Net change: an amount from its rate

Three quick questions from earlier lessons.

A bath holds 40 L. Over 10 minutes, $\displaystyle\int_0^{10} r(t)\,dt = 60$ L flow in. How much water is in the bath now?

The integral of a rate is the net change of the amount.

Net change:

F(b) = F(a) + \displaystyle\int_a^b F'(t)\,dt

. Starting value plus accumulated change.

A town has 5000 people; it grows at $r(t) = 200 + 30t$ people per year. Population after 4 years?

\displaystyle\int_0^4 (200 + 30t)\,dt = 800 + 240 = 1040

Accumulated change.

5000 + 1040 = 6040

Add the start.

Your turn: $F(2) = 7$ and $\displaystyle\int_2^5 F'(x)\,dx = -3$ . Find $F(5)$ .

F(5) = 4

7 + (−3).

$F' = f$ , $F(0) = -1$ and $\displaystyle\int_0^{ 3 } f(x)\,dx = 0$ . Find $F(3)$ .
$F' = f$ , $F(0) = -2$ and $\displaystyle\int_0^{ 6 } f(x)\,dx = -9$ . Find $F(6)$ .
$F' = f$ , $F(0) = 8$ and $\displaystyle\int_0^{ 4 } f(x)\,dx = -2$ . Find $F(4)$ .

A tank holds 37 L. Water flows in at $r(t) = 5 + 1t$ L/min. How much water is in the tank after 4 minutes, in liters?
A tank holds 57 L. Water flows in at $r(t) = 6 + 3t$ L/min. How much water is in the tank after 5 minutes, in liters?

$v(t)$ is a particle's velocity in m/s. What does $\displaystyle\int_0^{ 7 } v(t)\,dt$ give?
1. Its displacement (change in position) from $t = 0$ to $t = 7$ , in meters
2. The total distance it travels, in meters
3. Its velocity at $t = 7$

Three questions to check you have got it.

A tank holds 37 L. Water flows in at $r(t) = 5 + 1t$ L/min. How much water is in the tank after 4 minutes, in liters?
A tank holds 57 L. Water flows in at $r(t) = 6 + 3t$ L/min. How much water is in the tank after 5 minutes, in liters?
A tank holds 38 L. Water flows in at $r(t) = 4 + 1t$ L/min. How much water is in the tank after 2 minutes, in liters?

Net change

$F(b) = F(a) + \displaystyle\int_a^b F'(t)\,dt$ : start + accumulated change.

Watch out: a rate at one moment is not an amount.

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