# Find the total distance that the particle travels over the given interval?

I have 2 different functions but them same type of problem.

v(t) = 9t - 4, $0 \leq t \leq 3$

I was able to find displacement which is 57/2 by taking the anti derivative and integrating from 0 to 3. For finding total distance I split integration from 0 to 4/9 and 4/9 to 3. I got 745/18 and its wrong.

The second function is v(t) = 2 cos(t), $0 \leq t \leq 3\pi$

I got displacement same way as above and got 0 which is correct. I'm positive I got the total distance right too which is 4 but the program says its wrong. Urgent help needed

• Subtract the integral on the intervals where the function is negative, and add the integral on the intervals where the function is positive. You're trying to integrate the absolute value of your function over the interval. You might also check your arithmetic. I did the integrals quickly, so I could have messed up, too. I got 545/18 for the first, and 12 for the second. – Chris Leary Apr 11 '14 at 1:11

To find distance traveled, you can just do:

$$\int \big|v(t)\big|\; dt$$

If you have access to a calculator. You split it when you don't. So for your case, you had the right approach. Since $v(t)$ is negative from $t=0$ to $t=\frac{4}{9}$, we'll take the negative integral of the function in that interval, then add it to the rest of the area.

$$\text{distance}=-\int_0^\frac{4}{9}(9t-4)\;dt+\int_\frac{4}{9}^3(9t-4)\;dt$$

Easy integrals:

$$\text{distance}=-\left(\frac{9}{2}t^2-4t\Biggr|_{t=0}^{t=\frac{4}{9}} \right)+\left(\frac{9}{2}t^2-4t\Biggr|_{t=\frac{4}{9}}^{t=3} \right)=-\left(-\frac{8}{9}\right)+\left(28.5-\left(-\frac{8}{9}\right)\right)$$

Answer turns out to be $\approx 30.2778$.

For the second one, same exact process. There's a lot of negative to positive changes, so I suggest just graphing the absolute value of the velocity function and getting the area of that.

Answer is about $12$.