One Theorem, Two Different Jobs
A definite integral and a derivative answer different questions. A definite integral measures accumulated signed change across an interval. A derivative describes a function’s instantaneous rate of change. The Fundamental Theorem of Calculus connects these ideas, but its two parts are not interchangeable: one finds the derivative of an accumulation function, while the other evaluates a definite integral.
The distinction matters in practice. If a function is defined by an integral whose upper endpoint varies, the first part tells us its derivative. If we need the value of a definite integral and know an antiderivative of the integrand, the second part provides a way to calculate it.
Part I: Differentiating an Accumulation Function
Let \(f\) be continuous on an open interval \(I\), and choose a fixed point \(a\) in \(I\). For each \(x\) in \(I\), define the accumulation function \(A\) by integrating \(f\) from \(a\) to \(x\). The variable \(t\) is a dummy variable of integration: it identifies the input to \(f\) inside the integral.
The conclusion is about the derivative of the function \(A\), not the value of the integral at a particular endpoint. The fixed lower endpoint \(a\) sets where accumulation begins; the changing upper endpoint \(x\) determines the current rate of accumulation. The rate at \(x\) is the value \(f(x)\).
Continuity is the hypothesis that guarantees this conclusion at every point of the interval. The theorem as stated does not promise that \(A'(x)=f(x)\) at a point where \(f\) is not continuous. More detailed questions about what happens under weaker assumptions require separate conditions.
Part II: Evaluating a Definite Integral
The second part starts with an integrand and an antiderivative. An antiderivative of \(f\) on an interval is a function \(G\) whose derivative equals \(f\) throughout that interval. In the evaluation formula, the values of \(G\) at the endpoints determine the definite integral.
The notation \(G(b)-G(a)\) is sometimes written \(\left.G(x)\right|_a^b\). The order is upper endpoint minus lower endpoint. Any two antiderivatives of \(f\) on the interval differ by a constant, and that constant cancels when their endpoint values are subtracted. Thus, the result does not depend on which antiderivative is chosen.
Part II is an evaluation rule: it turns a definite integral into a difference of endpoint values. It does not say that the integral itself is a derivative. In particular, the result is a number once \(a\) and \(b\) are fixed.
How the Two Parts Connect
Part I creates an antiderivative from an accumulation function. Part II says that a definite integral can be evaluated from any antiderivative of its integrand. Together, they explain why differentiation and integration undo one another in this setting.
| Part | Given | Conclusion | Main question answered |
|---|---|---|---|
| Part I | A continuous function \(f\) and \(A(x)=\int_a^x f(t)\,dt\) | \(A'(x)=f(x)\) | What is the rate of change of this accumulation? |
| Part II | A continuous function \(f\) and an antiderivative \(G\) | \(\int_a^b f(x)\,dx=G(b)-G(a)\) | What is the value of this definite integral? |
Here is a derivation of Part II using Part I. This shows explicitly how the two statements fit together. Assume the hypotheses of Part II. Define \(A(x)=\int_a^x f(t)\,dt\) on \([a,b]\). Since \(f\) is continuous, Part I gives \(A'(x)=f(x)\) for \(a<x<b\). By hypothesis, \(G'(x)=f(x)\) on the same interval.
Define \(H(x)=A(x)-G(x)\). For every \(x\) in \((a,b)\), \(H'(x)=A'(x)-G'(x)=f(x)-f(x)=0\).
For any \(u,v\in[a,b]\) with \(u<v\), \(H\) is continuous on \([u,v]\) and differentiable on \((u,v)\). The Mean Value Theorem gives a \(c\in(u,v)\) such that \(H(v)-H(u)=H'(c)(v-u)=0\). Thus \(H\) has the same value at both endpoints.
Since \(A(a)=\int_a^a f(t)\,dt=0\), the equality \(H(b)=H(a)\) becomes \(A(b)-G(b)=-G(a)\). Rearranging gives \(\int_a^b f(t)\,dt=A(b)=G(b)-G(a)\), as claimed.
Worked Examples
Worked Example: Differentiate an Accumulation Function
Let \(A(x)=\int_0^x(2t+3)\,dt\). Find \(A'(x)\), and check the result by first writing \(A(x)\) explicitly.
The integrand \(f(t)=2t+3\) is continuous for every real \(t\), so Part I applies. It gives \(A'(x)=f(x)=2x+3\).
To check, use an antiderivative of \(2t+3\), namely \(t^2+3t\), to evaluate the integral: $$ A(x)=(x^2+3x)-(0^2+3\cdot0)=x^2+3x. $$ Differentiating this expression gives \(A'(x)=2x+3\), which agrees with Part I. For example, at \(x=2\), \(A'(2)=2\cdot2+3=7\).
Worked Example: Evaluate a Definite Integral
Evaluate \(\int_1^3(3x^2-4x+2)\,dx\).
The integrand is continuous on \([1,3]\). An antiderivative is \(G(x)=x^3-2x^2+2x\), since $$ G'(x)=3x^2-4x+2. $$ By Part II, $$ \begin{aligned} \int_1^3(3x^2-4x+2)\,dx &=G(3)-G(1)\\ &=(27-18+6)-(1-2+2)\\ &=15-1\\ &=14. \end{aligned} $$ Check: differentiating the antiderivative gives \(3x^2-4x+2\), the original integrand, so the endpoint evaluation used the correct antiderivative.
Worked Example: Distinguish an Accumulation Value from Its Derivative
Let \(B(x)=\int_{-1}^x(t^2+1)\,dt\). Find both \(B(2)\) and \(B'(2)\).
The integrand \(t^2+1\) is continuous, so Part I gives \(B'(x)=x^2+1\). Therefore, $$ B'(2)=2^2+1=5. $$ This is the derivative, or rate of change, at \(x=2\).
To find the accumulation value, use the antiderivative \(G(t)=\frac{t^3}{3}+t\). Its derivative is \(G'(t)=t^2+1\), as required. Part II gives $$ \begin{aligned} B(2) &=G(2)-G(-1)\\ &=\left(\frac{8}{3}+2\right)-\left(-\frac{1}{3}-1\right)\\ &=\frac{14}{3}-\left(-\frac{4}{3}\right)\\ &=6. \end{aligned} $$ Thus \(B(2)=6\), while \(B'(2)=5\). They are different quantities. As a check, differentiating \(G(x)-G(-1)\) gives \(x^2+1\), and its value at \(x=2\) is \(6\).
A Common Mix-Up
A frequent mistake is to use the statement of Part I as though it directly evaluated a definite integral. For example, someone might write $$ \int_1^3(3x^2-4x+2)\,dx=f(3)=3\cdot3^2-4\cdot3+2=17. $$ This is wrong: \(f(3)\) is the value of the integrand at one endpoint, not the accumulated value over the whole interval. Part I concerns the derivative of an accumulation function.
To evaluate the integral, use Part II instead. With \(G(x)=x^3-2x^2+2x\), $$ \int_1^3(3x^2-4x+2)\,dx=G(3)-G(1)=15-1=14. $$ The answer can also be distinguished by its role: \(14\) is the definite integral, whereas the derivative of \(x\mapsto\int_1^x(3t^2-4t+2)\,dt\) is \(3x^2-4x+2\).
Why the Hypotheses Matter
Each part has a specific input and conclusion. For Part I, continuity of \(f\) on an interval containing the fixed starting point and the variable endpoint ensures the accumulation function is differentiable there and has derivative \(f\). For Part II, continuity of \(f\) on the closed interval ensures the definite integral is defined in the usual setting, and the existence of an antiderivative \(G\) supplies the endpoint values used to evaluate it.
Do not switch the roles of the hypotheses and conclusions. Part I does not require you to begin with an antiderivative: it constructs a differentiable accumulation function from the integral. Part II does require an antiderivative in order to use the endpoint evaluation formula. Both parts concern a continuous integrand here, but their conclusions answer different questions.
Check Your Understanding
For each question, identify which part of the theorem applies and what its conclusion means.
- If \(A(x)=\int_2^x f(t)\,dt\) and \(f\) is continuous, what does Part I give for \(A'(x)\)?
- What additional function is needed to apply Part II to evaluate \(\int_a^b f(x)\,dx\)?
- In Part II, why is the endpoint difference written \(G(b)-G(a)\), rather than \(G(a)-G(b)\)?
- If \(A(x)=\int_0^x(t^2+1)\,dt\), what is \(A'(2)\), and what kind of quantity is it?
- For \(f(x)=2x\) on \([1,4]\), use an antiderivative to evaluate \(\int_1^4 2x\,dx\).
