## anonymous one year ago For an object whose velocity in ft/sec is given by v(t) = cos(t), what is its distance, in feet, travelled on the interval t = 1 to t = 5?

1. anonymous

@jim_thompson5910

2. anonymous

$\int\limits_{1}^{5} \cos(t) dt$

3. anonymous

Solve that ^

4. jim_thompson5910

what did you get

5. ganeshie8

careful, you need to put absolute value for "distance"

6. anonymous

yeah

7. anonymous

1.8

8. jim_thompson5910

good point, or break up the interval into smaller pieces

9. dan815

find the area under the curve

10. anonymous

So......1.8 it is

11. anonymous

And dan please give chris his maple syrup

12. anonymous

13. anonymous

@chris00 no man deserves to live without his syrup.

14. anonymous

should be around 2.2..

15. anonymous

2.2?

16. anonymous

Seriously?

17. anonymous

i just graphed it...lel

18. anonymous

Let's see what @jim_thompson5910 has to say.

19. ganeshie8

|dw:1444450101879:dw|

20. jim_thompson5910

$\Large \int_{1}^{5}|\cos(x)|dx$ $\Large \left|\int_{1}^{\pi/2}\cos(x)dx\right|+\left|\int_{\pi/2}^{3\pi/2}\cos(x)dx\right|+\left|\int_{3\pi/2}^{5}\cos(x)dx\right|$ $\Large \left|\color{blue}{\int_{1}^{\pi/2}\cos(x)dx}\right|+\left|\color{red}{\int_{\pi/2}^{3\pi/2}\cos(x)dx}\right|+\left|\color{blue}{\int_{3\pi/2}^{5}\cos(x)dx}\right|$ I'll let you finish

21. anonymous

|dw:1444450118173:dw|

22. jim_thompson5910

the blue areas on the graph can be treated as normal the red area must be forced to be positive

23. anonymous

if the question doesn't explicitly say to use algebra, by all means you can use graphical means (says time in an exam) . But always good to use calc techniques.

24. anonymous

saves*

25. anonymous

2.2 was right