# Basic Maths - Grand Test 1 -Q14

Which of the following integrals is/are convergent ?

$(1).\int_{0}^{a} \frac{1}{x^3} dx\text{ } (a>0)$                                $(2).\int_{0}^{a} \frac{1}{\sqrt x} dx\text{ } (a>0)$

$(3).\int_{0}^{\infty} \frac{1}{x\log x} dx\text{ } (a>0)$                       $(4).\int_{0}^{\infty} \frac{1}{x(\log x)^2} dx\text{ } (a>0)$

$(A).(2) \text{ and } (4)$

$(B).(1) \text{ and } (2)$

$(C).(2) \text{ and } (3)$

$(C).(3) \text{ and } (4)$

reshown Jun 23

+1 vote

$(1) \int_{0}^{a} \frac{1}{x^3} dx = \left [ -\frac{x^{-2}}{2} \right ]^a_0 = \left [ -\frac{1}{2x^2} \right ]^a_0 = \infty \longrightarrow divergent$

$(2) \int_{0}^{a} \frac{1}{\sqrt x} dx = \left [ \frac{x^{1/2}}{1/2} \right ]^a_0 = 2\sqrt a \text{ (finite)}\longrightarrow convergent$

$(3) \int_{a}^{\infty} \frac{1}{x\log x} dx = \int_{a}^{\infty} \frac{1/x}{\log x} dx = \left [ \log(\log x) \right ]^\infty_a = \infty \longrightarrow divergent$

$(4) \int_{a}^{\infty} \frac{1}{x(\log x)^2} dx$

$\text{Let } \log x =t \implies x = e^t \implies dx = e^t dt$

$\text{if } x = a \implies t = \log a, x = \infty\implies t = \infty$

$\int_{a}^{\infty} \frac{1}{x(\log x)^2} dx =\int_{\log a}^{\infty} \frac{e^t}{e^t t^2} dx = -\left [\frac{1}{t} \right ]_{\log a}^\infty = \frac{1}{\log 5} \text{ finite so convergent}$

(2) and (4) are convergent so Option A

answered Jun 24 by (21,290 points)