Fundamental Theorem of Calculus

1Introduction

This lecture proves that differentiating the accumulation of a continuous function recovers the original function and that a definite integral can be evaluated by the endpoint difference of an antiderivative. These are the two parts of the fundamental theorem of calculus.

2Part I: Differentiating an Accumulation Function

Let $f$f be continuous on a closed interval $[a,b]$[a,b], and define

Then $F$F is differentiable on $(a,b)$(a,b) and satisfies $F^{\prime }\left(x\right)=f\left(x\right)$F'(x)=f(x). At the endpoints, the analogous statement holds with one-sided derivatives.

2.1Proof

For $h\ne 0$h\ne0 with $x+h\in [a,b]$x+h\in[a,b], additivity of the definite integral gives

Therefore,

Let $M_h$M_h be the maximum of $\left|f\right(t)-f(x\left)\right|$|f(t)-f(x)| between $x$x and $x+h$x+h. Then

Continuity at $x$x gives $M_h\to 0$M_h\to0, proving $F^{\prime }\left(x\right)=f\left(x\right)$F'(x)=f(x). The estimate also applies when $h<0$h<0.

3Part II: Evaluation by an Antiderivative

Let $f$f be continuous on $[a,b]$[a,b]. Suppose that $G$G is continuous on $[a,b]$[a,b], differentiable on $(a,b)$(a,b), and satisfies $G^{\prime }=f$G'=f on $(a,b)$(a,b). Then

3.1Proof

The function $F\left(x\right)={\int }_a^{x}f\left(t\right)dt$F(x)=\int_a^x f(t)\,dt from Part I satisfies $F^{\prime }=f$F'=f. Hence $(F-G{)}^{\prime }=0$(F-G)'=0. The constant-difference theorem for antiderivatives shows that $F-G$F-G is constant on $[a,b]$[a,b]. Since $F\left(a\right)=0$F(a)=0, we have $F\left(x\right)=G\left(x\right)-G\left(a\right)$F(x)=G(x)-G(a). Substituting $x=b$x=b proves the formula.

4Variable Endpoints

For $p<q$p<q under the usual definition, extend the integral by setting

Let $f$f be continuous on an interval $J$J, and fix $a,b\in J$a,b\in J. Let $u,v$u,v be differentiable functions from an interval $D$D into $J$J. At interior points of $D$D, Part I and the chain rule give

and therefore

The lower-endpoint term has a minus sign because increasing the lower endpoint shortens the interval of integration.

5Total Change

If $g^{\prime }$g' is continuous on $[a,b]$[a,b], Part II applied to $f=g^{\prime }$f=g' gives

The left-hand side accumulates the instantaneous rate of change, while the right-hand side is the total change over the interval.

6Examples

For $$G\left(x\right)={\int }_0^x\cos tdt,$$G(x)=\int_0^x\cos t\,dt, Part I gives $G^{\prime }\left(x\right)=\cos x$G'(x)=\cos x. Also,

Here $\left[G\right(x){]}_{a^b}=G(b)-G(a)$[G(x)]_a^b=G(b)-G(a) denotes an endpoint difference.

7Scope of the Assumptions

This lecture assumes that $f$f is continuous. This hypothesis guarantees both Riemann integrability and that the accumulation function has derivative $f$f at every point. Under weaker hypotheses, one must adjust the set of points where the conclusion holds or the meaning of differentiation.

8Exercises

9Related Material