Answer

$100(e-1)$

Vedclass · Answer

(A) Let $f(x) = e^{x-[x]} = e^{\{x\}},$ where $\{x\}$ is the fractional part of $x.$Since the function $f(x) = e^{\{x\}}$ is periodic with period $1,$ we have $\int_{n-1}^{n} e^{\{x\}} dx = \int_{0}^{1} e^{\{x\}} dx$ for any integer $n.$In the interval $[0, 1],$ the fractional part $\{x\} = x.$Therefore,$\int_{0}^{1} e^{\{x\}} dx = \int_{0}^{1} e^x dx = [e^x]_0^1 = e^1 - e^0 = e - 1.$Now,the summation is $\sum_{n=1}^{100} \int_{n-1}^{n} e^{\{x\}} dx = \sum_{n=1}^{100} (e-1) = 100(e-1).$

The value of $\sum_{n=1}^{100} \int_{n-1}^{n} e^{x-[x]} dx,$ where $[x]$ is the greatest integer $\leq x,$ is

Similar Questions

$\int_{\pi /6}^{\pi /3} \frac{dx}{1 + \sqrt{\tan x}} = $

If $\int_{-\pi / 2}^{\pi / 2} \frac{8 \sqrt{2} \cos x \, dx}{(1+e^{\sin x})(1+\sin ^4 x)} = \alpha \pi + \beta \log _e(3+2 \sqrt{2})$,where $\alpha, \beta$ are integers,then $\alpha^2+\beta^2$ equals.....................

$\int_0^{\frac{\pi}{2}} \frac{dx}{1+(\cot x)^{101}} = $

$\int_{\frac{\pi}{4}}^{\frac{5 \pi}{4}} (|\cos t| \sin t + |\sin t| \cos t) dt =$

The value of the definite integral $\int_{\pi / 24}^{5 \pi / 24} \frac{d x}{1+\sqrt[3]{\tan 2 x}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module