Try this problem from ISI-MSQMS 2018 which involves the concept of Real numbers, sequence and series and Definite integral.

DEFINITE INTEGRAL | ISI 2018| MSQMS | PART A | PROBLEM 22


Let $I=\int_{0}^{1} \frac{\sin x}{\sqrt{x}} d x$ and $J=\int_{0}^{1} \frac{\cos x}{\sqrt{x}} d x,$ then which of the following is
true?

  • (a) $I<\frac{2}{3}$ and $J>2$
  • (b) $I>\frac{2}{3}$ and $J<2$ (c) $I>\frac{2}{3}$ and $J>2$
  • (d) $I<\frac{2}{3}$ and $J<2$

Key Concepts


REAL NUMBERS

REIMANN INTEGRATION

SEQUENCE AND SERIES

Check the Answer


But try the problem first…

Answer:(d) $I<\frac{2}{3}$ and $J<2$

Source
Suggested Reading

ISI 2018|MSQMS |QMA|PROBLEM 22

INTRODUCTION TO REAL ANALYSIS :BARTLE SHERBERT

Try with Hints


First hint

We know when $f(x)$>$g(x)$

$\int \limits_a^bf(x)$>$\int \limits_a^bg(x)$

We know for $0<x<1$, $ \cos x <1 $

Second hint

$ \frac{\cos x}{\sqrt x}$< $\frac{1}{\sqrt x}$ implies $\int \limits_0^1\frac{\cos x}{\sqrt x}\mathrm dx$<$\int \limits_0^1\frac{1}{\sqrt x}\mathrm dx $

$\int \limits_0^1\frac{1}{\sqrt x}\mathrm dx = 2$

$\int \limits_0^1\frac{\cos x}{\sqrt x}\mathrm dx $<$2$

$J$<$2$

Third hint

Again we claim $x-s\sin x$>$0$ for $0 \leq x\leq 1$

Let $f(x)=x-\sin x$

$f'(x)=1-\cos x\geq 0$

hence $f(x)$ is monotonic increasing.

Therefore $x-\sin x $> $0$, $x\epsilon [0,1]$

So,$x$>$sinx$

$\sqrt x$ > $\frac{\sin x}{\sqrt x}$ $x\epsilon [0,1]$

integrating both sides with limits $0$ to $1$ we get;

$\int \limits_0^1\frac{\sin x}{\sqrt x} \mathrm dx $<$\frac{2}{3}$

$I$<$\frac{2}{3}$

Final Step

Therefore,$I<\frac{2}{3}$ and $J<2$

Subscribe to Cheenta at Youtube