Finding a Final Quantity from Initial Data

From an Initial Amount to a Final Amount

A rate of change tells how quickly a quantity is changing, but it does not by itself tell you the quantity at a later time. To find that later quantity, you need an initial amount as well as the change accumulated over the time interval.

In an AP free-response solution, the calculation is only part of the response. You should also make clear what the initial value represents, why the integral has its particular bounds, and what units and contextual meaning the final result has.

Key idea: The quantity at a later time equals its known initial value plus the definite integral of its rate of change from the initial time to the later time.

The Initial-Value Accumulation Formula

Let \(Q(t)\) be a quantity at time \(t\), and let \(r(t)=Q'(t)\) be its rate of change. If \(Q(a)\) is known and \(r\) is continuous on the interval from \(a\) to \(b\), then the value at time \(b\) is

Q(b)=Q(a)+\int_a^b r(t)\,dt.

Here \(a\) is the time associated with the given initial amount, and \(b\) is the time whose quantity is requested. The word “initial” does not necessarily mean that the time is zero. If the prompt gives a quantity at time \(t=3\), that is the initial value for a calculation beginning at \(t=3\).

The integral represents the change in \(Q\) from time \(a\) to time \(b\). A positive rate contributes an increase; a negative rate contributes a decrease. The rate must be the rate of change of the same quantity whose initial and final values are being related.

Formula: If \(Q'(t)=r(t)\), \(r\) is continuous on \([a,b]\), and \(Q(a)\) is known, then \(Q(b)=Q(a)+\int_a^b r(t)\,dt\).

Why the Formula Works

The result follows directly from the Fundamental Theorem of Calculus. Because \(r(t)=Q'(t)\), the integral of the rate over the interval gives the difference between the endpoint values of \(Q\).

\int_a^b r(t)\,dt = \int_a^b Q'(t)\,dt = Q(b)-Q(a).

Rearranging this equality gives

Q(b)=Q(a)+\int_a^b r(t)\,dt.

The initial value is essential: the integral supplies the change, not the quantity itself. Many functions can have the same rate of change while differing by a constant, so knowing the rate alone does not determine the starting level. The initial datum fixes that level.

A Reliable Solution Structure

Before calculating, identify the quantity, the initial time, the requested time, and the units of the rate. Then write the formula with bounds that match the stated times.

Identify the quantity and rate.
Make sure the given rate describes the change in the quantity whose final value is requested.

Set the bounds.
Use the time of the known initial amount as the lower bound and the requested time as the upper bound.

Find the accumulated change.
Evaluate the definite integral, keeping the sign of the rate and the units in view.

Add the initial amount and state the conclusion.
Report the final quantity with appropriate units and explain what it represents.

Worked Examples

Worked Example: Finding a Later Volume

A tank contains \(250\) gallons of water at time \(t=2\) hours. Water enters the tank at a rate \(r(t)=6+2t\) gallons per hour. Find the amount of water in the tank at \(t=5\) hours.

Let \(V(t)\) be the volume of water in gallons. The initial value is \(V(2)=250\), and the rate is \(V'(t)=6+2t\). The required change is accumulated from \(t=2\) to \(t=5\), so

V(5)=V(2)+\int_2^5 (6+2t)\,dt.

An antiderivative of \(6+2t\) is \(6t+t^2\), since

\frac{d}{dt}(6t+t^2)=6+2t.

Evaluate the definite integral:

\begin{aligned} \int_2^5 (6+2t)\,dt &=[6t+t^2]_2^5\\ &=(6\cdot 5+5^2)-(6\cdot 2+2^2)\\ &=(30+25)-(12+4)\\ &=55-16\\ &=39. \end{aligned}

The accumulated change is \(39\) gallons, so

V(5)=250+39=289\text{ gallons}.

The units are consistent: gallons per hour integrated with respect to hours gives gallons. The antiderivative differentiates to the given rate, and substituting the calculated change into the initial-value formula gives \(250+39=289\). Therefore, the tank contains \(289\) gallons of water at \(t=5\) hours.

Worked Example: Using an Initial Value at a Nonzero Time

A quantity of liquid is \(480\) liters at \(t=1\) minute. Its rate of change is \(r(t)=20-2t\) liters per minute. Find the quantity at \(t=7\) minutes.

Let \(L(t)\) denote the quantity in liters. The initial time is \(1\), not \(0\), so the correct setup is

L(7)=L(1)+\int_1^7(20-2t)\,dt =480+\int_1^7(20-2t)\,dt.

An antiderivative is \(20t-t^2\), because

\frac{d}{dt}(20t-t^2)=20-2t.

Now evaluate the integral, showing both endpoint values:

\begin{aligned} \int_1^7(20-2t)\,dt &=[20t-t^2]_1^7\\ &=(20\cdot7-7^2)-(20\cdot1-1^2)\\ &=(140-49)-(20-1)\\ &=91-19\\ &=72\text{ liters}. \end{aligned}

Thus

L(7)=480+72=552\text{ liters}.

The derivative check confirms the integrand: differentiating \(20t-t^2\) returns \(20-2t\). The integral has units of liters because it combines liters per minute with minutes. The final quantity is the initial \(480\) liters plus \(72\) liters of change, so the amount at \(t=7\) minutes is \(552\) liters.

Worked Example: A Rate That Changes Sign

A reservoir contains \(300\) gallons at \(t=2\) hours. Its rate of change is \(r(t)=8-2t\) gallons per hour. Find the amount at \(t=5\) hours.

Let \(R(t)\) be the reservoir’s volume. Since \(R(2)=300\), apply the formula on the interval from \(2\) to \(5\):

R(5)=300+\int_2^5(8-2t)\,dt.

An antiderivative of the rate is \(8t-t^2\), as

\frac{d}{dt}(8t-t^2)=8-2t.

Evaluate:

\begin{aligned} \int_2^5(8-2t)\,dt &=[8t-t^2]_2^5\\ &=(8\cdot5-5^2)-(8\cdot2-2^2)\\ &=(40-25)-(16-4)\\ &=15-12\\ &=3\text{ gallons}. \end{aligned}

Therefore,

R(5)=300+3=303\text{ gallons}.

The rate is positive before \(t=4\), zero at \(t=4\), and negative after \(t=4\). The integral accounts for the rate’s sign over the whole interval; its value is a net increase of \(3\) gallons. The derivative of the antiderivative is the stated rate, and the units of the integral are gallons. Thus, at \(t=5\) hours the reservoir contains \(303\) gallons.

Bounds, Signs, and Units

The bounds are not decorative. They identify the exact period over which the rate changes the quantity. If the initial amount is known at time \(a\) and the desired amount is at time \(b\), the integral must run from \(a\) to \(b\). Reversing the bounds reverses the sign of the integral and gives the change in the opposite time direction.

Units provide a useful check. If \(Q\) is measured in liters and time in minutes, then \(r(t)\) has units of liters per minute. Multiplying by \(dt\), which represents a time increment measured in minutes, gives units of liters. The integral therefore has the same units as \(Q(a)\), as required before the two quantities can be added.

Check the dimensions: If \(Q\) has units of quantity and \(t\) has units of time, then \(Q'(t)\) has units of quantity per time, and \(\int_a^b Q'(t)\,dt\) has units of quantity.

Common Mistake: Reporting the Change as the Final Amount

Suppose a tank has \(75\) liters at \(t=0\), and its rate of change is \(r(t)=3t+2\) liters per hour from \(t=0\) to \(t=4\). A common error is to calculate the integral correctly and report that result as the final amount.

The integral is

\begin{aligned} \int_0^4(3t+2)\,dt &=\left[\frac{3}{2}t^2+2t\right]_0^4\\ &=\left(\frac{3}{2}\cdot16+8\right)-0\\ &=24+8\\ &=32\text{ liters}. \end{aligned}

The incorrect conclusion would be: “The final amount is \(32\) liters.” That is the change over the interval, not the final amount. The correction is to add the initial amount:

Q(4)=75+32=107\text{ liters}.

The antiderivative differentiates to \(3t+2\), and the integral’s units are liters. The final amount is \(107\) liters because it consists of the original \(75\) liters and the \(32\)-liter increase.

Writing a Complete Contextual Conclusion

A complete response connects the mathematical result to the situation. After the calculation, state the final quantity, its units, and the time at which it occurs. If useful, identify the integral as the change over the interval and the initial value as the quantity at the lower-bound time.

For example, “The integral gives an increase of \(72\) liters from \(t=1\) to \(t=7\). Adding this change to the \(480\)-liter amount at \(t=1\), the quantity at \(t=7\) is \(552\) liters.” This sentence makes the role of each value clear and avoids presenting a bare number without context.

Before finalizing an answer, check that the rate describes the correct quantity, the lower and upper bounds match the times in the prompt, the integral has been evaluated with the proper signs, and the change—not just the final amount—has the right units. Then add the initial value and state what the result means.

Check Your Understanding

For each question, identify the initial value, the correct integral bounds, and the contextual meaning of the resulting final quantity.

A container has \(90\) liters at \(t=3\) minutes, and its rate of change is \(5\) liters per minute. Write an expression for the amount at \(t=8\), including units.
A quantity is \(200\) grams at \(t=1\) hour and changes at \(r(t)=4t\) grams per hour. Set up, evaluate, and interpret the amount at \(t=4\) hours.
Why does the integral of a rate of change give a change in the quantity rather than the quantity itself?
If a rate is measured in gallons per minute and time is measured in minutes, what units does its definite integral have?
A student calculates the integral correctly but reports it as the final amount. What information is missing from that conclusion?

Finding a Final Quantity from Initial Data

What You'll Learn

From an Initial Amount to a Final Amount

The Initial-Value Accumulation Formula

Why the Formula Works

A Reliable Solution Structure