From a Starting Value to a Later Value
A rate describes how quickly a quantity changes, but it does not by itself tell us the quantity at a particular input. To determine that value, we also need a starting value. The definite integral connects the two: it accumulates the rate from the starting input to the input of interest.
For example, if \(Q(t)\) is a quantity and \(r(t)\) is its rate of change, knowing \(Q(a)\) lets us determine \(Q(x)\) by adding the change accumulated between \(a\) and \(x\). This combines an initial condition with an accumulation function based at \(a\).
The Initial-Value Accumulation Formula
Suppose \(r\) is continuous on an interval containing the initial input \(a\) and the input \(x\). Suppose a quantity \(Q\) has rate \(Q'(t)=r(t)\) and initial value \(Q(a)=Q_0\). Then its value at \(x\) is
The letter \(t\) is a dummy variable of integration: it indicates the input at which the rate is evaluated as the integral accumulates from \(a\) to \(x\). The output of the integral is a change in the quantity, not the quantity's starting value itself. Adding \(Q_0\) turns that change into the value \(Q(x)\).
The formula works whether \(x\) is greater than or less than \(a\). If \(x<a\), the integral's bounds are reversed relative to the forward direction, and the accumulated change has the corresponding sign. The formula does not require the rate to be positive.
Why the Formula Works
The reasoning combines additivity of definite integrals with the Fundamental Theorem of Calculus. If \(Q'=r\) and the rate is continuous, the theorem gives the change in \(Q\) between \(a\) and \(x\) as the integral of its derivative over that interval.
Since the initial condition says \(Q(a)=Q_0\), adding \(Q_0\) to both sides gives \(Q(x)=Q_0+\int_a^x r(t)\,dt\). This establishes the formula for every \(x\) in the interval.
There is also a direct way to check that the formula produces a quantity with the required rate and starting value. Define \(F(x)=Q_0+\int_a^x r(t)\,dt\). By the Fundamental Theorem of Calculus, \(F'(x)=r(x)\) at points where \(r\) is continuous. At the initial input, the integral has equal bounds, so \(F(a)=Q_0+\int_a^a r(t)\,dt=Q_0\). Thus \(F\) has the stated rate and initial value.
These conditions determine the quantity uniquely on the interval. If \(Q\) and \(F\) both have derivative \(r\) and the same value at \(a\), then the derivative of \(Q-F\) is zero. By the Mean Value Theorem, a differentiable function with zero derivative throughout an interval is constant there. Since \(Q(a)-F(a)=0\), that constant is zero, so \(Q(x)=F(x)\).
Worked Examples
Worked Example: Finding a Value from a Polynomial Rate
A quantity \(Q\) has rate \(Q'(t)=3t^2-4t+1\), and \(Q(1)=5\). Find \(Q(3)\).
Use the initial input \(a=1\), initial value \(Q_0=5\), and later input \(x=3\):
An antiderivative of the rate is \(t^3-2t^2+t\), since its derivative is \(3t^2-4t+1\). Evaluate at both bounds:
Therefore \(Q(3)=5+12=17\).
Check: The formula for the quantity is \(Q(x)=5+\int_1^x(3t^2-4t+1)\,dt=5+(x^3-2x^2+x)\). At \(x=1\), it gives \(5+(1-2+1)=5\). Its derivative is \(3x^2-4x+1\), the given rate. At \(x=3\), it gives \(5+(27-18+3)=17\), as calculated.
Worked Example: A Rate That Becomes Negative
A volume \(V(t)\), measured in liters, changes at the rate \(V'(t)=4-2t\) liters per hour. At \(t=0\), the volume is \(10\) liters. Find the volume at \(t=3\) hours.
Apply the formula with initial input \(0\) and initial value \(10\):
An antiderivative is \(4t-t^2\). Thus the change over the three hours is
So \(V(3)=10+3=13\) liters. The rate is positive for \(0\le t<2\) and negative for \(2<t\le3\), but the definite integral gives the total change across the entire interval, including both parts. A negative rate on part of the interval does not mean the formula should be changed; it is already included in the integral.
Check: The resulting function is \(V(x)=10+4x-x^2\). It gives \(V(0)=10+0-0=10\), and \(V'(x)=4-2x\), as required. At \(x=3\), \(V(3)=10+12-9=13\) liters.
Worked Example: Writing the Quantity as a Function
A quantity \(H\) has rate \(H'(x)=6x-2\), with initial value \(H(2)=4\). Write an accumulation formula for \(H(x)\), simplify it, and find \(H(5)\).
Start with the given input and value:
An antiderivative of \(6t-2\) is \(3t^2-2t\). Therefore, for \(x\) in the interval of the model,
Now substitute \(x=5\):
Check: At the initial input, \(H(2)=3(2)^2-2(2)-4=12-4-4=4\). Differentiating gives \(H'(x)=6x-2\), which matches the rate. The function therefore satisfies both the initial condition and the rate condition.
A Common Mistake: Forgetting the Initial Value
A definite integral of a rate gives the change from the initial input, not the final quantity. For instance, in the volume example, the calculation \(\int_0^3(4-2t)\,dt=3\) gives a change of \(3\) liters. Calling \(3\) liters the volume at \(t=3\) would ignore the initial \(10\) liters.
The correction is to keep the two roles distinct: the initial condition is the starting quantity, and the integral is the change measured from that starting quantity. In a function formula, the same idea appears as an added constant: the accumulation integral alone is zero at its base point, whereas the quantity at that point is \(Q_0\).
Choosing the Bounds and Checking the Result
Before evaluating, identify three pieces of information: the input \(a\) at which the quantity is known, its value \(Q_0\), and the input \(x\) being asked about. Then place \(a\) as the lower bound and \(x\) as the upper bound. A different choice of base point would require the quantity's value at that different point.
Write the known input \(a\) and value \(Q_0\) from \(Q(a)=Q_0\).
Write \(Q(x)=Q_0+\int_a^x r(t)\,dt\), using a dummy variable such as \(t\) inside the integral.
Use an antiderivative when available, substituting the upper bound and subtracting the value at the lower bound.
Substitute the initial input into the quantity formula and differentiate the formula to compare its derivative with the given rate.
Writing the formula before evaluating helps avoid confusing the accumulated change with the quantity itself. It also keeps the base point visible, which is especially useful when the requested input lies on the other side of that point.
Check Your Understanding
For each question, distinguish the initial value from the change supplied by the integral. Check a proposed function by testing its initial value and derivative.
- If \(P'(t)=2t+1\) and \(P(0)=6\), write an accumulation formula for \(P(x)\).
- For the conditions in question 1, evaluate \(P(2)\).
- If \(R'(t)=3-t\) and \(R(1)=8\), write an accumulation formula for \(R(0)\). How does the order of the bounds appear?
- A quantity has rate \(S'(x)=4x\) and initial value \(S(2)=7\). Find an accumulation formula for \(S(x)\), simplify it, and check the initial value.
- Suppose \(M(x)=5+\int_3^x r(t)\,dt\), where \(r\) is continuous. What is \(M(3)\), and what is \(M'(x)\)?
