Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Definite Integration

Question:

The value $\int\limits_1^a[x] f'(x) d x, a>1$, where $[x]$ denotes the greatest integer not exceeding $x$, is

Options:

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

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

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

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

Correct Answer:

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

Explanation:

Suppose $n \leq a<n+1$, where $n \in N$. Then, $[a]=n$.

∴  $I=\int\limits_1^a[x] f'(x) d x$

$\Rightarrow I=\sum\limits_{r=1}^{n-1} \int\limits_r^{r+1}[x] f'(x) d x+\int\limits_n^a[x] f'(x) d x$

$\Rightarrow I=\sum\limits_{r=1}^{n-1} \int\limits_r^{r+1} r f'(x) d x+\int\limits_n^a n f'(x) d x$

$\Rightarrow I=\sum\limits_{r=1}^{n-1} r[f(x)]_r^{r+1}+n[f(x)]_n^a$

$\Rightarrow I=\sum\limits_{r=1}^{n-1} r\{f(r+1)-f(r)\}+n\{f(a)-f(n)\}$

$\Rightarrow I=\{f(2)-f(1)\}+2\{f(3)-f(2)\}+3\{f(4)-f(3)\}+...+(n-1)\{f(n)-f(n-1)\}+n\{f(a)-f(n)\}$

$\Rightarrow I=-f(1)-f(2)-f(3)-f(4)-...-f(n-1)-f(n)+n f(a)$

$\Rightarrow I=[a] f(a)-\{f(1)+f(2)+f(3)+...+f[a]\}$