The velocity function, in feet per second, is given for a particle moving along a straight line. v(t) = t^3 − 9t^2 + 20t − 12, 1 ≤ t ≤ 7 Find the total distance that the particle travels over the given interval.

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The velocity function, in feet per second, is given for a particle moving along a straight line. v(t) = t^3 − 9t^2 + 20t − 12, 1 ≤ t ≤ 7 Find the total distance that the particle travels over the given interval.

Calculus1
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Distance is integral of velocity because velocity is the derivative of distance over time: $$ d=\int_1^7v(t)dt=\int_1^7\left (t^3 − 9t^2 + 20t − 12\right )~dt $$ Can you evaluate this integral?
yesssss I did I don't know why my answer is wrong
that will give you displacement think about something like: \(\int_{1}^{7}|v(t)|dt\)

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|dw:1444090723561:dw|
yes then it changes at (t-1) (t-2) (t-6)
it's in the interval \(1 ≤ t ≤ 7\) so the \(2 \le t \le 6\) is the bit that needs to be thought about |dw:1444091029721:dw|
okay.. what needs to be thought about it? I did all of the integrals
@IrishBoy123 is correct. Evaluate the "absolute" value of v(t) and split into three intervals rather than just a blanket 1-7. Otherwise, the way I did it, you will be subtracting distance rather than adding distance for that negative part -- not what you want.
yeah i split it into my intervals
and you got ... ?
-518 which doesn't make sense
\[\color{red} -\int_2^6 (t^3 − 9t^2 + 20t − 12 ) dt = 32\]
I'm going to attach a picture of my work
that's really hard to read \(\int \; t^3-9 t^2+20 t-12 \; dt = \dfrac{t^4}{4}-3 t^3+10 t^2-12 t \; [+C]\) \(\int_{1}^{2} \; |v(t)| \; dt = \dfrac{3}{4}\) \(\int_{2}^{6} \; |v(t)| \; dt = 32\) \(\int_{6}^{7} \; |v(t)| \; dt = \dfrac{53}{4}\)

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