calcul 1011

Elements of Mathematics: some embarrasignly simple (but

practical) tools for calculus

Jordi Villa i Freixa ([email protected])

November 23, 2011

Contents

1 To start with 3

2 Tools from calculus 5

2.1 Real numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.2 Real functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.3 Function limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.4 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

3 Derivatives 8

3.1 Differential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

3.2 Newton’s method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3.3 Chain rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3.4 Extreme values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3.5 Mean value theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

4 Taylor’s approximation 10

4.1 Taylor series for ND functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

5 Optimization 12

1

MAT: 2011-31035-T1 MSc Bioinformatics for Health Sciences

5.1 One-dimensional optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

5.1.1 Golden Search . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

5.2 Secant method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

5.3 Unconstrained optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

5.3.1 Gradient Search . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

5.3.2 The Newton-Raphson method . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

5.3.3 Conjugated gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

5.3.4 Quasi-Newton methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

5.4 Constrained optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

5.4.1 Lagrange multipliers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

5.5 Global Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

5.5.1 Stochastic methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

5.5.2 Simulated annealing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

5.5.3 Simplex method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

5.5.4 Genetic Algorithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

6 Integral calculus 17

6.1 Summatories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

6.2 Definite integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

6.3 Numerical integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

6.3.1 Trapezium formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

6.3.2 Simpson’s formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

7 Sources of information 23

Calculus 2


Figure 1: A very simple function and its two first derivatives

See also [1, 2].

1 To start with

Draw a function with these characteristics:

• Its limit when x→ −∞, is 5. The function approximates −∞ when x is arbitrarialy big.

• f(x) decreases for x ≤ −2, and increases in [1, 5].

• It has a local maxima at x = 5.

• f(−2) = 0 = f(6)

• When x left-approaches 1, the function approaches +∞, and when it right-approaches thatvalue the function approaches −1.

Now have a look at the plots in Figure 1 and try to determine to which functions they correspond(Trick: they correspond to succesive derivatives of an initial function)1

Exercise 1Draw a sketch of the function f(x) = 6x

1+x2 on the interval [−3, 3]

Exercise 2Draw a sketch of the function f(x) = x

1−x2 on the interval [−3, 3]

This course is focussed on optimization, although a final section is given on integration. The aimis that you realize what is the meaning of continuty, derivability and discrete character of functions.Let us start with stating the optimization problem

1Here are the solutions: f(x) = 3x3 − 10x2 − 56x+ 5; f ′(x) = 9x2 − 20x− 56; f ′′(x) = 18x− 20.

Calculus 3


Figure 2: Example of complex functions to optimize: a complex surface and the commercial agenttravelling problem

Figure 3: The structure of a given (bio)chemical system may have been associated to an energyfunction composed by a sum of all bonded (bonds, bends and tornions) and non bonded interactions(typically van der Waals and Coulomb interactions). Such a function yield a potential energysurface (PES) describing the energy of any given configuration of the system (the figure representsan oversimplification of this problem, by considering a PES depending on just 2 variables)

• An objective function which we want to minimize or maximize.

• A set of unknowns or variables which affect the value of the objective function.

• A set of constraints that allow the unknowns to take on certain values but exclude others.

Functions can be as simple as those in Figure 1 or as complex as those in Figure 2

Sometimes, for example, we may be interested in optimizing the structure of molecular structures,and for this we need to get good descriptions of the functions controling the energetics of suchmolecules (see Figure 3).

Calculus 4


2 Tools from calculus

2.1 Real numbers

Some useful issues:

• R and the real line

• Absolute value

|a| ={a if a ≥ 0;−a if a < 0.

• Triangle inequality|a+ b| ≤ |a| ± |b|

• Variable: leter we assign to any member of the set

• Domain: set of real numbers the variable represents

• Intervals; examples:

(a, b) = {x : a < x < b}[a, b) = {x : a ≤ x < b}[a, b] = {x : a ≤ x ≤ b}

(a,∞) = {x : a < x}

2.2 Real functions

Definition 1. A function f from D to E is a correspondence that assigns, to each element x ∈ D,a unique element y ∈ E that we call f(x).

The function is repreented by

Df→ E

f : D → E

D is the domain of the function f and the antidomain of f is the E-subset of all possible f(x)for x ∈ D.

Definition 2. Let be f a function such as when x is in D, −x it is also in D.

• f is even if f(−x) = f(x);∀x ∈ D.

• f is odd if f(−x) = −f(x);∀x ∈ D.

Definition 3. A function f with domain D and antidomain E, is a biunique function is whena 6= b in D, then f(a) 6= f(b) in E.

Calculus 5


Exercise 3Are these functions biunique?

f(x) = 3x+ 2

g(x) = x4 + 2x2

Definition 4. A function f is a polinomial if

f(x) = anxn + an−1x

n−1 + . . .+ a1x+ a0

where the coefficients a0, a1, . . . , an are real numbers and the powers are non-negative integers.

Definition 5. Let be f a function from D to E and let be g a function from E to K. The compositefunction g ◦ f is a function from D to K defined by

(g ◦ f)(x) = g(f(x));∀x ∈ D

2.3 Function limit

Let be a within an open interval, and let be f a function defined in the whole interval except,perhaps, in a, and L a real number. Then,

limx→a

f(x) = L

Definition 6. (informal) means that f(x) can arbitrarily approach L if x is chosen close to a (butx 6= a).

Definition 7. (formal) means that ∀ε > 0 ∈ R,∃δ > 0 ∈ R; si 0 < |x− a| < δ, then |f(x)− L| < ε.

Exercise 4Check that limx→4

12 (3x− 1) = 11

2

Theorem 1. Let be a a point contained in an open interval and f a function defined within thewhole interval excepts, perhaps, in a. Then, limx→a f(x) = L if and only if limx→a− f(x) = L andlimx→a+ f(x) = L.

Exercise 5Find limx→1− f(x), limx→1+ f(x) and limx→1 f(x) for

f(x) =

{2− x per x < 1x2 + 1 per x > 1

Some properties of the limit definition:

• limx→a c = c i limx→a x = a

Calculus 6


• If limx→a f(x) = L i limx→a g(x) = M :

limx→a

[f(x) + g(x)] = L+M

limx→a

[f(x) · g(x)] = L ·M

limx→a

[f(x)

g(x)] =

L

M, si M 6= 0

limx→a

[cf(x)] = cL,∀c ∈ R

limx→a

[f(x)− g(x)] = L−M

Theorem 2. Let us assume that for all x in an open interval containing a except, perhaps, forx = a, f(x) ≤ h(x) ≤ g(x). If limx→a f(x) = L = limx→a g(x), then limx→a h(x) = L.

Exercise 6Demonstrate, using the above theorem, that limx→0 x

2 sin 1x = 0.

It is simple to solve:

−1 ≤ sin1

x≤ 1

−x2 ≤ x2 sin1

x≤ x2

limx→0−x2 ≤ lim

x→0x2 sin

1

x≤ limx→0

x2

limx→0

x2 sin1

x= 0

2.4 Continuity

Theorem 3. We call a function f continuous at a if:

• f is defined in an open interval containing a;

• limx→a f(x) exists; and

• limx→a f(x) = f(a)

Theorem 4. (mean value theorem) If f is a continuous function within [a, b], and w is any valuebetween f(a) and f(b), then, there exists at least one value c ∈ [a, b] such that f(c) = w.

Theorem 5. If a function f is continuous in a given interval and does not have zeroes in it, thenf(x) > 0 or f(x) < 0 for all x in that interval.

Calculus 7


Exercise 7Find where P (x) = 1

2 (5x3 − 3x) is positive or negative.

3 Derivatives

Theorem 6. f is defined in an open interval containing a. The derivative of f in a, representedby f ′(a), is given by

f ′(a) = limh→0

f(a+ h)− f(a)

h

if this limit exists, or f ′(a) = limx→af(x)−f(a)

x−a .

It is easy to see that the derivative of a function is also a function. Notation:

f ′(x) = Dx[f(x)] = Dxy = y′ =dy

dx=

d

dx[f(x)]

Some rules worth having in mind:Dx(c) = 0

Dx(x) = 1

Dx(xn) = nxn−1si n ∈ Z

Dx[cf(x)] = cDx[f(x)]

Dx[f(x)± g(x)] = Dx[f(x)]±Dx[g(x)]

Dx[f(x)g(x)] = g(x)Dx[f(x)] + f(x)Dx[g(x)]

Dx[f(x)

g(x)] =

g(x)Dx[f(x)]− f(x)Dx[g(x)]

[g(x)]2, g(x) 6= 0

Exercise 8Use the linearization of f(x) = x2 at x = 5 to approximate (5.1)2.

3.1 Differential

Theorem 7. Be y = f(x), where f is a derivable function, and being ∆x and increment of x.

• The differential dx of x is dx = ∆x.

• The differential dy of the dependent variable y is dy = f ′(x)∆x = f ′(x)dx.

and one can write:dy

dx= f ′(x) = lim

∆x→0

∆y

∆x

Calculus 8


3.2 Newton’s method

Theorem 8. Being f a derivable function and being r a value where f is zero. If xn is an approxi-mation to r and f ′(xn) 6= 0, then the next approximation xn+1 is given by:

xn+1 = xn −f(xn)

f ′(xn)

Exercise 9Find the zeroes of f(x) = x3 + 3x2 − 9x− 29 and g(x) = x3 − 4x

Exercise 10Find where the graphs for the functions f(x) = 2x

1+x2 and g(x) = arctanx cross.

iteration xn f(x) f’(x) f”(x) dx0 0.0 +5.0 -56.0 -20.0 -2.81 -2.8 +17.5440 +70.560 -70.40 +1.00232 -1.7977 +55.9249 +9.0395 -52.3586 +0.17263 -1.6251 +56.7207 +0.2683 -49.2510 -0.00544 -1.6197 +56.7214 +0.0049 -49.1546 -0.0001

iteration xn f(x) f’(x) f”(x) dx0 +2.0 -123.0 -60.0 +16.0 +3.751 +5.750 -77.2969 +70.560 +83.50 -1.51572 +4.2343 -183.6597 +9.0395 +56.2174 -0.36783 +3.8665 -187.6118 +0.2683 +49.5967 -0.02464 +3.8419 -187.6268 +0.0049 +49.1551 -0.0001

3.3 Chain rule

Definition 8. If y = f(u), u = g(x), and the derivatives dydu and du

dx exist, then the functioncomposed by y = f(g(x)) has the derivative

dy

dx=dy

du

du

dx= f ′(u)g′(x) = f ′(g(x))g′(x)

Exercise 11Find the derivative of

√cos2(5x) + sin2(5x).

3.4 Extreme values

Theorem 9. If a function f is continuous in a closed interval [a, b], then f hits a minimum and amaximum at least once in [a, b].

Calculus 9


Theorem 10. Being c a number within the domain of f : f(c) is a local maximum if ∃(a, b) containingc in such a way that f(x) ≤ f(c) for all x ∈ (a, b). Local minimum is defined in an analogous way.

Theorem 11. A number c in the domain of a function f is called a critical point of f if f ′(c) = 0or f ′(c) does not exist.

Theorem 12. (Rolle) If a function f is continuous within the closed interval [a, b], derivable withinthe open interval (a, b) and f(a) = f(b), then there exists at least one number c within (a, b) suchthat f ′(c) = 0.

Exercise 12Verify Rolle’s thm for the function f(x) = 4x2 − 20x+ 29 at (1, 4).

3.5 Mean value theorem

Theorem 13. (Mean value) If a function f is continuous in a closed interval [a, b] and derivable inan open interval (a, b), then there exists a number c in (a, b) such that

f(b)− f(a) = f ′(c)(b− a)

Exercise 13Verify the mean value thm for f(x) = x3 − 8x− 5 at (1, 4).

Exercise 14Let f(x) = 9

x+1 . Find the point (c, d) on the graph of f , with c in the open interval (10, 12), suchthat

1. the tangent line to y = f(x) at (c, d) is parallel to the chord line joining (10, f(10)) and(12, f(12)), and

2. c is the largest value in (10, 12) that satisfies condition 1.

4 Taylor’s approximation

Theorem 14. Being f a function with n + 1 derivatives in a given interval containing c. If x is anumber of the interval, then there exists a value z within c and x such that

f(x) = f(c) +

n∑i=1

f (i)(c)

i!(x− c)i︸︷︷︸

Taylor’s polynomial

+f (n+1)(z)

(n+ 1)!(x− c)n+1︸︷︷︸

residual

Calculus 10


Figure 4: Taylor approximations for f(x) = ln(x+ 1)

Exercise 15Estimate the precission of approximating f(x) = lnx with a Taylor’s polynomial for n = 3 andc = 1.

Exercise 16Get the Taylor’s polynomial of degree n around x = 0 for f(x) = 1

1−x with x 6= 1.

Figure 4 shows the different Taylor approximations for f(x) = ln(x+ 1).

Taylor’s polynomials around c are called MacLaurin polynomials.

Exercise 17What is the bound on the error that the MacLaurin polynomial yields when trying to approximate

cos(x) by p4(π)?

4.1 Taylor series for ND functions

In a general multidimensional (N) case:

f(x) ≈ f(xk) + (x− xk)T · gk︸︷︷︸

linear term

+1

2(x− xk)

T ·Hk · (x− xk)︸︷︷︸quadratic term

where gk is the gradient vector of N dimensions and Hk is the N×N Hessian matrix at point xk:

gk =

∂f∂x1∂f∂x2

...∂f∂xN

xk

, Hk =

∂2f

∂x1∂x1

∂2f∂x1∂x2

· · · ∂2f∂x1∂xN

∂2f∂x2∂x1

∂2f∂x2∂x2

· · · ∂2f∂x2∂xN

...∂2f

∂xN∂x1

∂2f∂xN∂x2

· · · ∂2f∂xN∂xN

xk

Calculus 11


Figure 5: Schema for Golden Search

5 Optimization

5.1 One-dimensional optimization

5.1.1 Golden Search

See Figure 5.

• when bracketting a zero in a function we use intervals that are 12 the size of the two previous

values.

• From the Rolle thm we know we can bracket a minimum in a continuous function

• It can be shown that the best way to bracket a minimum is using the Golden Mean value.Thus: suppose, we search for the minimum of a given function f(x). The initial interval [x1, x4](the middle one in the figure below) is symmetrically divided into three subintervals so that(x2 − x1) = (x4 − x3) = g(x4 − x1) and (x3 − x1) = (1− g)(x4 − x1). Suppose, we know thatthe minimum is somewhere in the interval. If f(x2) < f(x3) then we can bracket the minimumby the interval [x1, x3] (see 5), so that xnew

1 = xold1 , xnew

3 = xold2 , xnew

4 = xold3 . The function

at xnew2 should be calculated. If we require that the new subintervals have the same relative

lengths then we come to the equation g(1− g) = (1− 2g) which solves for 3−√

52 ≈ 0.38197.2

See also http://www.shokhirev.com/nikolai/abc/optim/optim/optim.html and http://en.wikipedia.

org/wiki/Golden_section_search.

5.2 Secant method

• Assumes a function to be approximately linear in the region of interest.

• Each improvement is taken as the point where the approximating line crosses the axis.

• Retains only the most recent estimate!

2In mathematics and the arts, two quantities are in the golden ratio if the ratio of the sum of the quantities to thelarger quantity is equal to the ratio of the larger quantity to the smaller one: a+b

a= a

b= ψ = 1− g

Calculus 12


Figure 6: Secant method (left) and Gradient Search metghod (right)

5.3 Unconstrained optimization

5.3.1 Gradient Search

See Figure 6B. Starting at point x0. As many times as needed, move from point xi to xi+1 byminimizing along the line from xi in the direction of the local downhill gradient −∇f(xi).

Exercise 18Get the absolute maximum for f(x, y) = 2e−(x−1)2−(y−1)2 cos(x2 + y2) using a gradient search.

• select a point (x0, y0)

• Compute ∇f(x0, y0) and f(x0, y0)

• Get the maximum for the Davidon function: ψ(t) = f(x0 + fx(x0, y0)t, y0 + fy(x0, y0)t), andt0 will give you the new iteration point.

Nota de classe

mira http://linneus20.ethz.ch:8080/1_5_3.html#SECTION00253100000000000000 per de-tall del steepest descent, que no inclou la maximitzaci per Davidon function sino a pl.

5.3.2 The Newton-Raphson method

From the Taylor expansion of the function:

f(x) ≈ f(xk) + (x− xk)T · gk︸︷︷︸

linear term

+1

2(x− xk)

T ·Hk · (x− xk)︸︷︷︸quadratic term

we can take derivatives∇f(x) = gk + Hk · (x− xk)

Calculus 13


If we assume that f(x) takes its minimum at x = x∗, the gradient is zero:

Hk · (x∗ − xk) + gk = 0

which is a simple linear system. The Newton-Raphson considers x∗ to be the next point in theiterative formula:

xk+1 = xk −H−1k · gk

Exercise 19Get the absolute maximum for f(x, y) = 2e−(x−1)2−(y−1)2 cos(x2 + y2) using:

1. a gradient search;

2. a Newton Raphson approach.

You may use a table generated with a spread sheet or you may produce a short program in yourpreferred language.

Nota de classe

l’exercici proposat de maximitzar una funci de dues variables s difcil i cal treballar-lo. Mirar bel mtode d’steepest descent i el de Newton Raphson (els dos proposats a l’exercici). Mirar tambd’entendre correctament el mtode d’optimitzar la funci de Davidson. Mirar una bona explicacia http://linneus20.ethz.ch:8080/1_5_3.html#SECTION00253100000000000000. Fer un dibuixque mostri el concepte de ”constant norm”, entes com un radi determinat al voltant d’x, un radidonat per l’stepsize. s prou entendor aix

5.3.3 Conjugated gradient

• Let’s come back to the gradient search

• Let’s minimize f(x) over the hyperplane that contains all previous search directions.

x0+ < p0,p1,p2, . . . ,pi >

• If the vectors pi are chosen to be L.I. we should ideally perform only N searches.

f(x) ≈ c− g · x +1

2x ·H · x

• initial gradient g0 and an initial h0 = g0

• the CG method will construct gi+1 = gi − λH · hi and hi+1 = gi+1 + γhi

Calculus 14


• these vectors satisfy the orthogonality and conjugacy conditions:

gi · gj = 0

hi ·H · hj = 0

gi · hj = 0

and the scalars are given by:

λi =gi · gi

hi ·H · hiγi =

gi+1 · gi+1

gi · gi

5.3.4 Quasi-Newton methods

xk+1 = xk −H−1k · gk

• Davidon-Fletcher-Powell (DFP)

• Broyden-Fletcher-Goldfarb-Shanno (BFGS)

• Build an iterative approximation of the inverse of the Hessian matrix:

limi→inf

Ai = H−1

5.4 Constrained optimization

5.4.1 Lagrange multipliers

Theorem 15. Let be f(x, y) and g(x, y) two functions with continuous partial derivatives such thatf has a maximum or minimum f(x0, y0) when (x, y) is restricted by g(x, y) = 0. If ∇g(x0, y0) 6= 0then a value λ exists such that:

∇f(x0, y0) = λ∇g(x0, y0)

Typically we first build the Lagrangian function L = f − λg and calculate their critical pointsby making ∇L = 0. We find the values of x, y, ... as a function of λ and then substitute them intothe constraint equation. After that, we can come back to the critical points and find the point(s)that we were interested in.

Exercise 20Find the extreme values of f(x, y) = xy for (x, y) restricted to the elipse 4x2 + y2 = 4.

5.5 Global Optimization

We need this approach when we have complex functions to optimize. Complex means here that, forexample, they contain a huge number of minima or that they are discrete functions.

Calculus 15


Figure 7: Functins to be optimized can have complex shape

Figure 8: Simulated annealing: (a) f(x) = Ax2 + cos(x/n) with different ruggedness (b) Distri-bution of 10000 SA processes started at random initial positions for the PES with A=1 (left) andA=0.1(right) at the given T

5.5.1 Stochastic methods

• Fundamental challenge in stochastic optimization: to balance the number of downhill movesof the dynamical process against the number of uphill moved.

• Number of metastable states grows exponentially with degrees of freedom

5.5.2 Simulated annealing

• SA simulates the finite T dynamics of the system

• Starting from r with energy E(r) one generates a new r′ with energy E(r′) which replaces theoriginal configuration with probability:

P =

{exp (−β[E(r′)− E(r)]) if E(r′) > E(r)

1 otherwise

• At a given β SA samples the configurations r of the PES according to their thermodynamicprobability.

• basic hopping technique

Calculus 16


Figure 9: The simplex method

5.5.3 Simplex method

The method uses the concept of a simplex, which is a polytope of N + 1 vertices in N dimensions:a line segment in one dimension, a triangle in two dimensions, a tetrahedron in three-dimensionalspace and so forth.3

5.5.4 Genetic Algorithms

The results of a genetic algorithm, because of the implicit discreteness of the approximation and itsstochasticity, look like the graph in Figure 10.

6 Integral calculus

6.1 Summatories

n∑k=1

ak = a1 + a2 + a3 + · · ·+ an

Theorem 16. Let be n a positive integer and let be {a1, a2, . . . , an} i {b1, b2, . . . , bn} two sets ofreal numbers. Then

3http://mathworld.wolfram.com/SimplexMethod.html

Calculus 17


Figure 10: Genetic algorithms are good solutions to optimize complex functions. Lower right corner:typical results from a GA optimization. The fitness function improves in a discrete manner

Calculus 18


Figure 11: A graphical representation of a Riemann sum

•∑nk=1(ak + bk) =

∑nk=1 ak +

∑nk=1 bk

•∑nk=1 cak = c (

∑nk=1 ak) ;∀c ∈ R

•∑nk=1(ak − bk) =

∑nk=1 ak −

∑nk=1 bk

A Riemann sum (Figure 11) is defined as follows

Definition 9. Let be f a function defined in a closed interval [a, b] and let be P a partition in [a, b].A Riemann’s sum of f (or f(x)) for P is an expression RP obtained by

RP =

n∑k=1

f(wk)∆xk

where wk is a value in [xk−1, xk] and k = 1, 2, . . . , n.

6.2 Definite integral

Definition 10. Let be f a function defined in a closed interval [a, b]. The definite integral of fbetween a and b is given by: ∫ b

a

f(x)dx = lim||P ||→0

∑k

f(wk)∆xk

• If c > d, then∫ dcf(x)dx = −

∫ cdf(x)dx.

• If f(a) exists, then∫ aaf(x)dx = 0.

Calculus 19


If f is an integrable function and f(x) ≥ 0, for all x ∈ [a, b] then Area =∫ baf(x)dx.

Theorem 17. If a function f is continuous in [a, b], then f is integrable in [a, b].

Theorem 18. If f and g are integrable functions in [a, b] and c ∈ R:

•∫ bacdx = c(b− a)

•∫ bacf(x)dx = c

∫ baf(x)dx

•∫ ba

[f(x)± g(x)]dx =∫ baf(x)dx±

∫ bag(x)dx

Exercise 21Find

∫ π/20

(sinθ + cosθ)dθ

Theorem 19. (Mean value of the integral) If f is a continuous function in [a, b], then there existsa number z in (a, b) such that ∫ b

a

f(x)dx = f(z)(b− a)

and f(z) is called the mean value of f in [a, b].

Exercise 22If∫ 3

0x2dx = 9, find the value z that satisfies the above theorem

Theorem 20. (Fundamental calculus theorem) Let f be a continuous function in [a, b].

1. If the function G is defined by

G(x) =

∫ x

a

f(t)dt

∀x ∈ [a, b], G is an antiderivative of f in [a, b].

2. If F is whatever antiderivative function of f in [a, b], then∫ b

a

f(x)dx = F (x)]ba = F (b)− F (a)

Calculus 20


Exercise 23Check the above theorem for

∫ x1

3dt.

We can get,

G(x) =

∫ x

1

3dt = 3(x− 1) = 3x− 3

and we can see that for any z ∈ [a, b]

G′(z) = limh→0G(z + h)−G(z)

h= limh→0

3(z + h)− 3− (3z − 3)

h= 3

We should note that it is not necessary to evaluate the exact value of the integral, as any functionlike G(x) = 3x+K fullfills the theorem. In addition, the second result for this theorem says:∫ 2

1

3dt = (3x+K)]21 = (3 · 2 +K)− (3 · 1 +K) = 3

6.3 Numerical integration

Definition 11. If f is continuous in [a, b] and a = x0, x1, . . . , xn = b determines a uniform partitionof [a, b], then ∫ b

a

f(x)dx ≈ b− an

n∑i=1

f(xi)

where xk = (xk−1 + xk)/2 is the mid point of [xk−1, xk].

6.3.1 Trapezium formula

Definition 12. If f is a continuous function in [a, b] and if a = x0, x1, . . . , xn = b determines auniform partition of [a, b], then∫ b

a

f(x)dx ≈ b− a2n

[f(x0) +2f(x1) + 2f(x2) + · · ·

+2f(xn−1) + f(xn)]

If M > 0 ∈ R : |f ′′(x)| ≤M, ∀x ∈ [a, b], then the error when using the formula is less than or equalto M(b− a)3/(12n2).

Exercise 24Find the error in approximating the integral

∫ 2

1(1/x)dx with the trapezium formula with n = 10.

k xk f(xk) m mf(xk)

Calculus 21


0 1 1.0000 1 1.00001 1.1 0.9091 2 1.81822 1.2 0.8333 2 1.66673 1.3 0.7692 2 1.53854 1.4 0.7143 2 1.42865 1.5 0.6667 2 1.33336 1.6 0.6250 2 1.25007 1.7 0.5882 2 1.17658 1.8 0.5556 2 1.11119 1.9 0.5263 2 1.052610 2 0.5000 1 0.5000

The sum of the last column is 13.8754, that multiplied by b−a2n = 2−1

20 = 120 gives an approximation

to the integral ∫ 2

1

1

xdx =≈ 1

20(13.8754) ≈ 0.69377

To evaluate the error we should find a number M > 0 that is bigger than the second derivativeof the function in the completeconsidered interval. f(x) = (1/x), f ′′(x) = (2/x3), in the intervalx ∈ [a, b], reaches a maximum at x = 1 and, thus:

f ′′(x)| ≤ 2

(1)3= 2

Taking M = 2 and substituting in the formula for the error we get:

error =2(2− 1)3

12(10)2< 0.002

This is, the integral, corresponding to the evaluation of the value ln 2, has a signifficant error. Inorder to reduce it we could apply Simpson’s formula, that takes into account the approxiamtion tothe shape of the function to be integrated to the second order.

6.3.2 Simpson’s formula

Definition 13. Let f be a continuous function in [a, b] and n an even integer. If a = x0, x1, . . . , xn =b determines a uniform partition for [a, b], then∫ b

af(x)dx ≈ b−a

3n [f(x0) + 4f(x1) + 2f(x2) +

4f(x3) + · · ·+ 2f(xn−2) + 4f(xn−1) + f(xn)]

If M > 0 ∈ R : |f (iv)(x)| ≤M, ∀x ∈ [a, b], then the error is less than or equal to M(b−a)5/(180n4).

Exercise 25Find the error when evaluating

∫ 2

1(1/x)dx with n = 10 using Simpson’s formula.

Calculus 22


7 Sources of information

• http://www.nr.com

• http://www.math.dartmouth.edu/~klbooksite/

• http://www.math.temple.edu/~cow/

• http://www.math.ucdavis.edu/~calculus/

• http://www.math.scar.utoronto.ca/calculus/Redbook/

• http://www.math.utep.edu/Faculty/mabry/web/1411.htm

• http://www.math.dartmouth.edu/~klbooksite/all_exercises.htm

• Software:

– http://www.r-project.org/

– http://www.octave.org/

– http://www.gnuplot.info/

– http://demonstrations.wolfram.com/download-cdf-player.html

References

[1] A. Isaev. Introduction to mathematical methods in bioinformatics. Springer Verlag, 2004.

[2] C. Neuhauser. Calculus for biology and medicine. Prentice Hall Upper Saddle River (New Jersey),2000.

Calculus 23

calcul 1011

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