Fundamental Theorem of Calculus

Throughout this page, assume $f$ is continuous on the closed interval under consideration. In Part I, the ordinary derivative is asserted at interior points; endpoints have one-sided derivatives. In Part II, $F$ is continuous on $[a,b]$ and satisfies $F'=f$ on $(a,b)$. These sufficient hypotheses exclude integrating across a pole by merely subtracting endpoint values. For variable limits, $u,v$ must be differentiable and their values must stay in an interval where $f$ is continuous. The integrand here depends on $t$, not separately on $x$.

Part I

• Continuity: If $f(x)$ is continuous on $[a, b]$, then the function defined by the integral of $f$ from $a$ to $x$ is also continuous on $[a, b]$.

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

• Differentiation: The derivative of $g(x)$ with respect to $x$ is the original function $f(x)$.

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

3Blue1Brown’s velocity-and-area lesson follows a moving upper integration limit. Watch how the added thin strip relates the change in accumulated displacement to the current velocity, giving a geometric reading of Part I.

Variants of Part I

The variants involve differentiating an integral with variable limits of integration:

1. Upper Limit as a Function of $x$:

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

2. Lower Limit as a Function of $x$:

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

3. Both Limits as Functions of $x$:

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

Part II

• Anti-Derivative: If $F(x)$ is an anti-derivative of $f(x)$, meaning $F'(x) = f(x)$, then the integral of $f$ from $a$ to $b$ is the difference between the values of $F$ at these points.

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

Why differentiation recovers the integrand

Additivity gives

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

This is the average value of $f$ over a shrinking interval. Continuity at $x$ forces it to tend to $f(x)$. Thus $g'=f$. Any other antiderivative $F$ differs from $g$ by a constant, so $F(b)-F(a)=g(b)-g(a)=\int_a^bf$; this proves Part II under the stated assumptions.

For example, $\int_0^2 3x^2\,dx=[x^3]_0^2=8$. With variable limits,

$$\frac{d}{dx}\int_x^{x^2}e^{-t^2}\,dt =2x e^{-x^4}-e^{-x^2}.$$

No elementary antiderivative is needed for the second calculation. Continuity matters: if $f(t)=0$ for $t<0$ and $f(t)=1$ for $t\ge0$, then $\int_{-1}^x f(t)\,dt=\max(0,x)$ near zero. It is continuous but not differentiable at the jump.

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