# Thread: Distance traveled using integrals

1. ## Distance traveled using integrals

Ok, I know what to do (take antiderivative and plug in points from a certain interval) but the final answer i get is 10.5 (which is wrong to the system). Can someone please check my work and let me know the correct answer? it's driving me nuts

My work (the boxes show the antiderivative and the interval from 1 to 0. I know that you have to plug in 1 to the antiderivative and 0 and subtract the two. The same goes for the interval 3 to 1. You then would add the two together):

 3 |t2 + 3t − 4| dt 0
=
 1 −t2 − 3t + 4 dt 0
+
 3 t2 + 3t − 4 dt 1
Antiderivatve= -1/3 t^3 -3/2 t^2 +4t |1 and 0 intervals

Antiderivative= 1/3t^3 +3/2 t^2 -4t | 3 and 1 intervals

Interval 1 subtracted by interval of 0:
[-1/3 (1)^3 -3/2 (1)^2 +4(1)] - [0]

The interval of 3 subtracted by interval of 1:
[1/3(3)^3 +3/2(3)^2-4(3) ] - [1/3(1)^3 +3/2(1)^2-4(1)

2. ## Re: Distance traveled using integrals

I'd advise you to fix your post, seeing as it is indecipherable

3. ## Re: Distance traveled using integrals

Lol my fault. Fixed

anyone?

5. ## Re: Distance traveled using integrals

To be frank, I can't see anything amiss with your work.

You have the definite integral split at its proper place, so you'll always get a positive output when you evaluate the definite integral:

$\int_0^1(-t^2-3t+4)dt + \int_1^3(t^2+3t-4)dt$

Likewise, it seems that you proceeded with taking the antiderivative properly:

$\frac{-t^3}{3}- \frac{3}{2}t^2+4t \biggr|_0^1+ \frac{t^3}{3}+ \frac{3}{2}t^2 - 4t \biggr|_1^3$

Perhaps you've just had some sort of arithmetic error.

6. ## Re: Distance traveled using integrals

Originally Posted by Steelers72

 3 |t2 + 3t − 4| dt 0
=
 1 −t2 − 3t + 4 dt 0
+
 3 t2 + 3t − 4 dt 1
Antiderivatve= -1/3 t^3 -3/2 t^2 +4t |1 and 0 intervals

Antiderivative= 1/3t^3 +3/2 t^2 -4t | 3 and 1 intervals
You might find it helpful to take a look at our LaTeX forum. It may look hard at the start but it is reasonably easy to learn.

-Dan