Calculus / Unit 9: Applications of Integrals
9.4 Accumulation and Net Change
Notes
Net change theorem: $\displaystyle \int_a^b f'(x)\,dx = f(b) - f(a)$. The integral of a rate of change gives the total change.
- Example: water flows in at $r(t) = 6t$ liters per minute for $0 \leq t \leq 4$. The amount added is $\displaystyle \int_0^4 6t\,dt = \left[3t^2\right]_0^4 = 48$ liters.
- New amount = starting amount + net change.
- Units of the integral are (rate units) times (input units).
Practice
Water flows into a tank at $r(t) = 6t$ liters per minute. How much enters during the first 4 minutes?
$\displaystyle \int_0^4 6t\,dt = 3(16) = 48$.
A town's population grows at $P'(t) = 2t + 10$ people per year. How much does it grow in the first 5 years?
$\displaystyle \int_0^5 (2t + 10)\,dt = \left[t^2 + 10t\right]_0^5 = 25 + 50 = 75$.
A tank holds 100 gallons. Its volume changes at $5 - 2t$ gallons per minute. How many gallons are in it at $t = 3$?
$\displaystyle 100 + \int_0^3 (5 - 2t)\,dt = 100 + (15 - 9) = 106$.
Ready for more? Take the Unit 9 test.