Prove the Fundamental Theorem of Calculus

From Integrability to the Fundamental Theorem

The previous tutorial proved that every continuous function on a closed interval is Riemann integrable. The Fundamental Theorem of Calculus explains why this matters: integration and differentiation are linked, not merely as separate operations, but as ways of recovering one another. We will prove both directions of that link.

First, integrating a continuous function from a fixed point to a variable endpoint produces a function whose derivative is the original integrand. Second, if a function already has the integrand as its derivative, its change across an interval gives the integral. These statements have different hypotheses and different roles, so we will prove them separately.

Definition: Let $f$ be continuous on an interval containing $u$ and $v$. We use the signed-integral convention: if $u\leq v$, then $\int_u^v f(t)\,dt$ is the usual Riemann integral, and if $v<u$, then $$ \int_u^v f(t)\,dt=-\int_v^u f(t)\,dt. $$ In particular, $\int_u^u f(t)\,dt=0$. For a fixed $a$, the function $G(x)=\int_a^x f(t)\,dt$ is called the accumulated integral of $f$ from $a$.

We will use the basic properties of Riemann integrals established earlier, including the absolute-value bound. One additional property is central to comparing accumulated integrals at different endpoints: integrals add across adjacent intervals. We record its justification first.

Theorem (Additivity of the Integral): If $f$ is continuous on $[a,b]$ and $a\leq c\leq b$, then $$ \int_a^b f(t)\,dt=\int_a^c f(t)\,dt+\int_c^b f(t)\,dt. $$

Proof. The assertion is immediate if $c=a$ or $c=b$, since the integral over an interval of length zero is zero. Suppose $a<c<b$. The restrictions of $f$ to $[a,c]$ and $[c,b]$ are continuous, hence Riemann integrable by the theorem from the previous tutorial.

For a partition $P$ of either interval, its lower sum is at most the integral on that interval, and its upper sum is at least the integral. Given $\varepsilon>0$, the Darboux Criterion supplies partitions $P_1$ of $[a,c]$ and $P_2$ of $[c,b]$ such that each upper sum minus lower sum is less than $\varepsilon/2$. Write $L_1,U_1$ for the lower and upper sums of $P_1$, and $L_2,U_2$ for those of $P_2$. Joining the partitions gives a partition of $[a,b]$, whose lower and upper sums are $L_1+L_2$ and $U_1+U_2$.

Because $f$ is also integrable on $[a,b]$, its integral lies between the lower and upper sums of this joined partition. The sum of the two integrals over the smaller intervals lies between the same bounds. Thus the absolute difference between $\int_a^b f$ and $\int_a^c f+\int_c^b f$ is at most

(U_1+U_2)-(L_1+L_2) =(U_1-L_1)+(U_2-L_2)<\varepsilon.

Since this holds for every $\varepsilon>0$, the two quantities are equal. $\square$

Together with the signed-integral convention, additivity gives $\int_u^v f=\int_u^w f+\int_w^v f$ whenever all three endpoints lie in the interval. To see why the convention matters, when $v<u$, both the direction of integration and the sign of the integral reverse. This identity lets us express a small change in an accumulated integral as an integral over just the small interval between its endpoints.

The Derivative of an Accumulated Integral

Theorem (Fundamental Theorem of Calculus, Part I): Let $f:[a,b]\to\mathbb{R}$ be continuous, and define $$ G(x)=\int_a^x f(t)\,dt \qquad (x\in[a,b]). $$ Then $G$ is differentiable at every $c\in(a,b)$, and $G'(c)=f(c)$.

Proof. Fix $c\in(a,b)$. For any nonzero $h$ sufficiently close to zero that $c+h\in[a,b]$, additivity gives

\frac{G(c+h)-G(c)}{h} =\frac{1}{h}\int_c^{c+h}f(t)\,dt.

Subtract $f(c)$. By linearity of the integral and the signed-integral convention,

\frac{G(c+h)-G(c)}{h}-f(c) =\frac{1}{h}\int_c^{c+h}\bigl(f(t)-f(c)\bigr)\,dt.

The Absolute-Value Bound for an Integral implies

\left|\frac{G(c+h)-G(c)}{h}-f(c)\right| \leq \frac{1}{|h|}\int_{\min(c,c+h)}^{\max(c,c+h)} |f(t)-f(c)|\,dt.

Let $\varepsilon>0$. Continuity of $f$ at $c$ supplies a $\delta>0$ such that $|f(t)-f(c)|<\varepsilon$ whenever $t\in[a,b]$ and $|t-c|<\delta$. If $0<|h|<\delta$, every point between $c$ and $c+h$ satisfies that condition. The integrand in the last bound is therefore at most $\varepsilon$ throughout that interval, so

\left|\frac{G(c+h)-G(c)}{h}-f(c)\right| \leq \frac{\varepsilon |h|}{|h|} =\varepsilon.

Since this estimate holds for every positive $\varepsilon$ when $h$ is sufficiently small, the difference quotient tends to $f(c)$. Hence $G'(c)=f(c)$. $\square$

The proof depends on continuity at the point $c$, not on a formula for an antiderivative. It also applies whether $h$ is positive or negative. The signed-integral convention ensures that the difference quotient averages values of $f$ over the interval between $c$ and $c+h$ in either case.

Worked Example: Differentiating an Accumulated Integral

Let $G(x)=\int_2^x (1+t^2)\,dt$ for $x\in[0,4]$. The integrand $f(t)=1+t^2$ is continuous on this interval, so Part I gives

G'(x)=1+x^2 \qquad (0<x<4).

In particular, $G'(1)=1+1^2=2$. This conclusion does not require first evaluating $G(x)$; the theorem obtains the derivative directly from the integrand at the moving endpoint.

Evaluating an Integral with an Antiderivative

Part I constructs an antiderivative from an integral. The converse explains why a known antiderivative can be used to calculate an integral. We first make explicit the continuity needed at the endpoints when we apply the zero-derivative result from the previous tutorials.

Theorem (Fundamental Theorem of Calculus, Part II): Let $f:[a,b]\to\mathbb{R}$ be continuous. Suppose $F:[a,b]\to\mathbb{R}$ is continuous on $[a,b]$, differentiable on $(a,b)$, and satisfies $F'(x)=f(x)$ for every $x\in(a,b)$. Then $$ \int_a^b f(t)\,dt=F(b)-F(a). $$

Proof. Define $G(x)=\int_a^x f(t)\,dt$. By Part I, $G'(x)=f(x)$ for each $x\in(a,b)$. We also need $G$ to be continuous on the closed interval. Since $f$ is continuous on a compact interval, it is bounded: choose $M\geq0$ such that $|f(t)|\leq M$ for every $t\in[a,b]$. Additivity and the Absolute-Value Bound give, for $x,y\in[a,b]$,

|G(x)-G(y)| =\left|\int_y^x f(t)\,dt\right| \leq \int_{\min(x,y)}^{\max(x,y)}|f(t)|\,dt \leq M|x-y|.

This estimate implies that $G$ is continuous on $[a,b]$. Now set $H=F-G$. It is continuous on $[a,b]$, differentiable on $(a,b)$, and

H'(x)=F'(x)-G'(x)=f(x)-f(x)=0 \qquad (x\in(a,b)).

By the Zero Derivative Implies Constancy theorem from earlier in the course, $H$ is constant on $[a,b]$. Thus $H(b)=H(a)$, or $F(b)-G(b)=F(a)-G(a)$. Since $G(a)=\int_a^a f(t)\,dt=0$ and $G(b)=\int_a^b f(t)\,dt$, rearranging gives

\int_a^b f(t)\,dt=F(b)-F(a).

This proves Part II. $\square$

Worked Example: Evaluating a Polynomial Integral

On $[-1,2]$, let $f(t)=3t^2-4t+1$. The function $F(t)=t^3-2t^2+t$ satisfies

$$ F'(t)=3t^2-4t+1=f(t). $$

Both functions are continuous on the closed interval, and $F$ is differentiable in its interior. Part II therefore yields

\int_{-1}^{2}(3t^2-4t+1)\,dt =F(2)-F(-1) =(8-8+2)-(-1-2-1) =2-(-4) =6.

The endpoint values have been substituted into $F$ separately; the order $F(2)-F(-1)$ follows the orientation from the lower endpoint to the upper endpoint.

Variable Endpoints and the Chain Rule

Part I concerns an endpoint that moves at the same rate as the input variable. When both endpoints depend on $x$, the Chain Rule combines with Part I: the upper endpoint contributes its integrand value times its rate of change, while the lower endpoint contributes the corresponding term with a minus sign.

Worked Example: An Integral with Two Variable Limits

Define $J(x)=\int_x^{x^2+1} e^t\,dt$ for real $x$. The integrand is continuous, and $e^t$ is its own antiderivative. Part II gives an explicit expression:

J(x)=e^{x^2+1}-e^x.

Differentiating this expression verifies the rate of change:

J'(x)=2x e^{x^2+1}-e^x.

The two terms reflect the two moving endpoints. In particular, $J'(0)=0\cdot e^1-e^0=-1$. The upper endpoint has rate $2x$, which is zero at $x=0$; the lower endpoint has rate $1$, and its contribution is subtracted.

What the Theorem Does—and Does Not—Say

Part I needs continuity of the integrand to guarantee that its nearby values are close to $f(c)$. It then proves differentiability of the accumulated integral at interior points. The theorem does not claim that an arbitrary Riemann integrable function has this property at every point. Integrability alone is weaker than continuity, and the local estimate in the proof depends on continuity at the point where the derivative is taken.

Part II has a different starting point: a function $F$ whose derivative is the continuous integrand. It says that the integral is the net change in $F$. The continuity of $F$ on the closed interval matters in the proof: it allows the Zero Derivative Implies Constancy theorem to be applied on the whole interval, including its endpoints. In typical applications, $F$ is given by an elementary formula and this continuity is immediate.

The two parts can be used in sequence. Part I shows that $G(x)=\int_a^x f(t)\,dt$ is an antiderivative of $f$ in the interior. Part II then shows that any antiderivative satisfying its hypotheses evaluates the integral. This also explains why two antiderivatives of the same function differ by a constant: their difference has zero derivative on the interval.

For a derivative question, identify the moving endpoint.
Write the accumulated integral as a function of that endpoint and apply Part I at interior points.

For an evaluation question, find an antiderivative.
Verify that its derivative is the continuous integrand and that it is continuous on the closed interval.

Subtract endpoint values in the correct order.
For an integral from $a$ to $b$, use $F(b)-F(a)$; reversing the limits reverses the sign.

Check Your Understanding

Use the definitions and proofs above to answer these questions.

Why is a signed-integral convention useful when the variable endpoint lies to the left of the fixed endpoint?
In Part I, which continuity estimate makes the difference quotient converge to $f(c)$?
Where does additivity enter the proof that the accumulated integral is differentiable?
Why does Part II require continuity of $F$ on the closed interval, not only differentiability in its interior?
If $K(x)=\int_1^{x^3} f(t)\,dt$ and $f$ is continuous, what factors should appear when differentiating $K$ using Part I and the Chain Rule?