# Science:Math Exam Resources/Courses/MATH103/April 2017/Question 01 (c)/Solution 1

(i) Since ${\displaystyle f(x)={\frac {1}{x^{3}}}}$ is discontinuous at ${\displaystyle x=0}$ and unbounded in the sense of ${\displaystyle \lim _{x\to 0^{+}}{\frac {1}{x^{3}}}=+\infty }$, we can't directly evaluate this integral. Instead, we can write it as a limit and then get the value as follows.

${\displaystyle \int _{0}^{1}{\frac {1}{x^{3}}}dx=\lim _{a\to 0}\int _{a}^{1}{\frac {1}{x^{3}}}dx=\lim _{a\to 0}-{\frac {1}{2}}x^{-2}{\bigg |}_{x=a}^{x=1}=\lim _{a\to 0}-{\frac {1}{2}}\left(1-{\frac {1}{a^{2}}}\right)=\lim _{a\to 0}{\frac {1}{2a^{2}}}-{\frac {1}{2}}=+\infty .}$

Therefore, the integral diverges.

(ii) We notice that ${\displaystyle {\frac {1}{x(x+1)}}\leq {\frac {1}{x^{2}}}}$ for all positive x. The series converges by comparison to the integral ${\displaystyle \int _{e}^{\infty }{\frac {1}{x^{2}}}}$.

(iii) For any ${\displaystyle x>0}$, we have ${\displaystyle \ln x which implies that ${\displaystyle {\sqrt {\ln x}}<{\sqrt {x}}}$, therefore, ${\displaystyle {\frac {\sqrt {\ln(x)}}{x^{2}}}<{\frac {\sqrt {x}}{x^{2}}}={\frac {1}{x^{1.5}}}}$. Therefore, our integral converges by comparison with the integral ${\displaystyle \int _{1}^{\infty }{\frac {1}{x^{1.5}}}}$.

(iv) Since ${\displaystyle {\frac {\sqrt {x^{3}+1}}{x^{2}}}\leq {\frac {\sqrt {x^{3}}}{x^{2}}}}$, which simplifies to ${\displaystyle {\frac {x^{3/2}}{x^{2}}}}$ and further to ${\displaystyle {\frac {1}{x^{1/2}}}}$, the integral diverges by comparison to the integral ${\displaystyle \int _{1}^{\infty }{\frac {1}{x^{1/2}}}}$.

${\displaystyle \color {blue}Answer:}$

converging diverging
(i) ${\displaystyle \int _{0}^{1}{\frac {1}{x^{3}}}dx}$ no yes
(ii) ${\displaystyle \int _{e}^{\infty }{\frac {1}{x(x+1)}}dx}$ yes no
(iii) ${\displaystyle \int _{1}^{\infty }{\frac {\sqrt {\ln(x)}}{x^{2}}}dx}$ yes no
(iv) ${\displaystyle \int _{1}^{\infty }{\frac {\sqrt {x^{3}+1}}{x^{2}}}dx}$ no yes