Cfrtjryk

1. $$int_0^{0.5} cos(x^2),dx$$


2. $$int_0^{0.5} cos(sqrt{x}),dx$$

If $0le xle 0.5$, $x^2lesqrt{x}impliescos x^2ge cossqrt{x}$, so the first integral is greater.

At DavidG's suggestion, I'll mention the $implies$ uses the fact that $cos x$ decreases on this interval.

Make a change of variables $x^2=u$ in the first integral (hence $dx=frac{1}{2}u^{-1/2}du$) and $sqrt x=u$ in the second integral (hence $dx=2udu$). Note that the integration limits change, but it is then easy to find the solution. Isn't?

The integral with $cos x^2$ is slightly bigger than the one with $cos sqrt x$, simply for the property that x got a bigger exponent; you can verify this even on software like Wolfram Alpha Online.

edit: for the range considered this is true

Note that $sqrt{0.5} < pi/2$ and $0.5^2 < pi/2$. So, $cos(x^2)$ and $cos(sqrt{x})$ will be positive and decreasing for $xin(0,0.5)$. Now note that $x^2$ is increasing less quickly than $sqrt{x}$ on $xin(0,0.5)$. In a sense this verifies $0 < cos(sqrt{x}) < cos(x^2)$ for all $xin(0,0.5)$. Hence, the integral of $cos(x^2)$ will be greater than $cos(sqrt{x})$ over the interval $(0,0.5)$.

As an extra notion, it is not too difficult to extend this idea for the interval $(0,sqrt{pi/2})$. Going a bit further than this would require more work.

Hope this intuitive approach helps. If you have any questions feel free to ask them!

Just a small addition:

How do we know that for each $0le xle 0.5$ , $x^2lesqrt{x}$ ?

Two ways:

1. Plot the graphs of the two functions on the same coordinate system.
   You will see immediately that the only point the graphs cross each other is $(x=0, y=0)$.

For each point other than $x=0$ you'll find out that $x^2lesqrt{x}$ .

2. The easier way (or the "brutal" one...) :

Just assign the two edges of the range in the inequality equation:

Take a calculator.

Assign $x=0$ . You'll get an equality.

But if you assign higher values, including $x=0.5$ , you'll get through the values given by your calculator are higher for $sqrt{x}$ than the values of $x^2$.