How do you calculate the distance traveled using calculus?

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The velocity function is v(t) = - t^2 + 5 t - 6 for a particle moving along a line. Find the displacement and the distance traveled by the particle during the time interval [-3,6].

displacement = -67.5

distance traveled =

I was able to figure out the displacement, but I can't seem to get the distance traveled.

v(t) = - t^2 + 5 t - 6
s^1(t) = v(t) = - t^2 + 5 t - 6
s(t) = (-1/3)(t^3) + (5/2)(t^2) - 6x

(-1/3)(6^3) + (5/2)(6^2) - 6*6 - [(-1/3)((-3)^3) + (5/2)((-3)^2) - 6(-3)]
-18 - 49.5 = -67.5

Can someone please solve the distance traveled?
Which equation do I graph? What do you mean intergate to the x-axis border?
 
v(t) = - (t^2 - 5t + 6) = - (t - 2)(t - 3)
This shows that
v(t) < 0 for t < 2,
v(t) > 0 for 2 < t < 3 and
v(t) < 0 for t > 3

Thus, velocity changes direction when t = 2 and t = 3
=> to find the distance travelled, one should work out displacement in three parts and add them taking as +ve quantities
Thus,
distance travelled
= l ?(-3 to 2) (-t^2 + 5t -6) dt l + l ?(2 to 3) (-t^2 + 5t -6) dt l + l ?(3 to 6) (-t^2 + 5t -6) dt l
= l (-t^3/3 + 5t^2/2 - 6t) (t=-3 to t=2) l + l (-t^3/3 + 5t^2/2 - 6t) (t=2 to t=3) l + l (-t^3/3 + 5t^2/2 - 6t) (t=3 to t=6) l
= l (-8/3 + 10 - 12) - (9 + 45/2 + 18) l + l (-9 + 45/2 + 18) - (-8/3 + 10 - 12) l + l (- 72 + 90 - 36) - (-9 + 45/2 - 18) l
= l (-14/3) - (99/2) l + l (63/2) - (-14/3) l + l (-18) - (-9/2) l
= 325/6 + 217/6 + 135/6
= 677/6.
[Please recheck the calculations.]
 
Graph the function.

Integrate the function in parts using the x-axis as a border.
 
s ( t ) = - t ³ / 3 + 5 t ² / 2 - 6 t + k

s ( - 3 ) = 27 / 3 + 45/2 + 18 + k

s ( 6 ) = - 216 / 3 + 186 / 3 - 36 + k

s ( - 3 ) - s ( 6 ) = 243 / 3 + 45 / 2 - 186 / 3 + 18 + 36

s ( - 3 ) - s ( 6 ) = 57 / 3 + 45 / 2 + 54

s ( - 3 ) - s ( 6 ) = 114 / 6 + 135 / 6 + 324 / 6

s ( - 3 ) - s ( 6 ) = 573 / 6

s ( - 3 ) - s ( 6 ) = 96
 
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