The fundamental theorem of calculus is the bridge between the two halves of calculus. Differentiation measures how fast something changes. Integration adds up small pieces. The theorem says these are inverse processes, and it turns the hard job of computing a limit of Riemann sums into the easy job of finding an antiderivative. OpenStax Calculus Volume 1 calls Part 2 "perhaps the most important theorem in calculus".

Part 1: accumulation and its rate

If $f$ is continuous on $[a,b]$ and we define $$F(x) = \int_a^x f(t)\,dt,$$ then $F'(x) = f(x)$ on $(a,b)$. Read $F(x)$ as "the area accumulated from $a$ up to $x$". Push $x$ a little to the right by $h$ and you add a thin strip of width $h$ and height roughly $f(x)$. So the change in $F$ is about $f(x)\,h$, and the rate of change is $f(x)$. OpenStax makes this exact with the mean value theorem for integrals: the strip's area is exactly $f(c)\,h$ for some $c$ between $x$ and $x+h$, and $f(c) \to f(x)$ as $h \to 0$.

A consequence OpenStax points out: every continuous function has an antiderivative, because $F$ is one.

Part 2: the evaluation theorem

If $f$ is continuous on $[a,b]$ and $F$ is any antiderivative of $f$, then $$\int_a^b f(x)\,dx = F(b) - F(a).$$ OpenStax proves it by splitting $[a,b]$ into pieces, applying the mean value theorem to $F$ on each piece to get $F(x_i) - F(x_{i-1}) = f(c_i)\,\Delta x$, and adding. The sum telescopes to $F(b) - F(a)$ on one side and becomes a Riemann sum on the other. Take the limit and you have the theorem.

Worked example 1: evaluating a definite integral

OpenStax evaluates $\int_{-2}^{2} (t^2 - 4)\,dt$. An antiderivative is $\frac{t^3}{3} - 4t$. So $$\left[\frac{t^3}{3} - 4t\right]_{-2}^{2} = \left(\frac{8}{3} - 8\right) - \left(-\frac{8}{3} + 8\right) = -\frac{32}{3}.$$ The answer is negative because the parabola sits below the $x$ axis on that interval. A definite integral gives net signed area, not total area.

Worked example 2: differentiating an integral

For $g(x) = \int_1^x \frac{1}{t^3 + 1}\,dt$, Part 1 gives $g'(x) = \frac{1}{x^3+1}$ with no integration at all. When the upper limit is a function, combine Part 1 with the chain rule. OpenStax takes $F(x) = \int_1^{\sqrt{x}} \sin t\,dt$ and gets $$F'(x) = \sin(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}.$$

Worked example 3: who skated farther?

OpenStax's roller skating race: James skates at $5 + 2t$ feet per second and Kathy at $10 + \cos\left(\frac{\pi}{2}t\right)$ feet per second for 5 seconds. Distance is the integral of velocity. James covers $\left[5t + t^2\right]_0^5 = 50$ feet. Kathy covers $\left[10t + \frac{2}{\pi}\sin\left(\frac{\pi}{2}t\right)\right]_0^5 = 50 + \frac{2}{\pi} \approx 50.6$ feet. Kathy wins, narrowly.

Checkpoints

  1. Evaluate $\int_1^4 \frac{1}{2\sqrt{x}}\,dx$.
  2. Find $\frac{d}{dx}\int_0^{x^2} \cos t\,dt$.
  3. Find the average value of $f(x) = x^2$ on $[0,6]$ and a point $c$ where $f(c)$ equals it.
  4. Explain why $\int_0^{2\pi} \cos\theta\,d\theta = 0$ even though the graph encloses area.

Common mistakes

  • Subtracting in the wrong order. It is $F(\text{upper}) - F(\text{lower})$.
  • Forgetting the chain rule factor when the limit of integration is a function of $x$.
  • Reading a negative integral as an error. It is net signed area.
  • Integrating across a discontinuity. The theorem needs $f$ continuous on the whole interval. $\int_{-1}^{1} \frac{1}{x^2}\,dx$ is not $-2$.

Related reading

Sources

Last reviewed: 12 November 2026.

Common questions

What does the fundamental theorem of calculus say?

It has two parts. Part 1 says that if you accumulate a continuous function f from a fixed point a up to x, the rate of change of that accumulation is f(x). Part 2 says a definite integral of f from a to b equals F(b) minus F(a) for any antiderivative F.

Why don't you need the +C when evaluating a definite integral?

Any antiderivative works. The constant appears in both F(b) and F(a) and cancels when you subtract.

How does the chain rule combine with Part 1?

If the upper limit is a function u(x), the derivative of the integral from a to u(x) of f(t) dt is f(u(x)) times u'(x).

Can a definite integral be negative?

Yes. It measures net signed area, so area below the x axis counts as negative.