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

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

af( a ) - \ f( 1 ) + f( 2 ) + .............f( [ a ] ) \

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

af( [ a ] ) - \ f( 1 ) + f( 2 ) + .............f( a ) \

Definite Integration PYQ — JEE Main 2006 · Shift 0 · Q54 with Solution

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Correct answer: B

Let $n=[a].$ Then since $a>1$ , we split the integral over intervals where $[x]$ is constant:

I&=\int_1^a [x]f'(x)\,dx \\ &=\int_1^2 1\cdot f'(x)\,dx+\int_2^3 2\cdot f'(x)\,dx+\cdots+\int_n^a n\cdot f'(x)\,dx, \end{aligned}$$ where the last interval is $[n,a]$ if $a$ is not an integer, and this still works when $a=n$. 2. Evaluate each piece: $$\int_k^{k+1} k f'(x)\,dx = k\bigl(f(k+1)-f(k)\bigr), \quad k=1,2,\dots,n-1,$$ and $$\int_n^a n f'(x)\,dx = n\bigl(f(a)-f(n)\bigr).$$ So, $$I=\sum_{k=1}^{n-1} k\bigl(f(k+1)-f(k)\bigr)+n\bigl(f(a)-f(n)\bigr).$$ 3. Expand the sum: $$\begin{aligned} I&=\bigl(f(2)-f(1)\bigr)+2\bigl(f(3)-f(2)\bigr)+\cdots +(n-1)\bigl(f(n)-f(n-1)\bigr)+n\bigl(f(a)-f(n)\bigr). \end{aligned}$$ Collect coefficients of each term: - coefficient of $f(1)$ is $-1$ - coefficient of $f(2)$ is $1-2=-1$ - coefficient of $f(3)$ is $2-3=-1$ - ... - coefficient of $f(n)$ is $(n-1)-n=-1$ - coefficient of $f(a)$ is $n$ Hence, $$I=n f(a)-\{f(1)+f(2)+\cdots+f(n)\}.$$ Since $n=[a]$, $$\boxed{\int_1^a [x]f'(x)\,dx=[a]f(a)-\bigl(f(1)+f(2)+\cdots+f([a])\bigr).}$$ 4. Compare with the options: This matches **Option B**: $$[a]f(a)-\left\{f(1)+f(2)+\cdots+f([a])\right\}.$$

View written solutionFree

Correct answer: B

Let $n=[a].$ Then since $a>1$ , we split the integral over intervals where $[x]$ is constant:

I&=\int_1^a [x]f'(x)\,dx \\ &=\int_1^2 1\cdot f'(x)\,dx+\int_2^3 2\cdot f'(x)\,dx+\cdots+\int_n^a n\cdot f'(x)\,dx, \end{aligned}$$ where the last interval is $[n,a]$ if $a$ is not an integer, and this still works when $a=n$. 2. Evaluate each piece: $$\int_k^{k+1} k f'(x)\,dx = k\bigl(f(k+1)-f(k)\bigr), \quad k=1,2,\dots,n-1,$$ and $$\int_n^a n f'(x)\,dx = n\bigl(f(a)-f(n)\bigr).$$ So, $$I=\sum_{k=1}^{n-1} k\bigl(f(k+1)-f(k)\bigr)+n\bigl(f(a)-f(n)\bigr).$$ 3. Expand the sum: $$\begin{aligned} I&=\bigl(f(2)-f(1)\bigr)+2\bigl(f(3)-f(2)\bigr)+\cdots +(n-1)\bigl(f(n)-f(n-1)\bigr)+n\bigl(f(a)-f(n)\bigr). \end{aligned}$$ Collect coefficients of each term: - coefficient of $f(1)$ is $-1$ - coefficient of $f(2)$ is $1-2=-1$ - coefficient of $f(3)$ is $2-3=-1$ - ... - coefficient of $f(n)$ is $(n-1)-n=-1$ - coefficient of $f(a)$ is $n$ Hence, $$I=n f(a)-\{f(1)+f(2)+\cdots+f(n)\}.$$ Since $n=[a]$, $$\boxed{\int_1^a [x]f'(x)\,dx=[a]f(a)-\bigl(f(1)+f(2)+\cdots+f([a])\bigr).}$$ 4. Compare with the options: This matches **Option B**: $$[a]f(a)-\left\{f(1)+f(2)+\cdots+f([a])\right\}.$$