I was having a bit of trouble with this question.
For a) I found -3/1/3 for the displacement be evaluating the definite integral.
For b) I found 32/2/3 to be the distance travelled using the fact the below 4 the curve is negative and above 4 the curve is positive.
I would be great if some else could do these to and see if they get similar answers as I'm not sure what answers to expect.
For c) using the Mean Value Theorem f ave=1/b-a integral b a f(x) dx, I found the average velocity to be -6 but I'm not sure if I used the correct a and b values. I made a=1 and b=4 (as this is were the curve cuts the x axis)
If someone could show me how to do this it would be a great help.
Thanks for the help mr fantastic,
I'm not sure what you mean by Average velocity = displacement/time =. Doing this means the average velocity would be equal to -3/1/3 / 5 which is would be -2/3.
I was wondering if I could do this using the Mean Value Theorem? Its just I'm not sure how to find the definite integral to evaluate as Iím sure I have to go about the problem this way.
I was using this formula when I found the answer -6, but this was using a=1 and b=4, (b=4 because of the fact that the curve cut the x-axis at 4.)
I saw a similar problem with the function f=t-ľ*t^2 over the interval of 0=<t<=6 and the fact that this graph intercepts the x-axis at 4 a=0 and b=4. Not b=6 as what you are suggesting.
I should have said that this was the problem I was having with the question.
Using a=1 and b=6 gives an answer of -2/3 and using a=1 and b=4 gives an answer of -6.
I'm not sure that if the curve cuts the x-axis this needs to be taken into account as I have seen it possibly has in a similar question but Iím unsure.
mr fantastic and skeeter
Okay, thanks for the help. It seems that the similar question I mentioned may have been to find the average velocity over 0 to 4 seconds but I can't be sure.
I think using the interval seems making a=1 and b=6 seems like the most reasonable thing to do thing to do.