3 votes 3 votes For real $\alpha$, the value of $\int_{\alpha}^{\alpha+1} [x]dx$, where $[x]$ denotes the largest integer less than or equal to $x$, is $\alpha$ $[\alpha]$ $1$ $\dfrac{[\alpha] + [\alpha +1]}{2}$ Calculus isi2014-dcg calculus integration definite-integral + – Arjun asked Sep 23, 2019 • retagged Nov 12, 2019 by Lakshman Bhaiya Arjun 581 views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply `JEET commented Nov 19, 2019 reply Follow Share B? 0 votes 0 votes SatyamK commented Mar 28, 2020 i edited by SatyamK Mar 28, 2020 reply Follow Share Case 1:$\alpha$>=0 Let α=1.23, α+1=2.23, =$\int_{1.23}^{2} x dx$ +$\int_{2}^{2.23} x dx$ =$\int_{1.23}^{2} 1$ + $\int_{2}^{2.23} 2 dx$ ($ \because$ floor between $1.23, 2 = 1$, similarly floor of 2 and 2.23 is 2. ) = 0.77 + 2*(0.23) =0.77+0.46=1.23 ($\alpha$) Case 2:$\alpha$<0 Let α=-0.2, α+1=0.8, = $\int_{-0.2}^{0} x dx$+ $\int_{0}^{0.8} x dx$ = $\int_{-0.2}^{0} -1 dx$ + $\int_{0}^{0.8} 0 dx$ = -1(0-(-0.2) + 2*(0) =0.2 ($\alpha$) In both case, $\int_{\alpha}^{\alpha +1} x dx$ = $\alpha$ Answer : A. 2 votes 2 votes Please log in or register to add a comment.
0 votes 0 votes Hope, it helps. text might be small! neeraj_bhatt answered Sep 9, 2020 neeraj_bhatt comment Share Follow See all 0 reply Please log in or register to add a comment.