Sergei Yakovenko's blog: on Math and Teaching

Monday, February 8, 2010

Lecture 13 (Feb 2, 2010)

Integral: antiderivation and area

If f:[a,b]\to\mathbb R is a function, can we find another differential function F:[a,b]\to\mathbb R such that the derivative of F is f? 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 F_1(x),F_2(x) such that F'_{1,2}=f, then the derivative of their difference F=F_1-F_2 is identically zero. By the finite difference lemma, for any point x\in[a,b] there exists an intermediate point z\in[a,x] such that F(x)-F(a)=F'(z)=0, thus F is a constant.

Clearly, if F(x) is a solution, then F(x)+c,c\in\mathbb R, is also a solution. In particular, we can choose this constant so that F(a) takes any specified value.

Example. If f(x)\equiv\lambda\in\mathbb R, then F(x)=F(a)+\lambda (b-a), as one can check by the direct derivation.

Example. If f can be found in the right  hand side of any table of derivatives (eventually, with a constant coefficient), then F can be found in the same table. E.g., if f(x)=\mu x^{\mu-1},~\mu\ne 0, then F(x)=x^\mu,~\mu\ne 0. Denoting \mu-1=\nu and dividing both sides by \mu=\nu+1, we conclude that if f(x)=x^\nu,~\nu\ne -1, then F(x)=\frac1{\nu+1}\,x^{\nu+1}. The case of \nu=-1 occurs on another line in the table: if f(x)=\frac1x,~x>0, then F(x)=\ln x.

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

Example. Let [a,b]=[0,N] and assume that the function f takes the same constant value \lambda_i on the (semi)interval [i,i+1). Then one can easily check that the following function,

F(x)=\lambda_0+\lambda_1+\cdots+\lambda_{i-1}+\lambda_i(x-i),\qquad x\in [i,i+1),\quad i=0,\dots,N-1,\qquad(1)

is satisfying the following conditions:

  1. continuous on the entire segment [0,N],
  2. differentiable everywhere except the entire points 1,2,\dots,N-1,
  3. the derivative of F coincides with f,
  4. the graph of F is a broken line (קו שבור).

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

F(x)=area under the graph of f between the vertical segments a and x.\qquad (2)

Indeed, if h is small enough, then both x and x+h belong to the same interval (i,i+1) and hence F(x+h)-F(x)=\lambda_i h, thus the derivative F'(x)=\lambda_i=f(x).

Clearly, an arbitrary interval [a,b] can be subdivided into N 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 f(x) is piecewise constant  (“the step function”, פונקציית מדרגה, as above) and even when f 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 X\subseteq Y are two nested subsets of the plane, then the areas of the sets, if it is well defined, should satisfy the inequality s(X)\leq s(Y). For polygons this is an elementary theorem.

Let f:[a,b]\to\mathbb R be a bounded function on a finite interval. For any step function g_i(x):[a,b]\to\mathbb R which is everywhere less or equal to f, g_-(x)\le f(x), the area s(g_-) under the graph of g_- is called a lower sum for f on [a,b]. In a similar way, an upper sum for f is the area under the graph of any step function g_+(x) such that f(x)\le g_+(x) on [a,b]. Clearly, for any two such functions g_-,g_+, the inequality s(g_-)\le s(g_+) holds, thus any lower sum is less or equal to any upper sum.

Definition. A function  f:[a,b]\to\mathbb R is called integrable on the segment [a,b], if there exists a single number s=s(f) which separates the lower and the upper sums, so that s_-\le s \le s_+ for any pair of lower/upper sums s_\pm=s(g_\pm). This number is called the integral (האינטגרל המסוים) of the function f(x) on the interval [a,b] and denoted by \displaystyle \int_a^b f(x)\,dx.♦

An equivalent definition of integrability requires that for any positive \varepsilon>0 there exist an upper and a lower sum which differ by no more than \varepsilon:

\forall\varepsilon>0~~\exists g_-,g_+\text{ step functions }:g_-(x)\le f(x)\le g_+(x),~~s(g_+)-s(g_-)<\varepsilon.

Exercise. Prove that the two definitions are indeed equivalent.

Exercise. Prove that the function f(x)=px+q,~p,q\in\mathbb R, 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 [a,b] into N equal parts by the points a=x_0<x_1<\cdots<x_N=b and denote \lambda_i=f(x_i) the values of f at these points. Then the two step functions, g_-(x)=\lambda_{i},~~x\in [x_i,x_{i+1}) and g_+(x)=\lambda_{i+1},~~x\in [x_i,x_{i+1}) squeeze f between them. The lower sum is s_-=\frac1N(\lambda_0+\lambda_1+\cdots+\lambda_{N_1}) and s_+=\frac1N(\lambda_1+\cdots+\lambda_N). Their difference is less or equal to (\lambda_N-\lambda_0)/N and it  becomes less than any positive number \varepsilon when N 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 a=x_0<x_1<\cdots<x_N=b such that on each interval [x_i,x_{i+1}) these functions squeeze f between themselves and differ by less than a given $latex\varepsilon>0$. The points x_i should be close enough to each other so that the continuity of f implies the required proximity of the corresponding values.

More precisely, for any point c\in[a,b] there exists a small interval latex U_c\subseteq[a,b] containing c such that on this interval the function is squeezed between f(c)-\frac12\varepsilon and f(c)+\frac12\varepsilon.  The union of all these small intervals covers [a,b] which is compact. Hence there exists a finite covering of [a,b] by these intervals. The endpoints of these intervals subdivide [a,b] into finitely many intervals U_i in such a way that on each interval the function f is squeezed between two constants \lambda_i\le f(x) \le \lambda_+ such that \lambda_+-\lambda_-<\varepsilon. The corresponding upper and lower sum differ by no more than \sum (\lambda_+-\lambda_-)(x_{i+1}-x_i)\le \varepsilon \sum (x_{i+1}-x_i)\le \varepsilon (b-a). This difference can be as small as required if \varepsilon 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 [0,1]. 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.

  1. Prove that any function on [0,1] 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.
  2. Prove that a function that differs from identical zero at countably many points x_1,x_2,\dots,x_n,\dots, is integrable and the integral is zero if \lim_{k\to\infty}f(x_k)=0. Can one drop the limit assumption?
  3. Prove that the Riemann function equal to f(x)=1/q at the rational points x=p/q (assuming p,q mutually prime) and zero at irrational points, is integrable on [0,1] and its integral is zero.

Problem. Is the function f(x)=\sin \frac1x integrable on [0,1]? (Do not try to compute the integral :-))

Sin(1/x) for x near the origin

The Newton-Leibniz formula

Quite obviously, if f:[a,b]\to\mathbb R^1 is integrable on [a,b], then it is also integrable on any sub-interval [a,c], a\le c \le b. The formula (2) relating the area (definite integral) with the antiderivative F of an integrable function f will take the form

\displaystyle F(c)=F(a)+\int_a^c f(x)\,dx \iff \int_a^c f(x)\,dx=F(c)-F(a).

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 f is a step function with the partition points x_1,\dots,x_N, then the corresponding area function F(z)=\int_0^z f(x)\,dx is continuous but differentiable only everywhere except these points.

Theorem. If f(x) is continuous at a point c\in[a,b], then the area function F(z)=\int_a^z f(x)\,dx is differentiable at c and F'(c)=f(c). ♦

Change of the independent variable in the integrals

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

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

f(x)=g(h(x))=(g\circ h)(x).

The graphs of functions of f and g 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 x_1<x_2\cdots<x_N will be transformed into the step function with the partition points z_1<z_2<\cdots<z_N, z_i=h(x_i), with the same values. Moreover, if f(x) is squeezed between two step functions, f_-\le f\le f_+, then its transform is squeezed between the transforms g_\pm 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 \sum_1^N \lambda_i(x_{i}-x_{i-1}) and their transforms \sum_1^N \lambda_i(z_i-z_{i-1}). The heights \lambda_i are unchanged, and the widths z_i-z_i by the finite difference lemma are equal to the initial widths multiplied by the derivative h'(c_i) computed at some intermediate points c_i\in[x_{i-1},x_i]. The result is as if instead of the step function f(x) we would integrate the function f(x)\cdot h'(x). Passing to limit, we conclude that

\displaystyle \int_a^b g(h(x))\,h'(x)\,dx=\int_{h(a)}^{h(b)}g(z)\,dz.

Change of independent variable and integral of a step function

Of course, this is equivalent to the chain rule of differentiation for the primitive functions: if F(X)=\int_a ^X f(x)\,dx and G(Z)=\int_{h(a)}^Z g(z)\,dz, then F(X)=G(h(X)) and F'(X)=G'(h(X))\cdot h'(X).

The formula for change of variables of integrals can be easily memorized using the existing notation: in the formula \int g(z)\,dz one has to transform not just the integrand g(z) by substituting z=h(x), but also the differential dz should be transformed using the formula dz=h'(x)\,dx.

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