To compare an integral with an identity

[Hummingbird25]

Integral inequality and comparison

Homework Statement

Prove the inequality

\frac{2}{(n+1) \cdot \pi} \leq a_n \leq \frac{2}{n \pi}}

where a_n = \int_{0}^{\pi} \frac{sin(x)}{n \cdot \pi +x} dx

and n \geq 1

The Attempt at a Solution

Proof:

If n increased the left side of inequality because smaller every time it increases while the integral becomes larger, then by camparison the right side of the inequality always will become larger, than the integral.

Is this proof enough?

Sincerely
Maria.

 

[sutupidmath]

well try to use the fact that

n\pi\leq \pi n+x \leq n\pi+\pi, then

\frac{1}{n\pi+\pi}\leq\frac{1}{n\pi+x}\leq\frac{1}{n\pi}, multiply both sides by sinx, we get, notice that x goes only from 0 to pi, so sinx is always positive, so

\frac{sin x}{n\pi+\pi}\leq\frac{sin x}{n\pi+x}\leq\frac{sin x}{n\pi}, now

\int_0^{\pi}\frac{sin x }{n\pi+\pi}dx\leq\int_0^{\pi}\frac{sin x}{n\pi+x}dx\leq\int_0^{\pi}\frac{sin x}{n\pi}dx,
now
\frac{1}{n\pi+\pi}\int_0^{\pi}sin xdx\leq\int_0^{\pi}\frac{sin x}{n\pi+x}dx\leq\frac{1}{n\pi}\int_0^{\pi}\sin xdx,
I think this will work!

 

[sutupidmath]

I shouldn't have done the whole thing for you, since you are on the homework forum, and you are supposed to show your work, however, i hope u got it now!

 

[Hummingbird25]

I shouldn't have done the whole thing for you, since you are on the homework forum, and you are supposed to show your work, however, i hope u got it now!

I get it now. I will present something that I have been working later I hope you will review.

 

[sutupidmath]

I get it now. I will present something that I have been working later I hope you will review.

well, someone will!