The Fundamental Theorem of Calculus

Differentiation began with local rates. Integration began with sums of accumulation. The fundamental theorem explains why these operations are inverse in the appropriate sense. It is the reason an area problem can be solved by finding a function with a prescribed derivative.

Accumulation has the original rate

Let $f$ be continuous on $[a,b]$ and define $A(x)=\int_a^xf(t)\,dt$. For an interior $x$ and small nonzero $h$,

$$\frac{A(x+h)-A(x)}h=\frac1h\int_x^{x+h}f(t)\,dt.$$

Subtract $f(x)$ and use an integral bound. For either sign of $h$,

$$\left|\frac{A(x+h)-A(x)}h-f(x)\right| \leq\sup_{t\text{ between }x\text{ and }x+h}|f(t)-f(x)|.$$

Continuity at $x$ makes the right side tend to zero. Therefore $A'(x)=f(x)$. At endpoints use the matching one-sided version. The proof compares a short-interval average with the local value. It does not assume $f$ is constant on the short interval.

Continuity is sufficient, not necessary in every possible case. At a jump, however, the conclusion can fail. If $f(t)=0$ for $t<0$ and $1$ for $t\geq0$, the accumulation from $-1$ is zero for $x\leq0$ and $x$ for $x>0$. It has no derivative at zero. Substituting $f(0)=1$ into a theorem whose continuity hypothesis fails would be invalid.

Evaluation by an antiderivative

If $F'=f$ on the interval and $f$ is continuous, $F$ and $A$ have the same derivative. The constant-difference theorem gives $F(x)-A(x)=F(a)$, since $A(a)=0$. At $b$,

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

This is the evaluation form of the fundamental theorem. Different books interchange the labels “first” and “second” for its two parts; the statements, rather than the numbering, are what matter.

For example,

$$\int_0^2(3x^2-2x+1)\,dx =[x^3-x^2+x]_0^2=6.$$

The constant in an indefinite integral cancels in the endpoint difference. There is no need to write “$+C$” after the numerical value of a definite integral.

Variable bounds use the chain rule

For continuous $f$ and differentiable $g$ with values in its integration domain,

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

With two variable bounds, split at a fixed base point:

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

Thus $d[\int_1^{x^2}\cos t\,dt]/dx=2x\cos(x^2)$, and $d[\int_x^{2x}t^2\,dt]/dx=2(2x)^2-x^2=7x^2$. The integration variable $t$ is a dummy variable; its name prevents confusing it with the moving bound.

For $A(x)=\int_0^xe^{-t^2}\,dt$, $A'(x)=e^{-x^2}$ even though the integrand has no elementary antiderivative. An accumulation function can be mathematically well-defined and differentiable without having a formula in the elementary functions.

Net change and forbidden crossings

If a quantity $Q$ has continuous rate $Q'$, then $Q(b)-Q(a)=\int_a^bQ'(t)\,dt$. For $v(t)=2t-2$ on $[0,3]$, the displacement is $[t^2-2t]_0^3=3$. Distance is $-\int_0^1v+\int_1^3v=5$.

Do not apply the evaluation formula across a singularity without checking integrability. Writing $\int_{-1}^11/x^2\,dx=[-1/x]_{-1}^1=-2$ is invalid. The integrand is unbounded at zero and does not satisfy the stated theorem; the corresponding improper integral diverges positively. A negative answer for a nonnegative integrand is an immediate sign that a hypothesis has been violated.

Exercises

1. Evaluate $\int_1^3(2x+1)\,dx$.
2. Differentiate $\int_0^{\sin x}(1+t^2)\,dt$.
3. Differentiate $\int_{x^2}^{x^3}e^t\,dt$.
4. For $A(x)=\int_0^x\cos t\,dt$, identify intervals where $A$ is increasing.
5. Explain why $\int_0^xe^{-t^2}\,dt$ does not need an elementary closed form to have a derivative.
6. Identify the precise failure in evaluating $\int_{-1}^1(1/x)\,dx$ by the endpoint difference of $\log|x|$.

Worked solutions

1. $[x^2+x]_1^3=12-2=10$.
2. $(1+\sin^2x)\cos x$.
3. $3x^2e^{x^3}-2xe^{x^2}$.
4. Where $\cos x>0$, namely $(-\pi/2+2k\pi,\pi/2+2k\pi)$ for integers $k$.
5. Continuity of the integrand establishes the accumulation function and the derivative theorem applies directly.
6. The integrand is undefined and unbounded at zero. The two separate one-sided improper integrals do not both converge; endpoint cancellation is not an ordinary integral. A symmetric principal value is a distinct concept.