![Page 1: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/1.jpg)
Chapter 8
Integration Techniques
![Page 2: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/2.jpg)
8.1
Integration by Parts
![Page 3: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/3.jpg)
Slide 8 - 3
![Page 4: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/4.jpg)
Slide 8 - 4
Integrate by parts: Practice!
.
http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/intbypartsdirectory/IntByParts.html
![Page 5: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/5.jpg)
Slide 8 - 5
![Page 6: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/6.jpg)
Slide 8 - 6
![Page 7: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/7.jpg)
Slide 8 - 7
![Page 8: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/8.jpg)
8.2
Trigonometric Integrals
![Page 9: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/9.jpg)
Slide 8 - 9
![Page 10: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/10.jpg)
Slide 8 - 10
![Page 11: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/11.jpg)
Slide 8 - 11
![Page 12: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/12.jpg)
Slide 8 - 12
![Page 13: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/13.jpg)
8.3
Trigonometric Substitutions
![Page 14: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/14.jpg)
Slide 8 - 14
![Page 15: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/15.jpg)
Slide 8 - 15
Trigonometric Substitution
In finding the area of a circle or an ellipse, an integral of the form dx arises, where a > 0.
If it were the substitution
u = a2 – x2 would be effective but, as it stands,
dx is more difficult.
![Page 16: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/16.jpg)
Slide 8 - 16
Trigonometric Substitution
If we change the variable from x to by the substitutionx = a sin , then the identity 1 – sin2 = cos2 allows us to get rid of the root sign because
![Page 17: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/17.jpg)
Slide 8 - 17
Example 1
Evaluate
Solution:
Let x = 3 sin , where – /2 /2. Then dx = 3 cos d and
(Note that cos 0 because – /2 /2.)
![Page 18: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/18.jpg)
Slide 8 - 18
Example 1 – Solution
By Inverse Substitution we get:
cont’d
![Page 19: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/19.jpg)
Slide 8 - 19
Example 1 – Solution Since this is an indefinite integral, we must return to the
original variable x. This can be done either by using trigonometric identities to express cot in terms of sin = x/3 or by drawing a diagram, as in Figure 1, where is interpreted as an angle of a right triangle.
cont’d
sin =
Figure 1
![Page 20: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/20.jpg)
Slide 8 - 20
Example 1 – Solution
Since sin = x/3, we label the opposite side and the hypotenuse as having lengths x and 3.
Then the Pythagorean Theorem gives the length of the adjacent side as so we can simply read the value of cot from the figure:
cont’d
![Page 21: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/21.jpg)
Slide 8 - 21
Example 1 – Solution
Since sin = x/3, we have = sin–1(x/3) and so
cont’d
![Page 22: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/22.jpg)
Slide 8 - 22
![Page 23: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/23.jpg)
Slide 8 - 23
![Page 24: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/24.jpg)
Slide 8 - 24
Example 2
Find
Solution:Let x = 2 tan , – /2 < < /2. Then dx = 2 sec2 d and
=
= 2| sec |
= 2 sec
![Page 25: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/25.jpg)
Slide 8 - 25
Example 2 – Solution
Thus we have
To evaluate this trigonometric integral we put everything in terms of sin and cos :
cont’d
![Page 26: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/26.jpg)
Slide 8 - 26
Example 2 – Solution
=
Therefore, making the substitution u = sin , we have
cont’d
![Page 27: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/27.jpg)
Slide 8 - 27
Example 2 – Solutioncont’d
![Page 28: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/28.jpg)
Slide 8 - 28
Example 2 – Solution
We use the figure below to determine that csc = and so
cont’d
Figure 3
![Page 29: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/29.jpg)
Slide 8 - 29
Example 3
Find
Solution:First we note that (4x2 + 9)3/2 = so trigonometric substitution is appropriate.
Although is not quite one of the expressions in the table of trigonometric substitutions, it becomes one of them if we make the preliminary substitution u = 2x.
![Page 30: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/30.jpg)
Slide 8 - 30
Example 3 – Solution
When we combine this with the tangent substitution, we have x = which gives and
When x = 0, tan = 0, so = 0; when x = tan = so = /3.
cont’d
![Page 31: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/31.jpg)
Slide 8 - 31
Example 3 – Solution
Now we substitute u = cos so that du = –sin d.When = 0, u = 1; when = /3, u =
cont’d
![Page 32: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/32.jpg)
Slide 8 - 32
Example 3 – Solution
Thereforecont’d
![Page 33: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/33.jpg)
8.4
Partial Fractions
![Page 34: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/34.jpg)
Slide 8 - 34
![Page 35: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/35.jpg)
Slide 8 - 35
Integration of Rational Functions by Partial Fractions
To see how the method of partial fractions works in general, let’s consider a rational function
where P and Q are polynomials.
It’s possible to express f as a sum of simpler fractions provided that the degree of P is less than the degree of Q. Such a rational function is called proper.
![Page 36: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/36.jpg)
Slide 8 - 36
Integration of Rational Functions by Partial Fractions
If f is improper, that is, deg(P) deg(Q), then we must take the preliminary step of dividing Q into P (by long division) until a remainder R (x) is obtained such that deg(R) < deg(Q).
where S and R are also polynomials.
![Page 37: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/37.jpg)
Slide 8 - 37
Example 1
Find
Solution:
Since the degree of the numerator is greater than the degree of the denominator, we first perform the long division. This enables us to write:
![Page 38: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/38.jpg)
Slide 8 - 38
Integration of Rational Functions by Partial Fractions
If f(x) = R (x)/Q (x) is a proper rational function:
factor the denominator Q (x) as far as possible.
Ex: if Q (x) = x4 – 16, we could factor it as
Q (x) = (x2 – 4)(x2 + 4) = (x – 2)(x + 2)(x2 + 4)
![Page 39: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/39.jpg)
Slide 8 - 39
Integration of Rational Functions by Partial Fractions
Next: express the proper rational function as a sum of partial fractions of the form
or
A theorem in algebra guarantees that it is always possible to do this.
Four cases can occur.
![Page 40: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/40.jpg)
Slide 8 - 40
Integration of Rational Functions by Partial Fractions
Case I The denominator Q (x) is a product of distinct linear factors.
This means that we can write
Q (x) = (a1x + b1)(a2x + b2) . . . (akx + bk)
where no factor is repeated (and no factor is a constant multiple of another).
![Page 41: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/41.jpg)
Slide 8 - 41
Integration of Rational Functions by Partial Fractions
In this case the partial fraction theorem states that there exist constants A1, A2, . . . , Ak such that
These constants can be determined as in the next example.
![Page 42: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/42.jpg)
Slide 8 - 42
Example 2
Evaluate
Solution:
Since the degree of the numerator is less than the degree of the denominator, we don’t need to divide.
We factor the denominator as
2x3 + 3x2 – 2x = x(2x2 + 3x – 2)
= x(2x – 1)(x + 2)
![Page 43: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/43.jpg)
Slide 8 - 43
Example 2 – Solution
Since the denominator has three distinct linear factors, the partial fraction decomposition of the integrand has the form
To determine the values of A, B, and C, we multiply both sides of this equation by the product of the denominators, x(2x – 1)(x + 2), obtaining
x2 + 2x – 1 = A(2x – 1)(x + 2) + Bx(x + 2) + Cx(2x – 1)
![Page 44: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/44.jpg)
Slide 8 - 44
Example 2 – Solution
Expanding the right side and writing it in the standard form for polynomials, we get
x2 + 2x – 1 = (2A + B + 2C)x2 + (3A + 2B – C)x – 2A
These polynomials are identical, so their coefficients must be equal. The coefficient of x2 on the right side, 2A + B + 2C, must equal the coefficient of x2 on the left side—namely, 1.
Likewise, the coefficients of x are equal and the constant terms are equal.
cont’d
![Page 45: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/45.jpg)
Slide 8 - 45
Example 2 – Solution
This gives the following system of equations for A, B, and C:
2A + B + 2C = 1
3A + 2B – C = 2
–2A = –1
Solving, we get, A = B = and C = and so
cont’d
![Page 46: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/46.jpg)
Slide 8 - 46
Example 2 – Solution
In integrating the middle term we have made the mental substitution u = 2x – 1, which gives du = 2 dx and dx = du.
cont’d
![Page 47: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/47.jpg)
Slide 8 - 47
Note:We can use an alternative method to find the coefficients A, B and C. We can choose values of x that simplify the equation:
x2 + 2x – 1 = A(2x – 1)(x + 2) + Bx(x + 2) + Cx(2x – 1)
If we put x = 0, then the second and third terms on the right side vanish and the equation then becomes –2A = –1, or A = .
Likewise, x = gives 5B/4 = and x = –2 gives 10C = –1, so B = and C =
![Page 48: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/48.jpg)
Slide 8 - 48
Integration of Rational Functions by Partial Fractions
Case II: Q (x) is a product of linear factors, some of which are repeated.
Suppose the first linear factor (a1x + b1) is repeated r times; that is, (a1x + b1)r occurs in the factorization of Q (x). Then instead of the single term A1/(a1x + b1) in the equation:
we use
![Page 49: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/49.jpg)
Slide 8 - 49
Integration of Rational Functions by Partial Fractions
Example, we could write
![Page 50: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/50.jpg)
Slide 8 - 50
Example 3
Find
Solution:
The first step is to divide. The result of long division is
![Page 51: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/51.jpg)
Slide 8 - 51
Example 3 – Solution
The second step is to factor the denominator Q (x) = x3 – x2 – x + 1.
Since Q (1) = 0, we know that x – 1 is a factor and we obtain
x3 – x2 – x + 1 = (x – 1)(x2 – 1)
= (x – 1)(x – 1)(x + 1)
= (x – 1)2(x + 1)
cont’d
![Page 52: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/52.jpg)
Slide 8 - 52
Example 3 – Solution
Since the linear factor x – 1 occurs twice, the partial fraction decomposition is
Multiplying by the least common denominator, (x – 1)2(x + 1), we get
4x = A (x – 1)(x + 1) + B (x + 1) + C (x – 1)2
cont’d
![Page 53: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/53.jpg)
Slide 8 - 53
Example 3 – Solution
= (A + C)x2 + (B – 2C)x + (–A + B + C)
Now we equate coefficients:
A + C = 0
B – 2C = 4
–A + B + C = 0
cont’d
![Page 54: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/54.jpg)
Slide 8 - 54
Example 3 – Solution
Solving, we obtain A = 1, B = 2, and C = –1, so
cont’d
![Page 55: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/55.jpg)
Slide 8 - 55
Integration of Rational Functions by Partial Fractions
Case III: Q (x) contains irreducible quadratic factors, none of which is repeated.
If Q (x) has the factor ax2 + bx + c, where b2 – 4ac < 0, then, in addition to the partial fractions, the expression for R (x)/Q (x) will have a term of the form
where A and B are constants to be determined.
![Page 56: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/56.jpg)
Slide 8 - 56
Integration of Rational Functions by Partial Fractions
Example:f (x) = x/[(x – 2)(x2 + 1)(x2 + 4)] has the partial fraction decomposition:
Any term of the form: can be integrated by completing the square (if necessary) and using the formula
![Page 57: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/57.jpg)
Slide 8 - 57
Example 4
Evaluate
Solution:
Since the degree of the numerator is not less than the degree of the denominator, we first divide and obtain
![Page 58: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/58.jpg)
Slide 8 - 58
Example 4 – Solution
Notice that the quadratic 4x2 – 4x + 3 is irreducible because its discriminant is b2 – 4ac = –32 < 0. This means it can’t be factored, so we don’t need to use the partial fraction technique.
To integrate the given function we complete the square in the denominator:
4x2 – 4x + 3 = (2x – 1)2 + 2
This suggests that we make the substitution u = 2x – 1.
cont’d
![Page 59: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/59.jpg)
Slide 8 - 59
Example 4 – Solution
Then du = 2 dx and x = (u + 1), so
cont’d
![Page 60: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/60.jpg)
Slide 8 - 60
Example 4 – Solutioncont’d
![Page 61: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/61.jpg)
Slide 8 - 61
Note:Example 6 illustrates the general procedure for integrating a partial fraction of the form
We complete the square in the denominator and then make a substitution that brings the integral into the form
Then the first integral is a logarithm and the second is expressed in terms of
where b2 – 4ac < 0
![Page 62: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/62.jpg)
Slide 8 - 62
Integration of Rational Functions by Partial Fractions
Case IV: Q (x) contains a repeated irreducible quadratic factor.
If Q (x) has the factor (ax2 + bx + c)r, where b2 – 4ac < 0, then instead of the single partial fraction , the sum:
occurs in the partial fraction decomposition of R (x)/Q (x).
Each of the terms can be integrated by using a substitution or by first completing the square if necessary.
![Page 63: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/63.jpg)
Slide 8 - 63
Example 5
Evaluate
Solution:
The form of the partial fraction decomposition is
Multiplying by x(x2 + 1)2, we have
–x3 + 2x2 – x + 1 = A(x2 +1)2 + (Bx + C)x(x2 + 1) + (Dx + E)x
![Page 64: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/64.jpg)
Slide 8 - 64
Example 5 – Solution
= A(x4 + 2x2 +1) + B(x4 + x2) + C(x3 + x) + Dx2 + Ex
= (A + B)x4 + Cx3 + (2A + B + D)x2 + (C + E)x + A
If we equate coefficients, we get the system
A + B = 0 C = –1 2A + B + D = 2 C + E = –1 A = 1
which has the solution A = 1, B = –1, C = –1, D = 1 and E = 0.
cont’d
![Page 65: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/65.jpg)
Slide 8 - 65
Example 5 – Solution
Thus
cont’d
![Page 66: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/66.jpg)
Slide 8 - 66
![Page 67: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/67.jpg)
8.7
Improper Integrals
![Page 68: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/68.jpg)
Slide 8 - 68
Type 1: Infinite Intervals
![Page 69: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/69.jpg)
Slide 8 - 69
![Page 70: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/70.jpg)
Slide 8 - 70
Examples:
![Page 71: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/71.jpg)
Slide 8 - 71
Practice Example:
Determine whether the integral is convergent or divergent.
Solution:
According to part (a) of Definition 1, we have
The limit does not exist as a finite number and so the
Improper integral is divergent.
![Page 72: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/72.jpg)
Slide 8 - 72
![Page 73: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/73.jpg)
Slide 8 - 73
![Page 74: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/74.jpg)
Slide 8 - 74
Examples:
![Page 75: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/75.jpg)
Slide 8 - 75
Type 2: Discontinuous Integrands
Suppose that f is a positive continuous function defined on a finite interval [a, b) but has a vertical asymptote at b.
Let S be the unbounded region under the graph of f and above the x-axis between a and b. (For Type 1 integrals, the regions extended indefinitely in a horizontal direction. Here the region is infinite in a vertical direction.)
The area of the part of S between a and t is
Figure 7
![Page 76: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/76.jpg)
Slide 8 - 76
Type 2: Discontinuous Integrands
![Page 77: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/77.jpg)
Slide 8 - 77
Practice Example:
Find Solution:
We note first that the given integral is improper because has the vertical asymptote x = 2.
Since the infinite discontinuity occurs at the left endpoint of [2, 5], we use part (b) of Definition 3:
![Page 78: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/78.jpg)
Slide 8 - 78
Example – Solution
Thus the given improper integral is convergent and, since the integrand is positive, we can interpret the value of the integral as the area of the shaded region.
Figure 10
cont’d
![Page 79: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/79.jpg)
Slide 8 - 79
Gabriel’s Horn:
![Page 80: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/80.jpg)
8.8
Introduction to Differential Equations
![Page 81: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/81.jpg)
Slide 8 - 81
Definition
A differential equation is an equation containing an unknown function and its derivatives.
32 xdx
dy
032
2
aydx
dy
dx
yd
364
3
3
y
dx
dy
dx
yd
Examples:.
y is the dependent variable and x is independent variable.
1.
2.
3.
Ordinary Differential Equations
![Page 82: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/82.jpg)
Slide 8 - 82
Partial Differential Equation
Examples:0
2
2
2
2
y
u
x
u
04
4
4
4
t
u
x
u
t
u
t
u
x
u
2
2
2
2
u is the dependent variable and x and y are independent variables.
u is dependent variable and x and t are independent variables
1.
2.
3.
![Page 83: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/83.jpg)
Slide 8 - 83
Order of a Differential Equation
The order of the differential equation is the order of the highest derivative in the differential equation.
Differential Equation ORDER
32 x
dx
dy
0932
2
ydx
dy
dx
yd
364
3
3
y
dx
dy
dx
yd
1
2
3
![Page 84: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/84.jpg)
Slide 8 - 84
Degree of Differential Equation
Differential Equation Degree
032
2
aydx
dy
dx
yd
364
3
3
y
dx
dy
dx
yd
0353
2
2
dx
dy
dx
yd
1
1
3
The degree of a differential equation is the power of the highest order derivative term in the differential equation.
![Page 85: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/85.jpg)
Slide 8 - 85
Linear Differential Equation A differential equation is linear, if:
1. The dependent variable and its derivatives are of degree one,2. The coefficient of any term does not contain the dependent variable, y.
364
3
3
y
dx
dy
dx
yd is non-linear because the 2nd term is not of degree one.
.0932
2
ydx
dy
dx
ydExamples:is linear.
.0932
2
ydx
dy
dx
yd is non-linear because 3rd term contains y
![Page 86: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/86.jpg)
Slide 8 - 86
32
22 x
dx
dyy
dx
ydx
is non-linear because in the 2nd term the coefficient contains y.
3.
is non–linear because the coefficient on the left hand side contains y
ydx
dysin4.
![Page 87: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/87.jpg)
Slide 8 - 8787
Solving Differential Equations
![Page 88: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/88.jpg)
Slide 8 - 88
The most general first order differential equation can be written as:
There is no general formula for the solution. We will look at two types of these and how to solve them:Linear EquationsSeparable Equations
First Order Differential Equations
![Page 89: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/89.jpg)
Slide 8 - 89
Linear Differential Equations: Integrating Factor Method
If not already in the following form, re-express the equation in the form:
(1) where both p(t) and g(t) are continuous functions
Assume there is a function, , called an integrating factor.
Multiply each term in (1) by . This will give:
![Page 90: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/90.jpg)
Slide 8 - 90
NOW: assume that whatever is, it will satisfy the following :
Do not worry about how we can find a that will satisfy the above. As long as p(t) is continuous we will find it!
The equation becomes:
We recognize that the left side is nothing more than the following product rule:
![Page 91: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/91.jpg)
Slide 8 - 91
The equation becomes:
integrate both sides :
Finally, the solution y(t) is:
![Page 92: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/92.jpg)
Slide 8 - 92
What is for any given equation ?
We started with assuming:
So:
Finally:
![Page 93: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/93.jpg)
Slide 8 - 93
Solve with y(0) = 1.Practice Example:
Step1: Compare the equation with the standard form:
We identify: and .
Step2: Find the integrating factor
Step3: Multiplying through by the integrating factor, we get .
Step4: Rewrite as the derivative .
Step5: Integrate both sides with respect to x and get
Step6: Use the initial condition to find c. , gives: . So the solution to the problem is and finally:
![Page 94: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/94.jpg)
Slide 8 - 94
Linear Differential Equations: Separable Equations
A separable differential equation is any differential equation that we can write in the following form
Now rewrite the differential equation as:
Integrate both sides. Use the initial condition to find the constant of integration.
![Page 95: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/95.jpg)
Slide 8 - 95
Solve with y(0) = 1.Practice Example:
Step1: Divide through by y. We get:
Step2: Integrate both sides:
Step3: Solving for y gives: where
Step4: Use the initial condition: to get: A = 1 So the solution to the problem is:
![Page 96: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/96.jpg)
Slide 8 - 96
Examples from textbook:
![Page 97: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/97.jpg)
Slide 8 - 97
Example 2 (textbook) : Initial Value Problem Drug Dosing
![Page 98: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/98.jpg)
Slide 8 - 98
Example 3 (textbook) : Separable Equation
![Page 99: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/99.jpg)
Slide 8 - 99
Example 4 (textbook) : Separable Equation Logistic Population Growth
![Page 100: Chapter 8 Integration Techniques. 8.1 Integration by Parts](https://reader030.vdocuments.mx/reader030/viewer/2022012819/56649f285503460f94c40ac3/html5/thumbnails/100.jpg)
Slide 8 - 100
Directional Fields: Reading assignment (for your general Math knowledge).