The value of $\int_{1}^{a}[x] f^{\prime}(x) d x, a>1$, where $[x]$ denotes the greatest integer not exceeding $x$ is

(B) $[\mathrm{a}] \mathrm{f}(\mathrm{a})-\{\mathrm{f}(1)+\mathrm{f}(2)+\ldots+\mathrm{f}([\mathrm{a}])\}$

(A) $\mathrm{af}(\mathrm{a})-\{\mathrm{f}(1)+\mathrm{f}(2)+\ldots+\mathrm{f}([\mathrm{a}])\}$

(C) $[\mathrm{a}] \mathrm{f}([\mathrm{a}])-\{\mathrm{f}(1)+\mathrm{f}(2)+\ldots+\mathrm{f}(\mathrm{a})\}$

(D) $\mathrm{af}([\mathrm{a}])-\{\mathrm{f}(1)+\mathrm{f}(2)+\ldots+\mathrm{f}(\mathrm{a})\}$

JEE Main · 2006 · 30 Apr 2006 · Offline · Definite Integrals

The value of

\int_{1}^{a} [x] f^{'} (x) d x, a > 1

, where

[x]

denotes the greatest integer not exceeding

x

Answer: (B)

Split the range at the integers, because

[x]

is constant on each unit interval. Writing

[a] = k

, the integral is

\sum_{j = 1}^{k - 1} \int_{j}^{j + 1} j f^{'} (x) d x + \int_{k}^{a} k f^{'} (x) d x

. Each piece integrates to

j [f (j + 1) - f (j)]

for

j = 1, \dots, k - 1

and to

k [f (a) - f (k)]

for the last. When these are added the sum telescopes, leaving

k f (a) - [f (1) + f (2) + \dots + f (k)]

. Since

k = [a]

, the value is

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

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option