Introduction. Elements differential and integral calculations

Plan

Derivative of function Integration Indefinite integral Properties of indefinite integral Definite integral Properties of indefinite integral

Derivative of functionDerivative of function y=f(x) along argument х is called limit of ratio increase of function to increase of argument. Derivative of function y=f(x) is denoted by у, у(х), f, f(x),

dx

xdf

dx

df

dx

d;;

So, due to definition,

x

xfxxf

x

yxfy

xx

00

' limlim'

Derivative of derivative is called derivative of the second order or second derivative. It is denoted as y, y2, f(x), f2(x),

2

2

2

2

;dx

xfd

dx

yd

.

f x “f prime x” or “the derivative of f with respect to x”

y “y prime”

dy

dx“dee why dee ecks” or “the derivative of y with

respect to x”

df

dx“dee eff dee ecks” or “the derivative of f with

respect to x”

df x

dx“dee dee ecks uv eff uv ecks” or “the derivative

of f of x”( of of )d dx f x

 

Velocity of some arbitrary moving point is vector quantity, it is defined with the help of vector - displacement of point per some interval of time. However, if point is moving along the line, then its position, displacement, velocity, acceleration is given by numbers, i.e. scalar quantities. Let         is position function of some point, then s'(t) expresses velocity of movement as some moment t (instantateous velocity), i.e.         .

Physical sense of derivative

Differentiation Rules

If f(x) = x6 +4x2 – 18x + 90

f’(x) = 6x5 + 8x – 18

*multiply by the power, than subtract one from the power.

General differential formulas

Integration

Anti-differentiation is known as integration

The general indefinite formula is shown below,

Integrals of Rational and Irrational Functions

Definite integrals

1 3 x

yy = x2 – 2x + 5 Area under curve = A

A = ∫1 (x2-2x+5) dx = [x3/3 – x2 + 5x]1

= (15) – (4 1/3) = 10 2/3 units2

3

3

Integration – Area Approximation

The area under a curve can be estimated by dividing the area into rectangles.

Two types of which is the Left endpoint and right endpoint approximations.

The average of the left and right end point methods gives the trapezoidal estimate.

y

y = x2 – 2x + 5

x

y = x2 – 2x + 5

x

LEFT

RIGHT

Newton-Leibniz formulaThe formula expressing the value of a definite integral of a given function f over an interval as the difference of the values at the end points of the interval of any primitive F of the function f :

It is named after I. Newton and G. Leibniz, who both knew the rule expressed by (*), although it was published later. If f is Lebesgue integrable over [a,b] and F is defined by

where C is a constant, then F   is absolutely continuous,                 almost-everywhere on   [a,b]      (everywhere if f   is continuous on [a,b]  ) and (*) is valid. A generalization of the Newton–Leibniz formula is the Stokes formula for orientable manifolds with a boundary.

1. a

adxxf 0 ;

2. b

aabdx ;

3. b

a

a

bdxxfdxxf ;

4. b

a

b

a

b

adxxfdxxfdxxfxf 2121 ;

Properties of definite integral

Properties of definite integral

5. b

a

b

adxxfKdxxKf ;

6. b

a

c

a

b

cdxxfdxxfdxxf ;

7. b

adxxf 0 , если 0xf .

Integration by Substitution. Separable Differential Equations

M.L.King Jr. Birthplace, Atlanta, GA Greg KellyHanford High SchoolRichland, WashingtonPhoto by Vickie Kelly, 2002

The chain rule allows us to differentiate a wide variety of functions, but we are able to find antiderivatives for only a limited range of functions. We can sometimes use substitution to rewrite functions in a form that we can integrate.

Example 1:

52x dx Let 2u x

du dx5u du61

6u C

62

6

xC

The variable of integration must match the variable in the expression.

Don’t forget to substitute the value for u back into the problem!

Example 2:(Exploration 1 in the book)

21 2 x x dx One of the clues that we look for is if we can find a function and its derivative in the integrand.

The derivative of is .21 x 2 x dx

1

2 u du3

22

3u C

3

2 22

13

x C

2Let 1u x

2 du x dx

Note that this only worked because of the 2x in the original.Many integrals can not be done by substitution.

Example 3:

4 1 x dx Let 4 1u x

4 du dx

1

4du dx

Solve for dx.1

21

4

u du3

22 1

3 4u C

3

21

6u C

3

21

4 16

x C

Example 4:

cos 7 5 x dx7 du dx

1

7du dx

1cos

7u du

1sin

7u C

1sin 7 5

7x C

Let 7 5u x

Example 5: (Not in book)

2 3sin x x dx 3Let u x23 du x dx

21

3du x dx

We solve for because we can find it in the integrand.

2 x dx

1sin

3u du

1cos

3u C

31cos

3x C

Example 6:

4sin cos x x dx

Let sinu x

cos du x dx

4sin cos x x dx

4 u du51

5u C

51sin

5x C

Example 7:

24

0tan sec x x dx

The technique is a little different for definite integrals.

Let tanu x2sec du x dx

0 tan 0 0u

tan 14 4

u

1

0 u du

We can find new limits, and then we don’t have to substitute back.

new limit

new limit

12

0

1

2u

1

2We could have substituted back and used the original limits.

Example 8: (Exploration 2 in the book)

1 2 3

13 x 1 x dx

3Let 1u x

23 du x dx 1 0u

1 2u 1

22

0 u du

23

2

0

2

3u Don’t forget to use the new limits.

3

22

23

22 2

3 4 2

3

Separable Differential Equations

A separable differential equation can be expressed as the product of a function of x and a function of y.

dyg x h y

dx

Example:

22dy

xydx

Multiply both sides by dx and divide

both sides by y2 to separate the

variables. (Assume y2 is never zero.)

22

dyx dx

y

2 2 y dy x dx

0h y

Separable Differential Equations

A separable differential equation can be expressed as the product of a function of x and a function of y.

dyg x h y

dx

Example 9:

22dy

xydx

22

dyx dx

y

2 2 y dy x dx

2 2 y dy x dx 1 2

1 2y C x C

21x C

y

2

1y

x C

2

1y

x C

0h y

Combined constants of integration

Example 10:

222 1 xdyx y e

dx

2

2

12

1xdy x e dx

y

Separable differential equation

2

2

12

1xdy x e dx

y

2u x

2 du x dx

2

1

1udy e du

y

1

1 2tan uy C e C 21

1 2tan xy C e C 21tan xy e C Combined constants of integration