One Theorem, Two Ways to Describe Change
A definite integral and an antiderivative may seem to describe different things. The integral records accumulated signed change over an interval; an antiderivative is a function whose derivative is the given rate. The Fundamental Theorem of Calculus connects these descriptions.
The connection is especially clear when we begin with an accumulation function. Fix a starting point \(a\), and accumulate the values of a function \(f\) from \(a\) to a variable endpoint \(x\). If \(f\) is continuous, the resulting accumulation function has derivative \(f\). It is therefore an antiderivative of \(f\). Once we know that, the other part of the theorem evaluates the accumulated change using endpoint values of any antiderivative.
Accumulation Functions and Antiderivatives
The variable \(t\) in the integral below is a dummy variable: it indicates the input at which \(f\) is being accumulated. The endpoint \(x\), by contrast, determines how far the accumulation proceeds.
This defines a function \(A\), provided the integral exists for the values of \(x\) under consideration. In this tutorial we use the standard Fundamental Theorem of Calculus Part I: if \(f\) is continuous near an interior point \(x\), then the accumulation function is differentiable there and
Thus \(A\) is itself an antiderivative of \(f\). This is more than a way to compute \(A\): it shows that accumulation, which is defined by integration, creates a function whose instantaneous rate of change is the original integrand.
The theorem has a complementary evaluation statement. If \(F\) is any antiderivative of \(f\) on an interval containing \([a,b]\), meaning \(F'(x)=f(x)\) there, then
The hypotheses matter. The standard statement used here assumes \(f\) is continuous on an interval containing \([a,b]\), and that \(F\) is an antiderivative of \(f\) on that interval. Continuity ensures the accumulation function has the derivative required to connect the two statements. The evaluation formula then expresses the definite integral as a difference of endpoint values.
Why the Two Parts Fit Together
The key is that both \(A\) and \(F\) have the same derivative: \(f\). Two differentiable functions with equal derivatives on an interval can differ only by a constant. The accumulation function has a particularly useful value at its starting point: \(A(a)=0\). These facts determine which constant relates \(A\) and \(F\).
Here is the complete argument. By Part I, \(A'(x)=f(x)\) for interior points \(x\). By definition, \(F'(x)=f(x)\). Therefore the function \(A-F\) has derivative zero in the interior. The Mean Value Theorem implies that a continuous function on \([a,b]\) that is differentiable in \((a,b)\) and has derivative zero there must be constant on \([a,b]\).
For interior \(x\), \((A-F)'(x)=A'(x)-F'(x)=f(x)-f(x)=0\).
Since \(A-F\) is continuous on \([a,b]\) and differentiable on \((a,b)\), its zero derivative implies that \(A-F\) is constant on \([a,b]\).
Because \(A(a)=\int_a^a f(t)\,dt=0\), we have \((A-F)(a)=-F(a)\). The constant value of \(A-F\) is therefore \(-F(a)\).
At \(x=b\), the relation \(A(x)-F(x)=-F(a)\) gives \(A(b)=F(b)-F(a)\). Since \(A(b)=\int_a^b f(t)\,dt\), this is the endpoint evaluation formula.
The proof explains why an arbitrary antiderivative works. It need not equal the accumulation function itself: it differs from that function by the constant \(F(a)\). Subtracting endpoint values removes that constant.
Worked Examples
Worked Example: A Polynomial Accumulated from Zero
Let \(f(t)=3t^2-4t+1\), and define \(A(x)=\int_0^x f(t)\,dt\). Find \(A(x)\), then use the connection between accumulation and antiderivatives to evaluate \(\int_0^2 f(t)\,dt\).
An antiderivative of \(f\) is \(F(x)=x^3-2x^2+x\), since
Using the lower endpoint \(0\), the accumulation function is
Differentiate to check that it is an antiderivative:
At \(x=2\),
The endpoint formula gives the same value:
The two calculations agree because \(A(0)=0\), while \(F(0)=0\) in this example. In general, the accumulation function is \(F(x)-F(0)\), not necessarily \(F(x)\). Here differentiation confirms \(A'=f\), and the endpoint difference confirms the definite integral is \(2\).
Worked Example: A Nonzero Starting Point
Let \(f(t)=2t\), and accumulate from \(a=1\): \(A(x)=\int_1^x 2t\,dt\). Find \(A(x)\) and evaluate the accumulated change at \(x=3\), using an antiderivative with a nonzero constant.
Take \(F(x)=x^2+7\). Its derivative is
The accumulation function is the difference between the antiderivative at \(x\) and at the starting point:
Differentiate the result:
At the endpoint \(x=3\), the accumulation function gives
Using the endpoint difference directly gives
The constant \(7\) cancels, as it must. The derivative check confirms the accumulation function has derivative \(2x\), and both evaluations give \(8\).
Worked Example: A Trigonometric Integrand
Let \(f(t)=\cos t\), and define \(A(x)=\int_0^x \cos t\,dt\). Find \(A(x)\), then evaluate \(\int_0^{\pi/2}\cos t\,dt\).
An antiderivative is \(F(x)=\sin x\), because
The accumulation function is
Its derivative is \(A'(x)=\cos x=f(x)\), and it has the required starting value \(A(0)=\sin 0=0\). At the upper endpoint,
The endpoint formula independently gives
Thus the accumulation value and the antiderivative difference agree. The derivative and endpoint checks confirm the result.
A Common Mistake: Forgetting the Starting Value
A frequent error is to assume that the accumulation function \(A(x)=\int_a^x f(t)\,dt\) is exactly the same as a chosen antiderivative \(F(x)\). That would make \(A(a)=F(a)\). But the integral from \(a\) to itself is zero, so \(A(a)=0\); an arbitrary antiderivative can have any value at \(a\).
For instance, in the example with \(f(t)=2t\) and \(a=1\), choosing \(F(x)=x^2+7\) and incorrectly setting \(A(x)=F(x)\) would give \(A(1)=8\). This contradicts the definition:
The correction is to subtract the starting value of the antiderivative:
For a definite integral, the same correction appears as the endpoint difference \(F(b)-F(a)\). Keeping the lower endpoint in view prevents the constant from being lost.
Why This Connection Is Useful
The two parts of the theorem let us move between two perspectives. The accumulation function gives a precise way to represent total signed change up to a moving endpoint. Its derivative describes the rate at which that accumulated quantity changes. The antiderivative evaluation formula, in turn, gives an efficient way to compute the total change over a fixed interval.
The connection also guarantees that a continuous integrand has an antiderivative: the function \(A(x)=\int_a^x f(t)\,dt\) is one. This does not promise that the antiderivative can always be written using a familiar elementary formula. It does show that the integral-defined accumulation function has the required derivative.
The essential relationship can be remembered in this order: define the accumulation function from a fixed starting point; Part I identifies its derivative as the integrand; then compare it with any other antiderivative. Their difference is constant, and the starting value determines that constant. Evaluating at the upper endpoint yields the definite integral.
Check Your Understanding
Use the connection between accumulation and antiderivatives to answer these questions.
- If \(A(x)=\int_a^x f(t)\,dt\) and \(f\) is continuous, what is \(A'(x)\) at an interior point?
- If \(F'(x)=f(x)\), what expression in terms of \(F\) gives the accumulation function based at \(a\)?
- Why does the difference \(A-F\) have the same value at every point of the interval?
- For \(f(t)=4t\) and \(A(x)=\int_2^x 4t\,dt\), find \(A(x)\) and \(A(3)\).
- Why does adding a constant to an antiderivative not change the definite integral computed by endpoint evaluation?
