# Thread: May I ask one thing about a integration question?

1. ## May I ask one thing about a integration question?

Hi all

I am sorry to bother you.
Could you help me to solve this problem?

Question) particle's moves in a straight line, whose velocity can be modelled by the
following function
v(t) = tsin(t):
Find the distance s, covered by the particle from time t = 0 to t = 2 pie. Sketch
the graph for v(t) over the domain [0, 2 pi], and display what the value of s
represents on your sketch. Hint: recall s(t) = v(t), also think about the sign
of the function v(t) for t > pi.

My answer is that I need to find first the integration of xsin(x)dx which is sin(x)-xcos(x). Then, I put 2 pi into x. I can get -2 pi ?
Then, I have no idea..
Please help me out.

Thanks a lot.

2. ## Re: May I ask one thing about a integration question?

Originally Posted by yanirose
Hi all

I am sorry to bother you.
Could you help me to solve this problem?

Question) particle's moves in a straight line, whose velocity can be modelled by the
following function
v(t) = tsin(t):
Find the distance s, covered by the particle from time t = 0 to t = 2 pie. Sketch
the graph for v(t) over the domain [0, 2 pi], and display what the value of s
represents on your sketch. Hint: recall s(t) = v(t), also think about the sign
of the function v(t) for t > pi.

My answer is that I need to find first the integration of xsin(x)dx which is sin(x)-xcos(x). Then, I put 2 pi into x. I can get -2 pi ?
Then, I have no idea..
Please help me out.

Thanks a lot.
You've done everything correctly and obtained the correct answer. The key is to understand that both position and velocity can be negative. If your particle is moving on the real line, -velocity would generally correspond to moving to the left. Negative position would generally correspond to being left of 0.

So yes, your particle moved with more negative (leftward) velocity than positive (rightward) velocity during [0,2pi] and ended up 2pi left of where it started.