8.3 Using Accumulation Functions and Definite Integrals in Applied Contexts

Definite integrals as net accumulation

If a rate of change function r(t) is given, then the definite integral of r(t) from a to b gives the net change in the quantity over that interval — not the total amount itself. Add the starting value to the net change to find the amount at a later time: amount(b) = amount(a) + ∫ from a to b of r(t) dt.

Interpreting units and context

Always check the units of the rate function to interpret the integral correctly. If r(t) is in gallons/minute, ∫r(t)dt gives gallons (accumulated amount), not a rate. Contexts often use words like 'total,' 'net change,' or 'amount' — match the calculus operation (integral vs. derivative) to the question being asked.

Combining multiple rates

When a quantity has both an inflow rate and outflow rate, the net rate is (rate in) − (rate out). Integrating this net rate over [a,b] gives the net change; adding the initial amount gives the quantity at time b. Watch for times when the net rate is zero (max/min amount) or negative (amount decreasing).

Key terms

Formula for amount at time b using an accumulation function

Amount(b) = Amount(a) + ∫ from a to b of rate(t) dt

What does a definite integral of a rate function represent?

The net change in the quantity over that interval, not the total amount.

Net rate when there's inflow and outflow

Net rate = (rate in) − (rate out); integrate this to get net change.

How to find total distance vs. displacement

Displacement = ∫v(t)dt; Total distance = ∫|v(t)|dt.

Why check units in applied integral problems?

Units confirm whether the integral gives a total accumulated amount (matching the rate's numerator units) rather than a rate.

Try an AP-style question

A tank holds 50 L at t = 0, and water enters at r(t) L/min. Which expression gives the amount at t = 10?

1. 50 + ∫₀¹⁰ r(t) dt
2. 50 · r(10)
3. ∫₀¹⁰ r(t) dt
4. 50 + r(10)