Showing an Integral is increasing on a certain interval

Question

I'm having trouble with a proof showing a particular function is increasing on an interval. The particular function is an integral, the function being: $f(x) = \int_0^x sin(t^2)dt$. The interval in question is $[0, \sqrt{\pi}]$. I am really not sure where to start to show this. I considered if showing the integral was positive on this interval would suffice but of course it wouldn't since it could still be positive even if it decreased just slightly at some point in the interval and then resumed increasing. Unless I am overthinking it and this in fact does suffice. Any insight or guidance in how to start this proof would be appreciated.

Answer 1

Intuitively, the function $\sin{t^2}$ is positive on $[0, \sqrt{\pi}]$, so as $x$ increases in that interval, the area from $0$ to $x$ should increase. To prove this formally, you can use this result:

If the derivative of a function is positive on an interval, then the function is increasing.

Proof: Let $f$ be continuous on $[a, b]$ and differentiable on $(a,b)$, so $f$ is differentiable on the interval $(x,y)$ for all $x, y $, $y > x$ in $(a,b)$. By the mean value theorem, there exists $t$ in $(x,y)$ such that $$\frac{f(y) - f(x)}{y-x} = f'(t) > 0$$ so $$ f(y) > f(x)$$ and $f$ is increasing.

Now use the fundamental theorem of calculus on $f(x) = \int_0^x \sin{t^2dt}$ for $x$ in $[0, \sqrt{\pi}]$ and the result follows.