Fundamental Theorem of Calculus Part I

From Integrals to a Function

The Riemann integral assigns a number to a function on a fixed interval. We can also let one endpoint vary and ask how the resulting integral changes. This creates a new function from an existing one, and the Fundamental Theorem of Calculus, Part I, describes the connection between that new function’s derivative and the original integrand.

Let $f:[a,b]\to\mathbb{R}$ be continuous. For each $x\in[a,b]$, define

F(x)=\int_a^x f(t)\,dt.

The variable $t$ is a dummy variable of integration; the resulting value depends on the endpoint $x$. The function $F$ records the accumulated signed integral from the fixed starting point $a$ to $x$. When $f$ is positive, the accumulation increases; when $f$ is negative, the accumulation decreases. If $f$ changes sign, positive and negative contributions can cancel.

Definition (Accumulation Function): If $f$ is Riemann integrable on $[a,b]$, its accumulation function with base point $a$ is the function $F:[a,b]\to\mathbb{R}$ defined by $F(x)=\int_a^x f(t)\,dt$.

Continuity of $f$ guarantees Riemann integrability on the closed interval, by the Integrability of Continuous Functions theorem. In particular, the definition makes sense for every $x\in[a,b]$. The central question is how $F$ changes between two points. Additivity of the integral gives the answer before any differentiation is considered.

Increments and Their Size

For $a\leq x\leq y\leq b$, additivity across adjacent intervals gives

F(y)-F(x) =\int_a^y f(t)\,dt-\int_a^x f(t)\,dt =\int_x^y f(t)\,dt.

This identity says that the change in accumulated value is exactly the integral over the interval traversed. It also suggests an estimate: if the integrand is small in magnitude throughout that interval, then its accumulated contribution cannot be large.

Theorem (Increment Bound for an Accumulation Function): Suppose $f$ is Riemann integrable on $[a,b]$, and $|f(t)|\leq M$ for every $t\in[a,b]$, where $M\geq0$. For $F(x)=\int_a^x f(t)\,dt$ and any $x,y\in[a,b]$, $$ |F(y)-F(x)|\leq M|y-x|. $$

Proof. First suppose $x\leq y$. The increment identity and the Absolute-Value Inequality for the Riemann Integral imply

|F(y)-F(x)| =\left|\int_x^y f(t)\,dt\right| \leq\int_x^y |f(t)|\,dt.

Since $0\leq |f(t)|\leq M$, order preservation of the integral gives

\int_x^y |f(t)|\,dt \leq\int_x^y M\,dt =M(y-x).

Thus the claimed bound holds when $x\leq y$. If $y<x$, interchange $x$ and $y$: the absolute value is unchanged, and the already-proved case gives $|F(x)-F(y)|\leq M(x-y)=M|y-x|$. This proves the bound for every pair $x,y\in[a,b]$. $\square$

A function satisfying such a bound is called Lipschitz with constant $M$. In particular, every bounded integrand has a Lipschitz accumulation function. A continuous function on $[a,b]$ is bounded, so this applies to the integrands considered here. The estimate also directly proves continuity: given $\varepsilon>0$, if $M>0$, choose $\delta=\varepsilon/M$. Then $|x-y|<\delta$ implies $|F(x)-F(y)|<\varepsilon$. If $M=0$, the estimate says $F$ is constant, so it is continuous as well.

Worked Example: A Quadratic Integrand

Let $f(t)=3t^2$ on $[0,2]$, and define $F(x)=\int_0^x 3t^2\,dt$. Direct evaluation gives

F(x)=x^3,\qquad 0\leq x\leq2.

The integrand satisfies $0\leq f(t)\leq12$ on this interval, so the Increment Bound gives $|F(y)-F(x)|\leq12|y-x|$. For example, with $x=1$ and $y=3/2$, the actual increment is

F(3/2)-F(1)=\frac{27}{8}-1=\frac{19}{8},

while the estimate gives $|F(3/2)-F(1)|\leq12(1/2)=6$. The bound is not intended to be exact; it provides uniform control using only a bound on the integrand. Here direct differentiation also gives $F'(x)=3x^2=f(x)$, illustrating the derivative relationship that the Fundamental Theorem will assert in general.

Sign of the Integrand and Direction of Accumulation

The same increment identity gives a qualitative conclusion. If $f$ is nonnegative, every interval integral is nonnegative, so the accumulated value cannot go down. This fact does not require continuity; integrability and pointwise nonnegativity suffice.

Theorem (Monotonicity of the Accumulation Function): Suppose $f$ is Riemann integrable on $[a,b]$, with $f(t)\geq0$ for every $t\in[a,b]$. Then $F(x)=\int_a^x f(t)\,dt$ is nondecreasing on $[a,b]$.

Proof. Take any $x,y\in[a,b]$ with $x\leq y$. By the increment identity,

F(y)-F(x)=\int_x^y f(t)\,dt.

Order preservation and $f(t)\geq0$ imply $\int_x^y f(t)\,dt\geq0$. Hence $F(y)\geq F(x)$ whenever $x\leq y$, which is precisely that $F$ is nondecreasing. $\square$

If instead $f(t)\leq0$ everywhere, apply the result to $-f$, or use order preservation directly, to see that $F$ is nonincreasing. When $f$ takes both positive and negative values, neither direction of monotonicity is guaranteed; the interval integrals determine the changes.

Worked Example: A Nonnegative Integrand with a Flat Portion

On $[0,2]$, let $f(t)=0$ for $0\leq t\leq1$, and $f(t)=t-1$ for $1<t\leq2$. This function is continuous, nonnegative, and Riemann integrable. Its accumulation function is

F(x)= \begin{cases} 0, & 0\leq x\leq1,\\[4pt] \displaystyle\int_1^x(t-1)\,dt=\frac{(x-1)^2}{2}, & 1<x\leq2. \end{cases}

For $x\leq1$, the accumulated integral is zero. For $x>1$, it is nonnegative and increases as $x$ increases. For instance, $F(3/2)=1/8$ and $F(2)=1/2$, so $F(2)-F(3/2)=3/8\geq0$. The flat portion is consistent with the integrand being zero there: no signed area is accumulated over that part of the interval.

The Fundamental Theorem of Calculus, Part I

The increment bound gives continuity of $F$, but the fundamental theorem makes a stronger claim: at an interior point, the instantaneous rate of change of accumulated integral equals the value of the integrand at that point. The continuity hypothesis is essential to the standard statement below.

Theorem (Fundamental Theorem of Calculus, Part I): Let $f:[a,b]\to\mathbb{R}$ be continuous, and define $$ F(x)=\int_a^x f(t)\,dt. $$ Then $F$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $$ F'(x)=f(x)\qquad\text{for every }x\in(a,b). $$ At $a$ and $b$, the corresponding one-sided derivatives exist and equal $f(a)$ and $f(b)$, respectively.

The continuity assertion follows already from the Increment Bound: a continuous $f$ on a closed bounded interval is bounded. The derivative claim explains why integration can be used to construct antiderivatives. To see the shape of the argument, fix an interior point $x$ and take a nonzero increment $h$ small enough that $x+h\in[a,b]$. The increment identity rewrites the difference quotient as

\frac{F(x+h)-F(x)}{h} =\frac{1}{h}\int_x^{x+h}f(t)\,dt.

Thus the difference quotient is the average value of $f$ over the short interval between $x$ and $x+h$, with the oriented integral handling negative $h$. When $f$ is continuous at $x$, values of $f(t)$ on this shrinking interval are close to $f(x)$, so their average is close to $f(x)$. Establishing this carefully, including both signs of $h$ and the endpoint cases, is the central proof task.

Worked Example: A Linear Integrand and a Nonzero Base Point

Let $f(t)=2t+1$ on $[0,3]$, but choose base point $c=1$. Define $G(x)=\int_1^x(2t+1)\,dt$ for $x\in[0,3]$, using the usual oriented-integral convention when $x<1$. Evaluation gives

G(x)=\bigl(x^2+x\bigr)-\bigl(1^2+1\bigr)=x^2+x-2.

The constant term changes because the base point is $1$, not $0$, but differentiation removes that constant:

$$ G'(x)=2x+1=f(x). $$

For example, $G(0)=-2$, which agrees with the oriented integral $\int_1^0(2t+1)\,dt=-\int_0^1(2t+1)\,dt=-2$. The derivative relationship is therefore independent of the chosen fixed base point; changing the base point only adds a constant to the accumulation function.

Worked Example: Accumulating a Trigonometric Integrand

Let $f(t)=\cos t$ on $[0,\pi]$, and set $F(x)=\int_0^x\cos t\,dt$. Since $\cos t$ is continuous, the theorem applies. Evaluating the integral gives

F(x)=\sin x-\sin 0=\sin x.

Consequently, $F'(x)=\cos x=f(x)$ for $0<x<\pi$. The accumulation function increases on $[0,\pi/2]$ and decreases on $[\pi/2,\pi]$, matching the sign of the integrand: $\cos t\geq0$ up to $\pi/2$, and $\cos t\leq0$ from $\pi/2$ onward. In particular, $F(\pi)=0$, because the positive and negative contributions cancel:

\int_0^\pi\cos t\,dt=\sin\pi-\sin0=0.

Why the Theorem Matters—and a Hypothesis to Notice

Part I provides a systematic way to turn a continuous function into an antiderivative: integrate it from a fixed point, then differentiate the resulting accumulation function. It also clarifies why the integral over a short interval behaves like the integrand’s value times the interval length. The difference quotient is an average, and continuity makes that average approach the value at the point.

The continuity assumption should not be mistaken for a claim that every integrable function has an accumulation function differentiable everywhere. Integrability ensures the integral exists, and the increment estimate can ensure the accumulation function is continuous when the integrand is bounded. Differentiability at a point requires more: nearby averages must converge to the integrand’s value there. Continuity guarantees that behavior, while integrability alone does not.

Another common mistake is to omit the orientation when the endpoint moves to the left of the base point. The standard convention is $\int_c^x f=-\int_x^c f$ when $x<c$. With this convention, the same derivative formula holds on either side of the base point, and the difference quotient expression remains valid for positive and negative increments.

The next proof develops the derivative claim in full detail. The key steps are to express each difference quotient as an average over a short interval, control the difference between that average and $f(x)$ using continuity, and treat the two endpoint derivatives with one-sided quotients.

Check Your Understanding

Use the definitions, estimates, and theorem statement in this tutorial to answer the following questions.

Why does additivity of the integral imply $F(y)-F(x)=\int_x^y f(t)\,dt$ when $x\leq y$?
If $|f(t)|\leq 4$ on an interval, what increment bound follows for its accumulation function?
Why is the accumulation function nondecreasing when its integrand is nonnegative?
How can the difference quotient of an accumulation function be interpreted as an average value?
Which part of the Fundamental Theorem of Calculus, Part I, is supplied by the increment bound, and which part requires the continuity-based argument?