In fact, it is very simple.

Calculus Level 4

Evaluate the following integral. $\int_{8}^{13} \dfrac{\lfloor x^2 \rfloor}{\lfloor x^2 \rfloor + \lfloor x^2-42x+441 \rfloor}dx$

Notation: $\lfloor .. \rfloor$ denotes floor function .

2 solutions

Prasun Biswas
Nov 21, 2014

Let $\displaystyle \large \int_8^{13} \frac{\lfloor x^2 \rfloor}{\lfloor x^2 \rfloor + \lfloor x^2-42x+441 \rfloor} dx = I \ldots (i)$

Using the property $\displaystyle \large \int_a^b f(x) dx = \int_a^b f(a+b-x) dx$ , and a little calculation, we get,

$\displaystyle \large \int_8^{13} \frac{\lfloor x^2-42x+441 \rfloor}{\lfloor x^2 \rfloor + \lfloor x^2-42x+441 \rfloor} dx = I \ldots (ii)$

Adding (i) and (ii), we get,

$\displaystyle \large 2I=\int_8^{13} \frac{\lfloor x^2 \rfloor + \lfloor x^2-42x+441 \rfloor}{\lfloor x^2 \rfloor + \lfloor x^2-42x+441 \rfloor} dx$

$\implies \displaystyle \large 2I=\int_8^{13} dx$

$\implies \displaystyle \large 2I= [x]_8^{13} = 13-8 = 5$

$\implies \displaystyle \large I=\frac{5}{2} = \boxed{2.5}$

you can save your time in calculation if you in your mind orally thinks that:

$(x-21)^2 = x^2 - 42x + 441$ .

Karan Shekhawat - 6 years, 6 months ago

A Former Brilliant Member
Feb 3, 2017

highly-highly-highly overrated problem