A Rate Describes Change
A derivative measures how quickly a quantity changes. If \(Q(t)\) is a quantity and \(Q'(t)=f(t)\), then \(f(t)\) is its instantaneous rate of change. Over an interval from \(a\) to \(b\), the quantity’s net change is its ending value minus its starting value:
A definite integral of the rate over that interval gives the same net change. The reason is that adding the rate’s contributions over many small intervals approximates adding the quantity’s small changes. Those small changes telescope: every intermediate value is subtracted once and added once, leaving only the final value minus the initial value.
An antiderivative makes that net change easy to calculate. If \(F'(x)=f(x)\), then \(F\) can represent a quantity whose rate is \(f\), and its change from \(a\) to \(b\) is \(F(b)-F(a)\).
The Definite Integral and the Net-Change Theorem
Let \(f\) be continuous on \([a,b]\), and let \(F\) be a function differentiable on an interval containing \([a,b]\), with \(F'(x)=f(x)\) there. The Fundamental Theorem of Calculus, Part 2, says:
The expression on the right is read “\(F\) evaluated at \(b\), minus \(F\) evaluated at \(a\).” It is the net change in \(F\), not the sum of the endpoint values. Continuity of \(f\) guarantees that its Riemann sums converge to the definite integral. The condition \(F'=f\) links those sums to changes in \(F\).
This result explains why finding an antiderivative is useful for evaluating a definite integral: it replaces the limit of sums with a subtraction at the endpoints. It also explains the meaning of that subtraction. If \(f\) is the rate of a quantity, then integrating \(f\) gives the quantity’s net change over the interval.
Why the Formula Works: Add Small Changes
Consider a partition of \([a,b]\) into points \(a=x_0<x_1<\cdots<x_n=b\). Each small interval runs from \(x_{i-1}\) to \(x_i\), and has width \(\Delta x_i=x_i-x_{i-1}\). By the Mean Value Theorem, on each such interval there is a point \(\xi_i\) between its endpoints for which the change in \(F\) is exactly its derivative at that point times the interval width:
Now add these changes over all the subintervals. On the left, intermediate endpoint values cancel. On the right, the terms form a Riemann sum for \(f\). This proves the relationship as the partition is refined:
Because \(F\) is continuous on each closed subinterval and differentiable in its interior, the change in \(F\) has the form \(f(\xi_i)\Delta x_i\).
The intermediate values cancel, leaving \(F(b)-F(a)\), while the sum of rate contributions is \(\sum_{i=1}^{n} f(\xi_i)\Delta x_i\).
As the maximum subinterval width approaches zero, the Riemann sums approach \(\int_a^b f(x)\,dx\), because \(f\) is continuous. The left side remains \(F(b)-F(a)\).
In equation form, the cancellation and limit are:
The two expressions are equal for every partition, by the Mean Value Theorem. Taking the limit therefore gives \(F(b)-F(a)=\int_a^b f(x)\,dx\). This argument connects three ideas: a derivative is a rate, a rate times a short interval approximates a small change, and adding small changes produces the overall net change.
Worked Examples
Worked Example: Position Change from a Velocity Function
A particle has position \(p(t)=t^3-2t^2+t+4\) over the time interval \(0\leq t\leq 2\). Find its displacement using the velocity \(v(t)=p'(t)\), and verify the result with an antiderivative.
First differentiate the position to obtain velocity:
Because \(p\) is an antiderivative of \(v\), its change over the interval is:
The integral gives the same result. An antiderivative of \(v(t)\) is \(F(t)=t^3-2t^2+t\), so:
Check: \(F'(t)=3t^2-4t+1=v(t)\), and the displacement \(p(2)-p(0)=2\) agrees with the integral.
Worked Example: Net Change in Water Volume
A tank’s volume changes at rate \(r(t)=6t-2\) liters per minute, where \(t\) is time in minutes. Find the net change in volume from \(t=1\) to \(t=4\).
An antiderivative of the rate is \(V_0(t)=3t^2-2t\), since:
Evaluate the antiderivative at both endpoints, showing each value:
Subtract the initial endpoint value from the final endpoint value:
Check: Differentiating \(3t^2-2t\) recovers \(6t-2\), and its endpoint difference is \(39\). The net volume change is \(39\) liters; this difference alone does not specify the tank’s starting volume.
Worked Example: A Rate with Zero Net Change
A velocity \(w(t)=t^2-3t+5\) meters per second changes at rate \(a(t)=2t-3\) meters per second per second. Find the net change in velocity from \(t=0\) to \(t=3\).
Differentiate \(w\) to check the rate:
Since \(w\) is an antiderivative of \(a\), evaluate it at the endpoints:
Equivalently, evaluate the definite integral using \(W(t)=t^2-3t\), an antiderivative of \(a(t)\):
Check: \(W'(t)=2t-3\), and the velocity has the same value, \(5\) meters per second, at both endpoints. Thus its net change is zero, matching the integral.
Why Only the Endpoint Difference Matters
An antiderivative is not unique: if \(F'(x)=f(x)\), then \((F(x)+C)'=f(x)\) for any constant \(C\). But the constant disappears when endpoint values are subtracted:
This fits the net-change interpretation. Adding a constant changes the quantity’s chosen baseline but does not change how much it changes between two times or positions. The definite integral depends on the rate over the interval, not on the quantity’s starting amount.
For the same reason, the order of subtraction is important. The net change is final value minus initial value. The integral formula uses \(F(b)-F(a)\), not \(F(a)-F(b)\) or \(F(a)+F(b)\).
Common Mistake: Adding the Endpoint Values
A frequent error is to evaluate an antiderivative at both endpoints but add the results. For example, an incorrect evaluation might claim:
The error is that a net change subtracts the starting value. The correct calculation is:
Check: \(F'(x)=2x\), as required. The net change in \(F\) from \(x=1\) to \(x=3\) is \(9-1=8\), not \(10\).
When to Use the Relationship
Use the endpoint-difference formula when the integrand has an antiderivative on the interval and the definite integral is defined there. In the standard AP setting, if \(f\) is continuous on \([a,b]\) and \(F'=f\), evaluate \(F(b)-F(a)\). Interpret the result as net change when \(f\) is the rate of a quantity.
Keep the roles of the expressions distinct: \(f\) is the rate, \(\int_a^b f(x)\,dx\) is the accumulated rate over the interval, and \(F(b)-F(a)\) is the resulting net change. The equality follows because adding the rate’s small contributions adds the quantity’s small changes, which telescope to the endpoint difference.
Check Your Understanding
Use the net-change relationship to answer these questions.
- If \(F'(x)=f(x)\), what does \(F(b)-F(a)\) represent?
- What condition on \(f\) is sufficient in the standard theorem for \(\int_a^b f(x)\,dx=F(b)-F(a)\)?
- Why do the intermediate values cancel when changes in \(F\) are added across a partition?
- If \(Q'(t)=r(t)\), how would you express the net change in \(Q\) from \(t=a\) to \(t=b\) as a definite integral?
- For \(\int_0^2 3x^2\,dx\), use an antiderivative and compute the endpoint difference.
