5.3 The Fundamental Theorem of Calculus

Big idea

So far integration and differentiation have looked like separate subjects. Differentiation is local: it asks how a function behaves near one point. Integration is global: it adds contributions across a whole interval. The result in this section says the two operations undo each other, and that single fact turns a hard limit of sums into an evaluation you can do in two lines.

The bridge between them is the accumulation function. Fix a starting point $a$ and a continuous function $f$, and define a new function by integrating $f$ from $a$ up to a movable right endpoint $x$. As $x$ slides right, the accumulated amount changes, and the rate at which it changes is exactly the height of $f$ at the sliding edge. That is the first half of the theorem, and it is the reason every continuous function has an antiderivative: the accumulation function is one.

The second half runs the statement backwards and is the one you will use on nearly every problem. If you happen to know any antiderivative of $f$, the definite integral is just the change in that antiderivative across the interval. No partition, no sample points, no limit. Subtract two numbers.

It is worth pausing on how strange this is. The definition of the integral mentions rectangles and says nothing about derivatives. The evaluation rule mentions derivatives and says nothing about rectangles. The theorem is the claim that these describe the same number, and once you believe it, the entire toolkit you built for differentiating becomes a toolkit for integrating, read in reverse.

Decoder

If $f$ is continuous, then differentiating the integral of $f$ from a fixed point up to $x$ returns $f(x)$, and integrating $f$ across $[a,b]$ returns $F(b) - F(a)$ for any antiderivative $F$ of $f$.

The first half is a statement about a function, the second is a statement about a number, and the word “antiderivative” is what joins them. In the first half you start with $f$, build something new by accumulating, and differentiate back to $f$. In the second you start with something whose derivative is $f$ and read off the total accumulation as a difference of two of its values.

Notice what the second half does not say. It says “any antiderivative”, and that is not a loose end. Two antiderivatives of the same function differ by a constant, and a constant cancels in $F(b) - F(a)$, so the choice genuinely does not matter. This is why you may drop the $+C$ when evaluating a definite integral, and only then.

The classic mistake sits in the first half. When the upper limit is a function of $x$ rather than $x$ itself, the answer is not just $f$ evaluated at that expression. The accumulation function is being composed with something, so the chain rule applies and an extra factor appears. The second mistake is subtracting in the wrong order; the value at the top limit comes first.

Definitions and results

Accumulation function. For $f$ continuous on an interval containing $a$, define

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

The variable of integration must be a letter other than $x$, since $x$ is now the endpoint. Note $F(a) = 0$, and $F$ increases where $f$ is positive.

The derivative of an accumulation function. With $f$ continuous, $F$ is differentiable and

$$ \frac{d}{dx}\int_a^x f(t)\,dt = f(x) $$

Consequently every continuous function has an antiderivative, whether or not you can write it with familiar symbols.

Variable upper limit. If the top of the integral is $g(x)$, compose and apply the chain rule:

$$ \frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f\big(g(x)\big)\, g'(x) $$

Variable lower limit. Reverse the limits first, which flips the sign:

$$ \frac{d}{dx}\int_{g(x)}^{b} f(t)\,dt = -f\big(g(x)\big)\, g'(x) $$

Both limits moving. Split at any convenient constant $c$ and handle the pieces separately:

$$ \frac{d}{dx}\int_{u(x)}^{v(x)} f(t)\,dt = f\big(v(x)\big)v'(x) - f\big(u(x)\big)u'(x) $$

The evaluation rule. If $f$ is continuous on $[a,b]$ and $F' = f$ there, then

$$ \int_a^b f(x)\,dx = F(b) - F(a) $$

The standard shorthand for the right side is $\big[F(x)\big]_a^b$, sometimes written with a vertical bar.

Why the constant may be dropped. If $G(x) = F(x) + C$ is another antiderivative, then $G(b) - G(a) = F(b) - F(a)$. The constant cancels, so any antiderivative serves.

Net change reading. Written with $F$ in front, the rule says $\int_a^b F'(x)\,dx = F(b) - F(a)$: integrating a rate of change across an interval gives the net change in the quantity itself.

Worked examples

Reading a derivative off an integral

Let $G(x) = \displaystyle\int_2^{x}\sqrt{t^3+1}\,dt$ for $x \ge 0$. Find $G'(x)$, $G(2)$, and say where $G$ increases.

The integrand is continuous for $t \ge 0$, so the first half applies directly and $G'(x) = \sqrt{x^3+1}$. The integral over a degenerate interval is zero, so $G(2) = 0$. Since $\sqrt{x^3+1} > 0$ for every $x \ge 0$, the derivative is always positive and $G$ increases on its whole domain.

There is no elementary formula for $G$, and none is needed. The theorem hands you the derivative of a function you cannot write down.

An upper limit that is itself a function

Find $H'(x)$ for $H(x) = \displaystyle\int_0^{x^2}\cos t\,dt$.

The upper limit is $g(x) = x^2$, not $x$, so the chain rule factor appears:

$$ H'(x) = \cos\big(x^2\big)\cdot 2x = 2x\cos\big(x^2\big) $$

This one can be checked in full, because $\cos t$ has an obvious antiderivative. Evaluating gives $H(x) = \sin(x^2) - \sin 0 = \sin(x^2)$, and differentiating that with the chain rule gives $2x\cos(x^2)$. The two routes agree.

Both limits in motion

Find $K'(x)$ for $K(x) = \displaystyle\int_{2x}^{x^3}\frac{1}{1+t^2}\,dt$.

Split at $0$ to separate the moving ends:

$$ K(x) = \int_{0}^{x^3}\frac{dt}{1+t^2} - \int_{0}^{2x}\frac{dt}{1+t^2} $$

Each piece now has one variable limit, so

$$ K'(x) = \frac{3x^2}{1+x^6} - \frac{2}{1+4x^2} $$

Watch the substitutions: $t$ is replaced by $x^3$ in the first denominator, giving $1 + (x^3)^2 = 1 + x^6$, and by $2x$ in the second, giving $1 + 4x^2$.

Evaluating with an antiderivative

Evaluate $\displaystyle\int_1^4 (3x^2 - 2x + 1)\,dx$.

An antiderivative is $F(x) = x^3 - x^2 + x$, since differentiating returns the integrand term by term. Then

$$ \int_1^4 (3x^2-2x+1)\,dx = \big[x^3-x^2+x\big]_1^4 = (64-16+4) - (1-1+1) = 52 - 1 = 51 $$

For a second one with a check, take $\displaystyle\int_1^9 \frac{dx}{\sqrt{x}}$. Writing the integrand as $x^{-1/2}$, an antiderivative is $2x^{1/2}$, so the value is $2(3) - 2(1) = 4$.

Sanity check that number. On $[1,9]$ the integrand runs between $1/3$ and $1$, and the interval has width $8$, so the integral must lie between $8/3 \approx 2.67$ and $8$. The answer $4$ sits comfortably inside.

Practice

Start with the first half: differentiating accumulation functions, including the cases where a limit of integration is a function of $x$ and the chain rule contributes a factor.

Then the evaluation rule. These ask you to produce an antiderivative and subtract its values at the two endpoints.

Quiz

Five items on accumulation functions, the chain rule form, and evaluating definite integrals with antiderivatives.