Try this problem from ISI-MSQMS 2018 which involves the concept of Real numbers, sequence and series and Definite integral.
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?
REAL NUMBERS
REIMANN INTEGRATION
SEQUENCE AND SERIES
Answer:(d) $I<\frac{2}{3}$ and $J<2$
ISI 2018|MSQMS |QMA|PROBLEM 22
INTRODUCTION TO REAL ANALYSIS :BARTLE SHERBERT
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 $
$ \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 $<$2$
$J$<$2$
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}$
$I$<$\frac{2}{3}$
Therefore,$I<\frac{2}{3}$ and $J<2$
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