# Finding definite integral (trigonometric)

## Homework Statement

Hi guys ,, i have the following question (it's in the attachment) :
Find m and M such that m <= x sin x <= M if 0 <=x <= pi. (Any reasonably good bounds will do,
I am not asking for the best possible bounds.)
Hence find bounds on the value of the integral[x sin(x),0 to pi]

## The Attempt at a Solution

i have a formal solution (it's in the attachment) but the problem is i don't understand the question. Before i saw the question ,, i tried to solved it like this :
since (0 <=x <= pi) then 0 <= x sin x <= 0 ( i solved for zero and pi) ,, and stopped there.
can anyone tell me the problem in a way that i can understand ,, and in the answer it says "This is certainly not the best you could do, but it is definitely a bound" what does he mean ? is there many other ways to solve it ?? . (Thanks in advanced)

#### Attachments

• Question.JPG
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can anyone tell me the problem in a way that i can understand

If you graph the three functions y=x*sin(x), y=x, and y=-x using a window of xmin=0, xmax=30, ymin=-30, ymax=30, then I bet you will understand.

If you're still puzzled, just graph the two functions y=abs(x*sin(x)) and y=x using the same window. Ask again if you are still confused after graphing.

yep ,, got it :D ,, thanks very much mate,, but there is something else ,, i can get other bounds right ?? such as :
0<= integral[x*sin(x),0 to pi]<= ((pi)^2)/2 ,, is this right ?

yep ,, got it :D ,, thanks very much mate,, but there is something else ,, i can get other bounds right ?? such as :
0<= integral[x*sin(x),0 to pi]<= ((pi)^2)/2 ,, is this right ?

Make it 0<= | integral[x*sin(x),0 to pi] | <= ((pi)^2)/2 (with absolute value) and I'll agree.

Edited to add: Oh, never mind. x *sin(x) is positive there. Good job.

lol ,, thanks :D