Antiderivatives and Indefinite Integrals

1Introduction

This lecture defines an indefinite integral as an inverse problem for differentiation rather than as an area. It also explains why a constant of integration is necessary: functions with the same derivative on an interval differ by a constant.

On an interval $I$I, a function $F$F satisfying $F^{\prime }\left(x\right)=f\left(x\right)$F'(x)=f(x) is called an antiderivative of $f$f on $I$I.

2Constant of Integration

The mean value theorem states that if $H$H is continuous on $[u,v]$[u,v] and differentiable on $(u,v)$(u,v), then there is a point $c\in (u,v)$c\in(u,v) such that

Here is a proof. If a differentiable function has a local maximum or minimum at an interior point $c$c, comparison of the signs of the left and right difference quotients gives $H^{\prime }\left(c\right)=0$H'(c)=0. This is Fermat's stationary-point theorem.

Now suppose $H\left(u\right)=H\left(v\right)$H(u)=H(v). By the extreme value theorem introduced with limits and continuity, $H$H attains a maximum and a minimum on $[u,v]$[u,v]. If $H$H is not constant, at least one of these extrema occurs at an interior point, where Fermat's theorem gives $H^{\prime }\left(c\right)=0$H'(c)=0. If $H$H is constant, the same conclusion holds at every interior point. This proves Rolle's theorem.

For a general $H$H, define

Then $K\left(u\right)=K\left(v\right)$K(u)=K(v). Rolle's theorem gives a point $c\in (u,v)$c\in(u,v) with $K^{\prime }\left(c\right)=0$K'(c)=0, and rearranging this equality proves the mean value theorem.

Assume that $f$f has an antiderivative $F$F on the interval $I$I. Then $(F+C{)}^{\prime }=f$(F+C)'=f for every constant $C$C. Conversely, if $G^{\prime }=f$G'=f on $I$I, then $(G-F{)}^{\prime }=0$(G-F)'=0. Since differentiability implies continuity, the mean value theorem applies to $G-F$G-F on every closed subinterval $[u,v]\subset I$[u,v]\subset I. It gives a point $c\in (u,v)$c\in(u,v) such that

Hence $G-F$G-F is constant on $I$I, so $G=F+C$G=F+C for some constant $C$C. The indefinite integral is therefore written as the family

The constant $C$C is called the constant of integration.

If $f$f has no antiderivative on $I$I, its set of antiderivatives is empty and the representation $F+C$F+C is unavailable.

The interval assumption is essential. On a disconnected domain, a different constant can be added on each connected component, so one constant need not describe all antiderivatives on the entire domain.

3Difference from a Definite Integral

A definite integral is a single number obtained after specifying the endpoints of an interval. An indefinite integral is the set of all antiderivatives. The fundamental theorem of calculus connects the two concepts, but their definitions and outputs are different.

4Example

An antiderivative of $f\left(x\right)=2x$f(x)=2x is $F\left(x\right)=x^2$F(x)=x^2. Therefore,

The result is a family of functions that differ by constants, not a single function.

5Exercises

6Related Material