(adsbygoogle = window.adsbygoogle || []).push({}); Show that [tex]\int_0^1 {x^n \sin \left( {\frac{{\pi x}}{2}} \right)} {\kern 1pt} dx > 0[/tex] for all [tex]{\rm{n}} \ge {\rm{0}}[/tex]

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, [tex]x^n[/tex] and [tex]sin(\frac{\pi x}{2})[/tex] are both positive.

For the trig function, this is how i proceeded:

[tex]

\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}

[/tex]

Is that a valid way of doing it?

For the x^{n}function, i took the same approach:

[tex]

\begin{array}{l}

0 < x < 1 \\

0^n < x^n < 1^n \\

0 < x^n < 1 \\

\end{array}

[/tex]

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 cant use such a method?

My book has this for an answer:

Let [tex]g(x)=x^n[/tex] and [tex]h(x)=sin(\frac{\pi x}{2})[/tex].

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 dont like that method, since it really doesnt show much, and in exams, depending on the marker, i may lose marks for such a bland proof, which is why im wanting to use a more solid proof.

Thanks in advance,

Dan.

EDIT: Just realised 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?

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# Homework Help: Show that a definite integral is positive

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