Find $n$ such that $\int_1^n \lfloor{x}\rfloor\lfloor{\sqrt x}\rfloor dx>60$ 
Find the smallest positive integer $n$ such that $$\int_1^n \lfloor{x}\rfloor\lfloor{\sqrt x}\rfloor dx>60$$

where $\lfloor.\rfloor$ is the GIF. I couldn't find any decent method rather than explicitly evaluating it to a few terms by brute force.
Is there a better method out?

The function $f(x):=\lfloor x\rfloor\cdot\lfloor\sqrt{x}\rfloor$ satisfies $$f(x)=\left\{\eqalign{\lfloor x\rfloor\qquad&(1\leq x<4)\cr 2\lfloor x\rfloor\qquad&(4\leq x<9)\ .\cr}\right.$$
It follows that
$$\int_1^8 f(x)\>dx=1+2+3+8+10+12+14=50\>, $$
and $\int_1^9f(x)\>dx=50+\int_8^9f(x)\>dx=66$.
The answer to your question therefore is $9$. 
(Remark: If you had written $6382$ instead of $60$ I would have  set up a general scheme $\ldots$)
