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Antiderivative/
Indefinite Integral
Find all possible functions
F(x) whose derivative is
f(x) = 2x+1
F(x) =
x2 + x
F(x) = x2 + x + 5
F(x) = x2 + x -1000
F(x) = x2 + x + 1/8
F(x) = x2 + x - π
Definition
A function F is called an antiderivative (also an
indefinite integral) of a function f in the
interval I if
F '(x) f (x)
for every value x in the interval I.
The process of finding the antiderivative of a
given function is called antidifferentiation or
integration.
Find all antiderivatives
F(x) of f(x) = 2x+1
F(x) =
x2 + x
F(x) = x2 + x + 5
F(x) = x2 + x -1000
F(x) = x2 + x + 1/8
F(x) = x2 + x - π
In fact, any function of the form F(x) =
x2 + x + c where c is a constant is an
antiderivative of 2x + 1
Theorem
If F is a particular antiderivative of f on an
interval I, then every antiderivative of f on I
is given by
F(x) c
where c is an arbitrary constant, and all the
antiderivatives of f on I can be obtained by
assigning particular values for c. .
Notation
4 The symbol denotes the operation of
antidifferentiation, and we write
f (x)dx F(x) c
where F’(x)=f(x), and c is an arbitrary constant.
This is read “The indefinite integral of f(x)
with respect to x is F(x) + c".
In this notation,
is the integral sign;
c
f(x) is the integrand;
dx is the differential of x which denotes
the variable of integration; and
is called the constant of integration.
4 If the antiderivative of the function on interval
I exists, we say that the function is integrable
over the interval I.
f (x)dx F(x) c
Integration Rules
1. Constant Rule. If k is any real number, then
the indefinite integral of k with respect to x is
kdx kx C
2. Coefficient Rule. Given any real number
coefficient a and integrable function f,
af (x)dx a f (x)dx
Integration Rules
n
3. Sum and Difference Rule. For integrable
functions f and g,
[ f1(x) f2 (x)]dx f1(x)dx f2 (x)dx
4. Power Rule. For any real number n,
where n ≠ -1, the indefinite integral xn of is,
xn 1
x dx C
n 1
Example 1.
1 2
2
2
(5x 7)dx 5xdx 7dx
5 xdx 7dx
5(1
x2
C ) 7x C
5
x2
7x C
Integration Formulas for
Trigonometric Functions
sec x C
cscx C
sec x tan x dx
cscx cot x dx
cot x C
cos x C
sin x C
tan x C
sin x dx
cos x dx
sec2
x dx
csc2
x dx
Integration by Chain Rule/Substitution
For integrable functions f and g
f (g(x))[g '(x)dx] F(g(x)) C
where is an F antiderivative of f and C is an
arbitrary constant.
Example 3.
2
1 tan 2x C
2
1 cot2x
2
2
1 (tan2x) 1 tan 2x C
2 sec2
2xdx
1
2
1
2
sec2
2xdx
1)dx
sec2
2xdx
sec2
2x csc2
2xdx
sec2
2x(cot2
2x
sec2
2x cot2
2xdx
(tan2x) 2
sec2
2xdx
(tan2x) 2
2sec2
2xdx
1
Applications of
Indefinite Integrals
1. Graphing
Given the sketch of the graph of the function,
together with some function values, we
can sketch the graph of its antiderivative
as long as the antiderivative is
continuous.
Example 4. Given the sketch of the function f
=F’(x) below, sketch the possible graph of F if it is
continuous, F(-1) = 0 and F(-3) = 4.
4
5
2 3 4 5
1
-5 -4 -3 -2 -1
3
2
0 1
-1
-2
-3
-4
-5
F(x) F’(x) F’’(x) Conclusion
X<-3 + - Increasing,
Concave down
X=-3 4 0 - Relative maximum
-3<x<-2 - - Decreasing,
Concave down
X=-2 - 0 Decreasing,
Point of inflection
-2<x<-1 - + Decreasing
Concave up
X=-1 0 0 + Relative minimum
X>-1 + + Increasing,
Concave up
The graph of F(x)
4
5
-3 -2 -1 0 1 2 3 4 5
1
-4
-5
3
2
-1
-2
-3
-4
-5
1. Boundary/Initial Valued Problems
There are many applications of indefinite integrals
in different fields such as physics, business,
economics, biology, etc.
These applications usually desire to find particular
antiderivatives that satisfies certain conditions
called initial or boundary conditions, depending
on whether they occur on one or more than one
point.
Applications of
Indefinite Integrals
Example 5.
dy
6x 1
dx
Suppose we wish to find a particular
antiderivative satisfying the equation
and the initial condition y=7 when x =2.
Sol’n of Example 5
7 3(2)2
3
dy (6x 1)dx
dy (6x 1)dx
y 3x2
x C
but x 2 when
C
y 7, then
C 7
Thus the particular antiderivative desired,
y 3x2
x 7
The Differential Equations
Equation containing a function and its derivative or justits
derivative is called differential equations.
Applications occur in many diverse fields such as physics,
chemistry, biology, psychology, sociology, business,
economics etc.
The order of a differential equation is the order of the
derivative of highest order that appears in the equation.
The function f defined by y= f(x) is a solution of a
differential equation if y and its derivatives satisfy the
equation.
7 3(2)2
3
dy
6x 1
dx
dy (6x 1)dx
dy (6x 1)dx
y 3x2
x C
but x 2 when
C
y 7, then
C 7
Thus find the particular solution
y 3x2
x 7
If each side of the differential equations
involves only one variable or can be
reduced in this form, then, we say that these
are separable differential equations.
Complete solution (or general solution)
y = F(x) + C
Particular solution – an initial condition is
given
Example 6. Find the complete
solution of the differential equation
2
1
d2
y dy
4x 3
dx dx
dy (4x 3)dx
dy (4x 3)dx
y 2x2
3x C
1
1
1 2
dy 2x2
3x C
dx
dy (2x2
3x C )dx
y 2 x3 3 x2
3 2
C x C
d2 y d dy
dx dx
dy
dx
dx2
let y
4x 3
dx2
d2
y
integration-131127090901-phpapp01.pptx

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integration-131127090901-phpapp01.pptx

  • 2. Find all possible functions F(x) whose derivative is f(x) = 2x+1 F(x) = x2 + x F(x) = x2 + x + 5 F(x) = x2 + x -1000 F(x) = x2 + x + 1/8 F(x) = x2 + x - π
  • 3. Definition A function F is called an antiderivative (also an indefinite integral) of a function f in the interval I if F '(x) f (x) for every value x in the interval I. The process of finding the antiderivative of a given function is called antidifferentiation or integration.
  • 4. Find all antiderivatives F(x) of f(x) = 2x+1 F(x) = x2 + x F(x) = x2 + x + 5 F(x) = x2 + x -1000 F(x) = x2 + x + 1/8 F(x) = x2 + x - π In fact, any function of the form F(x) = x2 + x + c where c is a constant is an antiderivative of 2x + 1
  • 5. Theorem If F is a particular antiderivative of f on an interval I, then every antiderivative of f on I is given by F(x) c where c is an arbitrary constant, and all the antiderivatives of f on I can be obtained by assigning particular values for c. .
  • 6. Notation 4 The symbol denotes the operation of antidifferentiation, and we write f (x)dx F(x) c where F’(x)=f(x), and c is an arbitrary constant. This is read “The indefinite integral of f(x) with respect to x is F(x) + c".
  • 7. In this notation, is the integral sign; c f(x) is the integrand; dx is the differential of x which denotes the variable of integration; and is called the constant of integration. 4 If the antiderivative of the function on interval I exists, we say that the function is integrable over the interval I. f (x)dx F(x) c
  • 8. Integration Rules 1. Constant Rule. If k is any real number, then the indefinite integral of k with respect to x is kdx kx C 2. Coefficient Rule. Given any real number coefficient a and integrable function f, af (x)dx a f (x)dx
  • 9. Integration Rules n 3. Sum and Difference Rule. For integrable functions f and g, [ f1(x) f2 (x)]dx f1(x)dx f2 (x)dx 4. Power Rule. For any real number n, where n ≠ -1, the indefinite integral xn of is, xn 1 x dx C n 1
  • 10. Example 1. 1 2 2 2 (5x 7)dx 5xdx 7dx 5 xdx 7dx 5(1 x2 C ) 7x C 5 x2 7x C
  • 11. Integration Formulas for Trigonometric Functions sec x C cscx C sec x tan x dx cscx cot x dx cot x C cos x C sin x C tan x C sin x dx cos x dx sec2 x dx csc2 x dx
  • 12. Integration by Chain Rule/Substitution For integrable functions f and g f (g(x))[g '(x)dx] F(g(x)) C where is an F antiderivative of f and C is an arbitrary constant.
  • 13. Example 3. 2 1 tan 2x C 2 1 cot2x 2 2 1 (tan2x) 1 tan 2x C 2 sec2 2xdx 1 2 1 2 sec2 2xdx 1)dx sec2 2xdx sec2 2x csc2 2xdx sec2 2x(cot2 2x sec2 2x cot2 2xdx (tan2x) 2 sec2 2xdx (tan2x) 2 2sec2 2xdx 1
  • 14. Applications of Indefinite Integrals 1. Graphing Given the sketch of the graph of the function, together with some function values, we can sketch the graph of its antiderivative as long as the antiderivative is continuous.
  • 15. Example 4. Given the sketch of the function f =F’(x) below, sketch the possible graph of F if it is continuous, F(-1) = 0 and F(-3) = 4. 4 5 2 3 4 5 1 -5 -4 -3 -2 -1 3 2 0 1 -1 -2 -3 -4 -5 F(x) F’(x) F’’(x) Conclusion X<-3 + - Increasing, Concave down X=-3 4 0 - Relative maximum -3<x<-2 - - Decreasing, Concave down X=-2 - 0 Decreasing, Point of inflection -2<x<-1 - + Decreasing Concave up X=-1 0 0 + Relative minimum X>-1 + + Increasing, Concave up
  • 16. The graph of F(x) 4 5 -3 -2 -1 0 1 2 3 4 5 1 -4 -5 3 2 -1 -2 -3 -4 -5
  • 17. 1. Boundary/Initial Valued Problems There are many applications of indefinite integrals in different fields such as physics, business, economics, biology, etc. These applications usually desire to find particular antiderivatives that satisfies certain conditions called initial or boundary conditions, depending on whether they occur on one or more than one point. Applications of Indefinite Integrals
  • 18. Example 5. dy 6x 1 dx Suppose we wish to find a particular antiderivative satisfying the equation and the initial condition y=7 when x =2.
  • 19. Sol’n of Example 5 7 3(2)2 3 dy (6x 1)dx dy (6x 1)dx y 3x2 x C but x 2 when C y 7, then C 7 Thus the particular antiderivative desired, y 3x2 x 7
  • 20. The Differential Equations Equation containing a function and its derivative or justits derivative is called differential equations. Applications occur in many diverse fields such as physics, chemistry, biology, psychology, sociology, business, economics etc. The order of a differential equation is the order of the derivative of highest order that appears in the equation. The function f defined by y= f(x) is a solution of a differential equation if y and its derivatives satisfy the equation.
  • 21. 7 3(2)2 3 dy 6x 1 dx dy (6x 1)dx dy (6x 1)dx y 3x2 x C but x 2 when C y 7, then C 7 Thus find the particular solution y 3x2 x 7
  • 22. If each side of the differential equations involves only one variable or can be reduced in this form, then, we say that these are separable differential equations. Complete solution (or general solution) y = F(x) + C Particular solution – an initial condition is given
  • 23. Example 6. Find the complete solution of the differential equation 2 1 d2 y dy 4x 3 dx dx dy (4x 3)dx dy (4x 3)dx y 2x2 3x C 1 1 1 2 dy 2x2 3x C dx dy (2x2 3x C )dx y 2 x3 3 x2 3 2 C x C d2 y d dy dx dx dy dx dx2 let y 4x 3 dx2 d2 y