a 1, ; int_{1}^{a} [x] f'(x) dx = | vedclass

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

Question

(B) Let $n = [a]$,where $n$ is an integer such that $n \leq a < n+1$.The integral can be split as:$\int_{1}^{a} [x] f'(x) dx = \int_{1}^{2} 1 f'(x) dx

Vedclass · Accepted Answer

$[a] f(a) - \{f(1) + f(2) + \dots + f([a])\}$

Vedclass · Answer

(B) Let $n = [a]$,where $n$ is an integer such that $n \leq a The integral can be split as:$\int_{1}^{a} [x] f'(x) dx = \int_{1}^{2} 1 f'(x) dx + \int_{2}^{3} 2 f'(x) dx + \dots + \int_{n-1}^{n} (n-1) f'(x) dx + \int_{n}^{a} n f'(x) dx$Evaluating each integral:$= 1[f(2) - f(1)] + 2[f(3) - f(2)] + \dots + (n-1)[f(n) - f(n-1)] + n[f(a) - f(n)]$Rearranging the terms:$= -f(1) - f(2) - f(3) - \dots - f(n) + n f(a)$Since $n = [a]$,we get:$= [a] f(a) - \{f(1) + f(2) + \dots + f([a])\}$

$a > 1, \; \int_{1}^{a} [x] f'(x) dx = $

Similar Questions

If $f : R \to R$ is a continuous function such that $f(x) = \int\limits_1^x {tf(t)dt}$,then the correct statement is -

Considering four sub-intervals,the value of $\int_{0}^{1} \frac{1}{1+x} d x$ by Trapezoidal rule,is

$\int_2^3 \frac{d x}{x^2-x}$ is equal to

If $I$ is the greatest of $I_1=\int_0^1 e^{-x} \cos ^2 x \, dx, I_2=\int_0^1 e^{-x^2} \cos ^2 x \, dx, I_3=\int_0^1 e^{-x^2} \, dx, I_4=\int_0^1 e^{-x^2 / 2} \, dx$, then

$\int_{0}^{\pi} \sqrt{\frac{1 + \cos 2x}{2}} \, dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module