Introduction. Elements differential and integral calculations
Introduction. Elements differential and integral calculations.
Transcript of Introduction. Elements differential and integral calculations.
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Introduction. Elements differential and integral calculations
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Plan
Derivative of function Integration Indefinite integral Properties of indefinite integral Definite integral Properties of indefinite integral
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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
.
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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
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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
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Differentiation Rules
If f(x) = x6 +4x2 – 18x + 90
f’(x) = 6x5 + 8x – 18
*multiply by the power, than subtract one from the power.
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General differential formulas
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Integration
Anti-differentiation is known as integration
The general indefinite formula is shown below,
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Integrals of Rational and Irrational Functions
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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
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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
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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.
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1. a
adxxf 0 ;
2. b
aabdx ;
3. b
a
a
bdxxfdxxf ;
4. b
a
b
a
b
adxxfdxxfdxxfxf 2121 ;
Properties of definite integral
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Properties of definite integral
5. b
a
b
adxxfKdxxKf ;
6. b
a
c
a
b
cdxxfdxxfdxxf ;
7. b
adxxf 0 , если 0xf .
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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.
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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!
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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.
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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
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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
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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
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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
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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.
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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
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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
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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
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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