From Instantaneous Rate to Total Change
A derivative describes an instantaneous rate of change. A definite integral records accumulated change across an interval. The second part of the Fundamental Theorem of Calculus connects these two ideas: when a function is the derivative of an antiderivative, its integral is the change in that antiderivative.
The formula is useful for evaluating integrals, but it also has a reason behind it. The accumulation function and the antiderivative have the same derivative. Their difference must therefore remain constant, so their changes from one endpoint to the other agree.
The function \(F\) is called an antiderivative of \(f\) on the interval. The continuity condition on \(f\) ensures that the accumulation function used in the proof has derivative \(f\), as stated by the first part of the Fundamental Theorem of Calculus. The continuity of \(F\) at the endpoints lets us apply the Mean Value Theorem on the closed interval.
The Accumulation Function
Fix the starting point \(a\) and define an accumulation function \(A\) by integrating \(f\) from \(a\) to a variable point \(x\):
The variable \(t\) is used inside the integral to track the values being accumulated. The endpoint \(x\) determines where the accumulation stops. Since \(f\) is continuous, the first part of the Fundamental Theorem of Calculus gives \(A'(x)=f(x)\) for \(a<x<b\). By assumption, \(F'(x)=f(x)\) there as well.
Thus \(F\) and \(A\) have the same derivative. To turn that observation into a statement about their endpoint values, consider their difference:
The derivative of this difference is zero throughout the interior of the interval. The Mean Value Theorem then shows that \(H\) has the same value at both endpoints. This is the key step: matching derivatives means the two functions differ by a constant, and constants cancel when changes are calculated.
Why the Endpoint Difference Equals the Integral
Here is the complete argument. First, \(A\) is continuous on \([a,b]\) and differentiable on \((a,b)\), and \(F\) has those same regularity properties by assumption. Therefore \(H=F-A\) is continuous on \([a,b]\) and differentiable on \((a,b)\). For every \(x\) in the interior,
Take any \(u,v\in[a,b]\) with \(u<v\). The Mean Value Theorem applied to \(H\) on \([u,v]\) gives some \(c\in(u,v)\) such that
Hence \(H(v)=H(u)\) for any such pair of points. In particular, \(H(b)=H(a)\). Substituting the definition of \(H\) gives
By definition, \(A(a)=\int_a^a f(t)\,dt=0\), and \(A(b)=\int_a^b f(t)\,dt\). Rearranging the preceding equality now yields
This proves the theorem. It is not necessary for \(F\) and \(A\) to have the same starting value: the proof shows that their difference is constant, and the change from \(a\) to \(b\) is unaffected by that constant.
Worked Examples
Worked Example: A Polynomial Integrand
Evaluate \(\int_0^2(3x^2-4x+1)\,dx\) using an antiderivative, and check that its derivative matches the integrand.
An antiderivative is \(F(x)=x^3-2x^2+x\). Differentiate it term by term:
Thus \(F'(x)\) is exactly the given integrand. The integrand is a polynomial, so it is continuous on \([0,2]\); the theorem applies. Evaluate \(F\) at each endpoint:
Subtract the starting value from the ending value:
Check: differentiating the stated antiderivative recovers \(3x^2-4x+1\), and the endpoint difference is \(2\). In the proof’s terms, the accumulation function starts at zero and has the same derivative as \(F\), which also has value zero at \(x=0\). Their difference is therefore zero, so the accumulation at \(x=2\) is \(F(2)=2\).
Worked Example: A Trigonometric Integrand
Evaluate \(\int_0^{\pi/2}\cos x\,dx\). Use \(F(x)=\sin x\) and verify the derivative and endpoint values.
First check the antiderivative:
The function \(\cos x\) is continuous on \([0,\pi/2]\), so the theorem applies. At the upper endpoint, \(F(\pi/2)=\sin(\pi/2)=1\). At the lower endpoint, \(F(0)=\sin(0)=0\). Therefore
Check: the derivative of \(F(x)=\sin x\) is the original integrand, and substituting both endpoints gives the displayed difference \(1\). Since the accumulation function and \(F\) have equal derivatives, the integral’s accumulated change must equal this endpoint change.
Worked Example: A Reciprocal Integrand
Evaluate \(\int_1^e \frac{1}{x}\,dx\), where \(e\) is the base of the natural logarithm.
On \([1,e]\), \(x\) is positive, so \(\ln x\) is defined and differentiable, with
The integrand \(1/x\) is continuous on \([1,e]\), so the theorem’s continuity condition is satisfied. At the upper endpoint, \(F(e)=\ln e=1\). At the lower endpoint, \(F(1)=\ln 1=0\). Hence
Check: \(F'(x)=1/x\) on the entire interval, and the endpoint values give a difference of \(1\). The theorem identifies that difference with the accumulated integral.
A Common Mistake: Subtracting the Wrong Values
The theorem calls for subtracting the values of an antiderivative, not the values of the integrand. For instance, consider \(f(x)=2x\) on \([1,3]\). A common incorrect calculation is
That difference compares the integrand at the endpoints; it does not give the integral. Choose \(F(x)=x^2\), for which \(F'(x)=2x\). The correct calculation uses the endpoint values of \(F\):
Check: differentiating \(x^2\) gives \(2x\), and its endpoint change is \(8\). The wrong calculation gives \(4\) because it uses \(f\) where the theorem requires \(F\). The two functions play different roles: \(f\) is the rate being accumulated, while \(F\) records a quantity whose change represents that accumulation.
Conditions and Meaning
The theorem is a precise link, not a rule to apply without checking its assumptions. In the version used here, the integrand must be continuous on the interval, and an antiderivative must be defined and continuous at the endpoints and differentiable in the interior, with derivative equal to the integrand. Continuity of \(f\) is what permits the accumulation function to have derivative \(f\) in the interior.
The integral represents net accumulation. If \(f\) is negative on part of the interval, that portion contributes negatively; the theorem still relates the resulting signed total to \(F(b)-F(a)\). No assumption that the integrand is positive is needed.
The practical evaluation rule and the justification should be kept together. To use the rule, identify an antiderivative, verify its derivative, confirm the relevant conditions, and compute the endpoint difference. The proof explains why that calculation measures the integral: both describe the same accumulated change.
Check Your Understanding
Try each question by connecting the derivative condition to the endpoint change.
- State the continuity and differentiability conditions in the theorem as presented here.
- Why does the accumulation function \(A(x)=\int_a^x f(t)\,dt\) have derivative \(f(x)\) in the interval’s interior?
- If \(F'(x)=A'(x)\) on the interval, what can be concluded about \(F(x)-A(x)\), and what result justifies that conclusion?
- For \(f(x)=4x^3\) on \([0,1]\), use an antiderivative to evaluate the integral and check the derivative.
- What is wrong with using \(f(b)-f(a)\) in place of \(F(b)-F(a)\)?
