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

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

Question
The value of ∫0n2[x ] dx\displaystyle\int_{0}^{n^{2}}\big[\sqrt{x}\,\big]\,dx, where [ ⋅ ][\,\cdot\,] denotes the greatest integer function and n∈Nn\in\mathbb{N}, is
An2(n+1)24\dfrac{n^{2}(n+1)^{2}}{4}
B16 n(n−1)(4n+1)\dfrac{1}{6}\,n(n-1)(4n+1)correct
C∑n2\displaystyle\sum n^{2}
Dn(n+1)(n+2)6\dfrac{n(n+1)(n+2)}{6}
Solution
Step 1: Break [0,n2][0,n^{2}] at the perfect squares. On each subinterval [k2,(k+1)2)\big[k^{2},(k+1)^{2}\big) the value x\sqrt{x} lies in [k,k+1)[k,k+1), so [x]=k\big[\sqrt{x}\big]=k. The length of this subinterval is (k+1)2−k2=2k+1(k+1)^{2}-k^{2}=2k+1. Step 2: Sum the contributions for k=0,1,2,…,n−1k=0,1,2,\dots,n-1 (the upper endpoint x=n2x=n^{2} contributes nothing as a single point):
∫0n2[x] dx=∑k=0n−1k (2k+1)=∑k=0n−1(2k2+k).\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 ∑k=0n−1k2=(n−1)n(2n−1)6\displaystyle\sum_{k=0}^{n-1}k^{2}=\dfrac{(n-1)n(2n-1)}{6} and ∑k=0n−1k=(n−1)n2\displaystyle\sum_{k=0}^{n-1}k=\dfrac{(n-1)n}{2}:
∑k=0n−1(2k2+k)=2⋅(n−1)n(2n−1)6+(n−1)n2=(n−1)n[2n−13+12].\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:
2n−13+12=2(2n−1)+36=4n+16 ⇒ ∫0n2[x] dx=16 n(n−1)(4n+1).\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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