## Integral: antiderivation and area

If is a function, can we find another differential function such that the derivative of is ? If yes, then how many such functions exist? If we have a complete record of the car speedometer, can we trace the route? compute the end point of the travel?

“Uniqueness” of solution is easy: if there are two solutions such that , then the derivative of their difference is identically zero. By the finite difference lemma, for any point there exists an intermediate point such that , thus is a constant.

Clearly, if is a solution, then , is also a solution. In particular, we can choose this constant so that takes any specified value.

**Example**. If , then , as one can check by the direct derivation.

**Example**. If can be found in the right hand side of any table of derivatives (eventually, with a constant coefficient), then can be found in the same table. E.g., if , then . Denoting and dividing both sides by , we conclude that if , then . The case of occurs on another line in the table: if , then .

What to do if cannot be so easily found? A good idea is to try approximating by some simple functions and see what happens.

**Example**. Let and assume that the function takes the same constant value on the (semi)interval . Then one can easily check that the following function,

is satisfying the following conditions:

*continuous*on the entire segment ,*differentiable*everywhere except the entire points ,- the derivative of coincides with ,
- the graph of is a broken line (קו שבור).

Instead of the cumbersome (מורכות) formula (1), one can use the equivalent verbal description:

area under the graph of between the vertical segments and

Indeed, if is small enough, then both and belong to the same interval and hence , thus the derivative .

Clearly, an arbitrary interval can be subdivided into equal subintervals: the formula (1) will be changed, yet its meaning (2) would obviously remain the same.

This observation suggests that the formula (2) gives the answer also in the general case, provided that the expression “the area under the graph” makes sense. Clearly, this is the case when is piecewise constant (“the step function”, פונקציית מדרגה, as above) and even when is piecewise linear (קו שבור).* Note that we allow only for step functions having only finitely many intervals of continuity, to avoid summing infinitely many areas of rectangles! *

To define the area in the general case, we appeal to the following intuitively clear observation: if are two nested subsets of the plane, then the areas of the sets, if it is well defined, should satisfy the inequality . For polygons this is an elementary theorem.

Let be a bounded function on a finite interval. For any step function which is everywhere less or equal to , , the area under the graph of is called a lower sum for on . In a similar way, an upper sum for is the area under the graph of any step function such that on . Clearly, for any two such functions , the inequality holds, thus **any lower sum is less or equal to any upper sum**.

**Definition**. A function is called * integrable* on the segment , if there exists a single number which separates the lower and the upper sums, so that for any pair of lower/upper sums . This number is called the integral (האינטגרל המסוים) of the function on the interval and denoted by .♦

An equivalent definition of integrability requires that for any positive there exist an upper and a lower sum which differ by no more than :

**Exercise**. Prove that the two definitions are indeed equivalent.

**Exercise**. Prove that the function , is integrable (in the above sense) on any interval and its integral is equal to the (geometrically calculated) area of the corresponding trapeze (טרפז).

**Proposition**. Any monotone (bounded) function is integrable.

**Proof**. Consider the partition of into equal parts by the points and denote the values of at these points. Then the two step functions, and squeeze between them. The lower sum is and . Their difference is less or equal to and it becomes less than any positive number when is large enough.

**Theorem**. Any continuous function (on the closed interval) is integrable.

**Proof**. The idea is to construct two step functions with the common set of jump points such that on each interval these functions squeeze between themselves and differ by less than a given $latex\varepsilon>0$. The points should be close enough to each other so that the continuity of implies the required proximity of the corresponding values.

More precisely, for any point there exists a small interval latex containing such that on this interval the function is squeezed between and . The union of all these small intervals covers which is compact. Hence there exists a finite covering of by these intervals. The endpoints of these intervals subdivide into finitely many intervals in such a way that on each interval the function is squeezed between two constants such that . The corresponding upper and lower sum differ by no more than . This difference can be as small as required if is chosen small enough. ♦

Clearly, continuity is sufficient but not necessary assumption for integrability: the step functions are by definition integrable, though they are discontinuous. The accurate description of all integrable functions goes beyond the scope of these notes, yet there are certainly many non-integrable functions.

**Example**. The Dirichlet function is non-integrable on . Indeed, any upper sum must be at least 1, and any lower sum at most 0, hence the gap between these values cannot be bridged (complete all details of this proof!).

**Problems**.

- Prove that any function on which differs from identical zero at only finitely many points, is integrable and its integral is zero, no matter what are the values at these points.
- Prove that a function that differs from identical zero at countably many points , is integrable and the integral is zero if . Can one drop the limit assumption?
- Prove that the Riemann function equal to at the rational points (assuming mutually prime) and zero at irrational points, is integrable on and its integral is zero.

**Problem**. Is the function integrable on ? (Do not try to compute the integral :-))

## The Newton-Leibniz formula

Quite obviously, if is integrable on , then it is also integrable on any sub-interval , . The formula (2) relating the area (definite integral) with the antiderivative of an integrable function will take the form

This allows to express the area under the graph of a function in terms of its antiderivative (if the latter exists and is known).

However, integrability is a weaker condition than existence of the antiderivative: if is a step function with the partition points , then the corresponding area function is continuous but differentiable only everywhere except these points.

**Theorem**. If is continuous at a point , then the area function is differentiable at and . ♦

## Change of the independent variable in the integrals

If is a function defined on the interval , and is a new variable which is obtained from by a monotone differentiable transformation, , then this transformation maps bijectively (1-1-way) the interval into the interval . The function $f(x)$ becomes after such change a new function of the new variable: if and only if , i.e., the two functions take equal values at the two points “connected” by the transformation .

The “formal” relationship between these functions is easier written “in the opposite direction”, expressing via :

The graphs of functions of and are obtained from each other by a “non-uniform stretch along the horizontal axis” which keeps the vertical direction. In particular, any step function with the partition points will be transformed into the step function with the partition points , , *with the same values*. Moreover, if is squeezed between two step functions, , then its transform is squeezed between the transforms of these functions.

Thus it is sufficient to study how the change of variables affects areas under step functions, which are equal to finite sums and their transforms . The heights are unchanged, and the widths by the finite difference lemma are equal to the initial widths multiplied by the derivative computed at some intermediate points . The result is as if instead of the step function we would integrate the function . Passing to limit, we conclude that

Of course, this is equivalent to the chain rule of differentiation for the primitive functions: if and , then and .

The formula for change of variables of integrals can be easily memorized using the existing notation: in the formula one has to transform not just the integrand by substituting , but also the differential should be transformed using the formula .

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