1.) v(t)=sint, 0<=t<=5pi/4

<= (less than or equal to) *don't know how else to put*

1.) v(t)=sint, 0<=t<=5pi/4

<= (less than or equal to) *don't know how else to put*

<= (less than or equal to) *don't know how else to put*

- stapel_eliz
**Posts:**1628**Joined:**Mon Dec 08, 2008 4:22 pm-
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junsta12 wrote:1.) v(t)=sint, 0<=t<=5pi/4

<= (less than or equal to) *don't know how else to put*

Using "<=" for "less than or equal to" is

Please be complete. Thank you!

Find the total distance. I know you're suppose to take the absolute value of the function when it is negative and add to the positive. i split the integral into two, from 0 to 3pi/4 amd from 3pi/4 to 5pi/4. and solved but answer was wrong. i think i may have my limits of integration wrong when splitting into two.

- stapel_eliz
**Posts:**1628**Joined:**Mon Dec 08, 2008 4:22 pm-
**Contact:**

junsta12 wrote:Find the total distance.

Find the total distance

junsta12 wrote:I know you're suppose to take the absolute value of the function when it is negative and add to the positive.

Are you maybe saying that you need to take the absolute value of something, perhaps then splitting one function into pieces, and then adding the results of whatever you're doing to the pieces? Or are you really adding two different functions, one of which is an absolute value?

junsta12 wrote:i split the integral into two, from 0 to 3pi/4 amd from 3pi/4 to 5pi/4. and solved but answer was wrong. i think i may have my limits of integration wrong when splitting into two.

You split

Please reply with the full and exact text of the exercise, the complete instructions, and a clear listing of your efforts so far, so we can tell what is going on. Thank you.