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8.3 Using Accumulation Functions and Definite Integrals in Applied Contexts

Syllabus
2020
Topic
8.3
Level
—

Learning objectives

—CHA-4.D—Interpret the meaning of a definite integral in accumulation problems• CHA-4.D Interpret the meaning of a definite integral in accumulation problems.• CHA-4.D.1 A function defined as an integral represents an accumulation of a rate of change.• CHA-4.D.2 The definite integral of the rate of change of a quantity over an interval gives the net change of that quantity over that interval.• Enduring understanding CHA-4: Definite integrals allow us to solve problems involving the accumulation of change over an interval.—CHA-4.E—Determine net change using definite integrals in applied contexts• CHA-4.E Determine net change using definite integrals in applied contexts.• CHA-4.E.1 The definite integral can be used to express information about accumulation and net change in many applied contexts.• Enduring understanding CHA-4: Definite integrals allow us to solve problems involving the accumulation of change over an interval.

An Accumulation Function Adds Up a Rate

If r(t)r(t) is the rate of change of a quantity, then A(x)=∫axr(t) dtA(x)=\int_a^x r(t)\,dt records the quantity’s net change from input aa to input xx. The lower bound fixes the starting point; the upper bound makes the accumulated value change with xx.

A(x)=\int_a^x r(t),dt,\qquad A(a)=0,\qquad A'(x)=r(x)

Feature Interpretation
r(t)>0r(t)>0 positive contributions make AA increase
r(t)<0r(t)<0 negative contributions make AA decrease
∫abr(t) dt\int_a^b r(t)\,dt net change from aa to bb
rate units QQ/time integral units QQ

If the actual quantity is QQ and Q(a)=Q0Q(a)=Q_0, then Q(x)=Q0+A(x)=Q0+∫axr(t) dtQ(x)=Q_0+A(x)=Q_0+\int_a^x r(t)\,dt. Thus the integral alone is the change since the start; adding the initial amount gives the current quantity.

Net change is signed. Positive and negative rate contributions can cancel, so it is not automatically the total amount of activity. Also, A(a)=0A(a)=0 does not mean the original quantity was zero; it means no change has accumulated over an interval of zero length.

Initial Amount Plus Integrated Net Rate Gives Final Amount

In an applied accumulation problem, first identify the rate that changes the target quantity. When material enters and leaves, use net rate = incoming rate −- outgoing rate. Integrating that net rate gives the signed change, not the final amount by itself.

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

  1. Define the target quantity and its units.
  2. Build the signed net rate in quantity/time units.
  3. Integrate over the exact time interval.
  4. Interpret the integral as net change.
  5. Add the initial quantity if the question asks for the amount at the end.

A tank initially contains 100100 liters and has illustrative net inflow r(t)=12−2tr(t)=12-2t liters/minute for 0≤t≤40\le t\le4. Its net change is ∫04(12−2t) dt=[12t−t2]04=32\int_0^4(12-2t)\,dt=[12t-t^2]_0^4=32 liters. Therefore the amount after 44 minutes is 100+32=132100+32=132 liters.

Do not add the initial amount when only net change is requested, and do not omit it when the final amount is requested. Check that rate and integration-variable units match; integrating liters/minute with respect to minutes produces liters.