Show that a definite integral is positive

[danago]

Show that $$\int_0^1 {x^n \sin \left( {\frac{{\pi x}}{2}} \right)} {\kern 1pt} dx > 0$$ for all $${\rm{n}} \ge {\rm{0}}$$

Basically, what i want to show is that over the interval (0,1), the integrand is above the x-axis, or positive. I will show that over the interval, both functions, $$x^n$$ and $$sin(\frac{\pi x}{2})$$ are both positive.

For the trig function, this is how i proceeded:
$$\begin{array}{l} 0 < x < 1 \\ 0 < \frac{{\pi x}}{2} < \frac{\pi }{2} \\ \sin 0 < \sin \left( {\frac{{\pi x}}{2}} \right) < \sin \left( {\frac{\pi }{2}} \right) \\ 0 < \sin \left( {\frac{{\pi x}}{2}} \right) < 1 \\ \end{array}$$

Is that a valid way of doing it?

For the xⁿ function, i took the same approach:

$$\begin{array}{l} 0 < x < 1 \\ 0^n < x^n < 1^n \\ 0 < x^n < 1 \\ \end{array}$$

Since, over the interval of (0,1) of which the function is being integrated, the integrand is positive, the definite integral will be positive for all values of n.

My main question though; Is my approach a good way of doing such a problem? It seemed to work fine for this question, but are there cases where i can't use such a method?

My book has this for an answer:

Let $$g(x)=x^n$$ and $$h(x)=sin(\frac{\pi x}{2})$$.

On the interval 0 < x < 1 , g(x)>0. On the same interval, h(x)>0, therefore, the definite integral is greater than zero.

That is basically the concept i used, but i really don't like that method, since it really doesn't show much, and in exams, depending on the marker, i may lose marks for such a bland proof, which is why I am wanting to use a more solid proof.

Thanks in advance,
Dan.

EDIT: Just realized that this method using the inequalities does not work for all cases. But it should work if the function is strictly increasing, or strictly decreasing over the interval, right?

 

[nrqed]

Show that $$\int_0^1 {x^n \sin \left( {\frac{{\pi x}}{2}} \right)} {\kern 1pt} dx > 0$$ for all $${\rm{n}} \ge {\rm{0}}$$

Basically, what i want to show is that over the interval (0,1), the integrand is above the x-axis, or positive. I will show that over the interval, both functions, $$x^n$$ and $$sin(\frac{\pi x}{2})$$ are both positive.

For the trig function, this is how i proceeded:
$$\begin{array}{l} 0 < x < 1 \\ 0 < \frac{{\pi x}}{2} < \frac{\pi }{2} \\ \sin 0 < \sin \left( {\frac{{\pi x}}{2}} \right) < \sin \left( {\frac{\pi }{2}} \right) \\ 0 < \sin \left( {\frac{{\pi x}}{2}} \right) < 1 \\ \end{array}$$

Is that a valid way of doing it?

No. Consider using your logic to get the following obviosuly wrong result:


$$\begin{array}{l} 0 < x < 1 \\ 0 < \pi x < \pi \\ \sin 0 < \sin \left( \pi x\right) < \sin \left( \pi \right) \\ 0 < \sin \left( \pi x \right) < 0 \\ \end{array}$$

Don't try to do a proof with usual manipulation of inequalities when you have trig functions like this because trig functions are not monotonous, they increase and then decrease and so on.

You should simply state it as an obvious fact that for an argument between 0 and pi/2, the sin function is always positive.