Let $f$ continuous on $[0,1]$, $f(0)=0$ and $f'(0)$ exists. Show that $f(x)x^{-3/2}$ is integrable on $[0,1]$.
($f(x)x^{-3/2}$ integrable if and only if $|f(x)|x^{-3/2}$ integrable on $[0,1]$. )
I have this. $f'(0)$ exists then $\lim_{x\to 0^{+}} \frac{f(x)}{x}=L$ then forall $\epsilon>0$, exists $\delta>0$ such that $|x|<\delta$ implies $|\frac{f(x)}{x}-L|<\epsilon.$ Now, $|x|<\delta$ implies $|\frac{f(x)}{x}|\leq |L|+\epsilon$, implies $|f(x)|\leq (|L|+\epsilon)x$. Therefore, on $[0,\delta]$, $\int_{0}^{\delta}|f(x)|x^{-3/2}\,dx\leq (|L|+\epsilon)\int_{0}^{\delta}x^{1+-3/2}\,dx\leq (|L|+\epsilon)\int_{0}^{1} x^{-1/2}\,dx=(|L|+\epsilon)2<\infty$
Therefore, $\int_{0}^{1} |f(x)|x^{-3/2}=\int_{0}^{\delta}|f(x)|x^{-3/2}\,dx+\int_{\delta}^{1}f(x)x^{-3/2}\,dx $with $\int_{0}^{\delta}|f(x)|x^{-3/2}\,dx<\infty$.
Why $\int_{\delta}^{1}f(x)x^{-3/2}\,dx<\infty$?
$f$ is continuous on $[0,1]$, so $x\mapsto f(x)x^{-3/2}$ is continuous on $[\delta,1]$, so integrable on $[\delta,1]$.