Question

We have the function of f(x)= sin(x^2)for x from 0 to Pi. Find a one-unit interval so the area under between sin(x^2)and x-axis is greatest. We have the function of f(x)= sin(x^2)for x from 0 to Pi. Find a one-unit interval so the area under between sin(x^2)and x-axis is greatest. @Mathematics

Answer 1

we need to integrate f(x) over some unit interval. lets say it starts at "a", so:\[\int\limits_{a}^{a+1}\sin x^{2}dx=\left[\frac{x}{2}-\frac{\sin 2x}{4}\right]_a^{a+1}\]\[=\frac{a+1}{2}-\frac{\sin 2(a+1)}{4}-\frac{a}{2}+\frac{\sin 2a}{4}=\frac{\sin(2a+2)-\sin2a-2}{4}\]so I assume what you need to do is find a maximum for this expression.
will you be able to continue from here?

Answer 2

But how will we find maximum for this?

Answer 3

how would you find the maximum of:\[y=3-x^{2}\]
use the same principal.

Answer 4

Should we find derivative of this and put it equal to zero?

Answer 5

bingo!

Answer 6

I did that but I did not get the correct answer:(

Answer 7

the derivative of the above comes to:\[-\sin(1)\sin(2a+1)\]so setting this to zero we get:-\[\sin(1)\sin(2a+1)=0\]so\[\sin(2a+1)=0\]so\[2a+1=0,\pi,2\pi,...,n\pi,...\]so\[a=-1/2,(\pi-1)/2,(2\pi-1)/2,...\]since original question said 0 to PI, then we must have:\[a=(\pi-1)/2\]

Answer 8

so the interval with maximum area is:\[\frac{\pi}{2}-\frac{1}{2} to \frac{\pi}{2}+\frac{1}{2}\]

Answer 9

Thank you so much:)