Question

So I've got this question (screenshot attached below) and I've got a feeling on how to solve it but I'm not very confident on it. Here's what I've got:

\[\int \sin (2t) dt = \frac{-cos(2t)}{2} + C = x(t)\]

\[\frac{1}{2}\left( -\cos \frac{2 \pi}{2}\right) + C = 3\]

\[C = \frac{5}{2}\]

\[\int_0^\pi \frac{1}{2}\left( -\cos\ t\right) + \frac{5}{2} dt = \frac{5 \pi}{2}\]

Any assistance is appreciated.

Answer 1

how r u doing so far?

Answer 2

Well I think I'm going in the right direction but I'm not sure. Are you able to see my work on the original post?

Answer 3

All of your work is correct :)
good!

Answer 4

Woop woop *dance*

Thanks hartnn!

Answer 5

welcome ^_^
*dances with ya*

Answer 6

i think there is one typo!

Answer 7

\(\int_0^\pi \frac{1}{2}\left( -\cos \Large{2} t\right) + \frac{5}{2} dt = \frac{5 \pi}{2}\)

Answer 8

@crumply_paper

but that will not change your answer :)

Answer 9

Woops, you're totally right. I typed it wrong when I was posting the question, but I've got in right in my notebook. Thanks for catching that :)

Answer 10

:)

Answer 11

i think you should get total distance travelled = 2

Answer 12

distance travelled in \(0 \le t\le \pi\) :
\[\int\limits_0^{\pi} |v(t)|\, dt = \int\limits_0^{\pi} |\sin(2t)|\, dt = 2\int\limits_0^{\pi/2}\sin(2t)\, dt = 2 \]

Answer 13

I thought that the definite integral of a velocity function only gives you the net velocity on the interval? I didn't know it would give distance travelled

Answer 14

so what we found out was displacement ?

Answer 15

definite integral of velocity gives you "net displacement"

definite integral of | velolocity | gives you total "distance travelled"

Answer 16

Hmmm, okay. If that's the case, then what would the definite integral of a position function be?

Answer 17

there is no explicit name for that afaik :

```
velolcity = integral of acceleration
position = integral of velocity
? = integral of position
```

Answer 18

but yeah nobody gona stop us from integrating position ;)

Answer 19

Haha you know it. Well thanks again ganeshie (and thank you too hartnn.)