If we plug in the information that at time t = 1, x(1) = 9, we can solve for C.
2(1)^3 − 9(1)^2 + 12(1) + C = 9
5 + C = 9
C = 4
so x(t) = 2t^3 − 9t^2 + 12t + 4
(d) Find the total distance traveled by the particle from t = to t = 6.
Step 1: Normally, all that we have to do to find the distance traveled is to integrate the
velocity equation from the starting time to the ending time. But we have to watch out for
whether the particle changes direction. If so, we have to break the integration into two parts—a
positive integral for when it is traveling to the right, and a negative integral for when it is
traveling to the left.
One way to solve this is to find two integrals and add them together. Because you can use a
calculator, it is simpler to use the fnInt calculation of the absolute value for t = and t = 6,
using the function of velocity 6x^2 − 18x + 12.
Graphing Calculator (TI-83 and TI-84)
Press MATH and select 9: fnInt from the list.
Press MATH then select the NUM menu, and choose 1: abs(
Enter the function 6x^2 − 18x + 12 and follow with the closing parentheses. List the variable and
low and high values for t, separated by commas, and follow with final closing parentheses so
your expression looks like the following:
fnInt(abs(6x^2 − 18x + 12), x, 3/2, 6)
Press ENTER
The result is 176.500
- Let R be the region enclosed by the graphs of y = 2 ln x and y = and the lines x = 2 and x = 8.
(a) Find the area of R.
Step 1: If there are two curves, f(x) and g(x), where f(x) is always above g(x), on the interval
[a, b], then the area of the region between the two curves is found by
