Calculus 1-Net Change Integrating the Derivative, guaranteed 100% Pass

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Net Change: Integrating the Derivative


If 𝑠(𝑡) is the position of an object moving along a line, then 𝑠 (𝑏) − 𝑠(𝑎) is the
displacement of the object for 𝑎 ≤ 𝑡 ≤ 𝑏 (displacement to the right/up is
positive, to the left/down is negative).

Recall that the velocity at time 𝑡 is 𝑣 (𝑡) = 𝑠′(𝑡). Thus we have:
𝒃 𝒃
∫𝒂 𝒗(𝒕)𝒅𝒕 = ∫𝒂 𝒔′ (𝒕)𝒅𝒕 = 𝒔(𝒃) − 𝒔(𝒂) = 𝒅𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕
= 𝑵𝒆𝒕 𝑪𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝑷𝒐𝒔𝒊𝒕𝒊𝒐𝒏 𝒐𝒗𝒆𝒓 [𝒂, 𝒃].
Displacement can be positive or negative. Distance is always non-negative. To
find the distance travelled we need to integrate the speed |𝑣 (𝑡)|.
𝒃
∫𝒂 |𝒗(𝒕)|𝒅𝒕 = 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒕𝒓𝒂𝒗𝒆𝒍𝒆𝒅 𝒇𝒐𝒓 𝒂 ≤ 𝒕 ≤ 𝒃.


Ex. A particle moves along a straight line so that its velocity is

𝑣(𝑡) = 𝑡 2 − 𝑡 − 6 𝑚/𝑠𝑒𝑐
a. Find the displacement during 1 ≤ 𝑡 ≤ 4.
b. Find the distance traveled during 1 ≤ 𝑡 ≤ 4.


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a. Displacement= ∫1 𝑣(𝑡)𝑑𝑡
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= ∫1 (𝑡 2 − 𝑡 − 6)𝑑𝑡
1 3 1 2 𝑡=4
= 𝑡 − 𝑡 − 6𝑡|
3 2 𝑡=1
1 1 1 1
= ( (4)3 − (4)2 − 6(4)) − ( (1)3 − (1)2 − 6(1))
3 2 3 2
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= − 𝑚 (to the left).
2

, 2


4 4
b. Distance Traveled= ∫1 |𝑣 (𝑡)|𝑑𝑡 = ∫1 |(𝑡 2 − 𝑡 − 6)|𝑑𝑡
To integrate the absolute value of a function we need to know where the
function is positive and where it’s negative. We then use the fact that:
|𝑓 (𝑡)| = 𝑓(𝑡) if 𝑓(𝑡) ≥ 0
= −𝑓(𝑡) if 𝑓(𝑡) ≤ 0.

𝑡 2 − 𝑡 − 6 = (𝑡 − 3)(𝑡 + 2) = 0 ⟹ 𝑡 = 3, −2.
By testing the sign of this function on the intervals:
𝑡 < −2, − 2 < 𝑡 < 3, 3 < 𝑡, we get:


sign of 𝑡 2 − 𝑡 − 6 ______+_______|________−_________|_____+____

−2 3

𝑡 2 − 𝑡 − 6 ≥ 0 when 𝑡 ≤ −2 or 𝑡 ≥ 3
𝑡 2 − 𝑡 − 6 ≤ 0 when −2 ≤ 𝑡 ≤ 3.

So when 1 ≤ 𝑡 ≤ 4 we have:
𝑡 2 − 𝑡 − 6 ≥ 0 when 3 ≤ 𝑡 ≤ 4
𝑡 2 − 𝑡 − 6 ≤ 0 when 1 ≤ 𝑡 ≤ 3.

So |(𝑡 2 − 𝑡 − 6)| = 𝑡 2 − 𝑡 − 6 when 3 ≤ 𝑡 ≤ 4
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= −(𝑡 − 𝑡 − 6) when 1 ≤ 𝑡 ≤ 3

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Distance Traveled= ∫1 |(𝑡 2 − 𝑡 − 6)|𝑑𝑡
3 4
= − ∫1 (𝑡 2 − 𝑡 − 6)𝑑𝑡 + ∫3 (𝑡 2 − 𝑡 − 6)𝑑𝑡
1 1 𝑡=3 1 1 𝑡=4
= − ( 𝑡 3 − 𝑡 2 − 6𝑡)| + ( 𝑡 3 − 𝑡 2 − 6𝑡)|
3 2 𝑡=1 3 2 𝑡=3
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= 𝑚.
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