Consider the integral I = int_{0}^{10} frac{[ | vedclass

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

Question

(C) The integral is given by $I = \int_{0}^{10} [x] e^{[x]-x+1} dx$. Since $[x] = n$ for $x \in [n, n+1)$,we can split the integral as: $I = \sum_{n=0

Vedclass · Accepted Answer

$45(e-1)$

Vedclass · Answer

(C) The integral is given by $I = \int_{0}^{10} [x] e^{[x]-x+1} dx$. Since $[x] = n$ for $x \in [n, n+1)$,we can split the integral as: $I = \sum_{n=0}^{9} \int_{n}^{n+1} n \cdot e^{n-x+1} dx$. Evaluating the integral for each term: $I = \sum_{n=0}^{9} n \left[ -e^{n-x+1} ight]_{n}^{n+1} = \sum_{n=0}^{9} n \left( -e^{0} + e^{1} ight) = (e-1) \sum_{n=0}^{9} n$. Using the sum formula $\sum_{n=0}^{9} n = \frac{9 	imes 10}{2} = 45$,we get: $I = 45(e-1)$.

Consider the integral $I = \int_{0}^{10} \frac{[x] e^{[x]}}{e^{x-1}} dx$,where $[x]$ denotes the greatest integer less than or equal to $x$. Then the value of $I$ is equal to:

Similar Questions

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin (x-[x]) \, dx=$

Define a function $f: R \rightarrow R$ by $f(x) = \max \{|x|, |x-1|, \ldots, |x-2n|\}$,where $n$ is a fixed natural number. Then,$\int_0^{2n} f(x) dx$ is

The value of $12 \int \limits_0^3 \left| x^2 - 3x + 2 \right| dx$ is $.............$

The value of $\int_{3}^{4} \sqrt{(4-x)(x-3)} d x$ is

The integral $\int_0^{\frac{\pi}{4}} \frac{136 \sin x}{3 \sin x+5 \cos x} dx$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module