Question

For a real number $x,$ let $\left[x\right]$ denote the largest integer less than or equal to $x$. The smallest positive integer $n$ for which the integral ${\int }_1^n\left[x\right]\left[\sqrt{x}\right]dx$ exceeds $60$ is:

• A. $8$
• B. $9$
• C. $10$
• D. $\left[{60}^{\frac{2}{3}}\right]$

Answer 1

Answer: (B)

$9$

Let us evaluate the integral $I_k=\underset{k}{\overset{k+1}{\int }}\left[x\right]\left[\sqrt{x}\right]dx$ for $k\in Z$

By the properties of $[.]$, we know that $\left[x\right]=k$ for $x\in [k,k+1)$

Similarly, $\left[\sqrt{x}\right]=\left[\sqrt{k}\right]$ for $x\in [k,k+1)$

Thus, $I_k=k\left[\sqrt{k}\right]$

Therefore, ${\int }_1^n\left[x\right]\left[\sqrt{x}\right]dx=\sum _{k=1}^{n-1}{I}_k$

First few $I_k$ are listed below.

$I_1=1I_2=2I_3=3I_4=8I_5=10I_6=12I_7=14I_8=16I_9=27$.

Observe that $\sum _{k=1}^8{I}_k>60$

Thus, least $n-1$ possible is $8$

$n=9$