For an Integral over a function f(x) which is continuous and bounded in an interval [a, b] there exists always amean valueξ,so thatRb
adxf(x) =f(ξ)(b−
a) for ξ∈(a, b).
Because a continuous and bounded function in an interval passes through its extrema and all the values in between.
Figure 7.8: Mean value theorem of the integral calculus
7.4 Fundamental Theorem of Differential and Inte-
The procedure is fully analogous to the expansion step of the differential calculus from the gradientf0(x0) of a function f(x) at a certain point x0 to the first derivative f0(x) as a function of the variable x.
7.4.2 Differentiation with Respect to the Upper Border
In order to study the functional dependence of the indefinite integral from the variable upper border, we are at first interested in the gradient of the function Fa(y) :
( d
dy)Fa(y) := ( d dy)
Z y a
dxf(x) = lim
∆y→0
Ry+∆y
a dxf(x)−Ry
a dxf(x)
∆y =
if we insert the definition of the derivative. Because of the interval addition we get:
= lim
∆y→0
Ry
a dxf(x) +Ry+∆y
y dxf(x)−Ry
a dxf(x)
∆y = lim
∆y→0
Ry+∆y
y dxf(x)
∆y =
According to the mean value theorem of the integral calculus there exists in the interval [y, y+ ∆y] a mean valuey+θ∆y with 0≤θ≤1, so that the following holds true:
= lim
∆y→0
f(y+θ∆y)∆y
∆y = lim
∆y→0f(y+θ∆y) =f(y).
Altogether we get the
First part of the Fundamental Theorem: Fa0(y) := (dyd) Z y
a
dxf(x) = f(y),
i.e. the differential quotient of a indefinite integral with respect to its upper border is the integrand taken at the upper border.
Exactly in this sense is differentiation the reversal of integration. For example (dyd)Ry
0 dtsin(ωt+
α) = sin(ωy+α) holds.
7.4.3 Integration of a Differential Quotient
After having learned to differentiate an integral, we are inquisitive about the reverse process, namely about the integral of a differential quotient: We start from the well-known continuous differential quotient F0(x) = f(x) which may be given to us as a continuous function f(x), i.e. really from a differential equation of the first order for F(x). We want to integrate this differential quotient over the interval [a, b]:
Z b a
dxF0(x) := lim
m→∞
m
X
n=1
∆xnF0(ξn)
with the nodes ξn ∈ [xn−1, xn] within the interval of length ∆xn = xn − xn−1, if we insert the definition of the integral. According to theMean Value Theorem ofDifferential Calculus the gradient at the nodes ξn can be replaced by the gradient of the secant, meaning the replacement of the differential quotient by the difference quotient:
F0(ξn) = F(xn)−F(xn−1) xn−xn−1
.
Written in detail this yields:
m→∞lim (F(x1)−F(x0))+(F(x2)−F(x1))+. . .+(F(xm−1)−F(xm−2))+(F(xm)−F(xm−1)) .
We easily see that all terms are cancelling in pairs except the second and the second last one, which do not at all depend onmand thus will not be affected by the limiting process:
...=F(xm)−F(x0) =F(b)−F(a) =:F(x)
b a. We therefore get altogether as
Second Part of the FUNDAMENTAL THEOREM:
Rb
a dxF0(x) = F(b)−F(a) =: F(x)
b a,
i.e. the definite integral of the differential quotient of a continuous differentiable function over an interval is equal to the difference of the function values at the upper and lower border of the interval.
Also in this sense integration is the reversal of differentiation.
For instance we get again the result: F(x2; 0, b) =Rb
0 dxx2 =b3/3,which we were hard put to derive from the definition of the integral, but now effortlessly from the differentiation (dxd)x3 = 3x2.
This second part of the Fundamental Theorem is the crucial step to the solution of our integration problem: Because we are now able to calculate all definite integrals of all the functions which we find in the second column of our Differentiation Table in Chapter 5. We simply have to read the TABLE backward from right to left and complete the heading in the following accordingly:
TABLE FOR DIFFERENTIATION AND INTEGRATION Line F(x) =R
dxf(x) F0(x)≡(dxd)F(x) = f(x) Comments:
1 const 0
2 xr rxr−1 r∈R
3 xr+1/(r+ 1) xr −16=r∈R
4 sinx cosx
5 cosx −sinx
6 tanx 1/cos2x x6= (z+ 1/2)π, z∈Z
7 cotx −1/sin2x x6=zπ, z ∈Z
8 −π/2<arcsinx < π/2 1/p
(1−x2) |x|<1 9 0<arccosx < π −1/p
(1−x2) |x|<1 10 −π/2<arctanx < π/2 1/(1 +x2)
11 0<arccotx < π −1/(1 +x2)
12 ex ex
13 rx rxlnr 0< r∈R
14 ln|x| 1/x x6= 0
15 logb|x| 1/xlnb x6= 0, 0< b∈R, b6= 1
16 sinhx coshx
17 coshx sinhx
18 tanhx 1/cosh2x
19 cothx −1/sinh2x x6= 0
20 arsinhx 1/p
(x2+ 1)
21 0<arcoshx 1/p
(x2−1) x >1
22 artanhx 1/(1−x2) |x|<1
23 arcothx −1/(x2−1) |x|>1
Our example out of the insert above, can be obtained e.g. from line two for r= 2.
A further example from this line is Rb
adx x3 = (b4 −a4)/4 with the borders a and b, generally for an arbitrary real r ∈R follows:
F(xr;a, b) = Z b
a
dx xr= br+1−ar+1 r+ 1 ,
which we have filled in the third line of the TABLE being empty until now, because it occurs very often.
From the fourth line we find for instance:
F(sinx;a, b) = Z b
a
dxsinx=−cosb+ cosa, from the twelfth line: F(ex;a, b) =
Z b a
dx ex =eb −ea and analogously many further integrals.
Exercise 7.1 Calculate the following examples of integrals:
a) Z 3
1
1
xdx, b) Z 1
−1
dx
1 +x2, c) Z b
0
√ dx
1 +x2 , d) Z 1/√
2
−1/√ 2
√ dx
1−x2 , e)
Z a
−a
dx coshx , f ) Z π/4
0
dx
cos2x , g) Z 2
1
dx
x1+a , h) Z a
−a
dx x2n+1 for n ∈Z
7.4.4 Primitive Function
Although we are able to calculate a considerable number of definite integrals over a large number of intervals, the question for theindefinite integral of a differential quotient F0(x) = f(x) remains. We once again replace the constant upper border b of the definite integral through avariable y and as above we get
Z y a
dxf(x) = Z y
a
dxF0(x) = F(y)−F(a).
This we rewrite in the following way:
F(y) = Z y
a
dxf(x) +F(a) =:
Z y a
dxf(x) +c,
because with respect to the variabley F(a) is indeed a constant, although it depends on the starting point of the interval a. Since we want to have the usual letter x as symbol for the independent variable in the functionF(y), an extraordinary sloppy way of writing has sneaked into the above equation world-wide: symbolically it is written as:
F(x) = Z
dxf(x) +cand F(x) is called the primitive function of f(x).
Thexon the left side serves only as a hint, that it is a function of an independent variable, and has obviously nothing to do with the really arbitrarily denotable integration variable xon the right hand side, which obviously does not occur anymore after the integration on the right side. Once we have cleared up this sloppy mess, it is quite a comfortable matter and acknowledged world-wide.
Written in such a sloppy way, the primitive function is actually a whole family of functions with the family parameter c. The primitive function F(x) of f(x) is exactly the family of functions which solves our original differential equation F0(x) = f(x) and that is the reason why it is so important for physicists. Out of this family with pre-given gradient f(x), the physicist has to only pick out the solution by choosing the constant c so that the function fulfils the correct boundary conditionc=F(a), and the problem is solved.
For example, the primitive function F(x) of f(x) = 3x is searched for, which fulfils the boundary condition F(1) = 2. Out of the function family F(x) = 3x2/2 +c we have to choose the function which fulfills F(1) = 3/2 +c = 2, meaning c = 1/2: consequently F(x) = (3x2+ 1)/2 is the desired solution.
Exercise 7.2 Determine the primitive functions of the following functions:
a) f(x) =x3 , b) f(x) = 1
√x2−1 , c) f(x) = sinhx and d) f(x) = 2x
Exercise 7.3 Determine the primitive functions of the following functions with the given boundary conditions:
a) f(x) = sinx with F(π) = 1, b) f(x) = 1
√x with F(4) = 1 and c) f(x) = 1
cosh2x with F(a) = 12