Definite IntegrationhardFree

∫[√x] dx from 0 to n² (greatest integer) | JEE

JEE Maths question with a full step-by-step solution.

Question

The value of

\displaystyle\int_{0}^{n^{2}}\big[\sqrt{x}\,\big]\,dx

, where

[\,\cdot\,]

denotes the greatest integer function and

n\in\mathbb{N}

, is

\dfrac{n^{2}(n+1)^{2}}{4}

\dfrac{1}{6}\,n(n-1)(4n+1)

correct

\displaystyle\sum n^{2}

\dfrac{n(n+1)(n+2)}{6}

Solution

Step 1: Break

[0,n^{2}]

at the perfect squares. On each subinterval

\big[k^{2},(k+1)^{2}\big)

the value

\sqrt{x}

lies in

[k,k+1)

, so

\big[\sqrt{x}\big]=k

. The length of this subinterval is

(k+1)^{2}-k^{2}=2k+1

. Step 2: Sum the contributions for

k=0,1,2,\dots,n-1

(the upper endpoint

x=n^{2}

contributes nothing as a single point):

\int_{0}^{n^{2}}\big[\sqrt{x}\big]\,dx=\sum_{k=0}^{n-1}k\,(2k+1)=\sum_{k=0}^{n-1}\big(2k^{2}+k\big).

Step 3: Use the standard sums

\displaystyle\sum_{k=0}^{n-1}k^{2}=\dfrac{(n-1)n(2n-1)}{6}

and

\displaystyle\sum_{k=0}^{n-1}k=\dfrac{(n-1)n}{2}

\sum_{k=0}^{n-1}\big(2k^{2}+k\big)=2\cdot\dfrac{(n-1)n(2n-1)}{6}+\dfrac{(n-1)n}{2}=(n-1)n\left[\dfrac{2n-1}{3}+\dfrac12\right].

Step 4: Combine the bracket over a common denominator:

\dfrac{2n-1}{3}+\dfrac12=\dfrac{2(2n-1)+3}{6}=\dfrac{4n+1}{6}\ \Rightarrow\ \int_{0}^{n^{2}}\big[\sqrt{x}\big]\,dx=\dfrac{1}{6}\,n(n-1)(4n+1).

Correct answer: (2)

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