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Национальный авиационный университет

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DDR 3 p.132-189

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arcsin

x 1

arcsin

C .

2(x 1)

Trigonometric substitution

16.

3 2x x 2

dx .

Solution. Completing the square of the radicand: 3 2x x 2

4 (x 1)2 .

Letting

x 1 2 sin t , dx 2 cos tdt , we obtain then

1)2

4 4 sin 2 t

2 cos tdt

2 cos t

2 cos tdt

(x 1)

4 sin

cos22 tdt

1 sin2

tdt ò

-ò dt =-ctg t -t +C .

sin

Replacing t by arcsin

x 1

. Thus

1 sin 2 t

1) 2

3 2x x 2

cos t

ctg t sin t

sin t

x 1

Consequently,

3 2x x 2

arcsin

x 1

C .

17.

xdx

9 x

x 3tg t

3tg t

3dt

Solution. I

cos2 t

9 9 tg2 t

cos

t arctg

142

sin tdt

z cost

cos2 t cost 1

dz sin tdt

z2 z 1

This integral can be found by the partial fractions:

B C

B C z2 A B z A

z2 z 1

z 1

z2 z 1

B C 0

C 1;

B 0 B

then

I 3

z 1

A 1

A 1;

3ln

cost

3ln

cost 1

cost

t arctg

cos t

9 x2

3ln

9 x

1 tg

3ln

9 x2

C C

9 x

3ln 3 3ln

9 x

3ln

9 x

9 x2

3ln

3ln 3.

T 5. Exercises for class and homework

Find the integrals

1.				dx				.
				x 2 6x 11

4.		(6x 1)dx					.
				x x 2

7.	3			x 1			dx			.
				x 1		(x 1)		3

10.					dx				.
			x( x 5
							x 2 )

2.		dx			.		3.		(3x 5)dx					.

	3	2x x 2						2x 2 12x 15
5.	(2 sin x 1) cos xdx					.	6.			dx		.
								4
	5 4 sin x sin 2 x								x		x
8.	6 xdx		.				9.			x 3	dx .
	3
	1	x								x		x
11.		1		x dx .			12.		3 4 x 1				dx .
		1
				x						x

143

13.		3	x		dx .				14.				x7			dx .
	1 3 x 2												1 x 2
16.			dx						. 17.				x (1 2 6 x )3dx .
16.	x		x 2 x 4						. 17.				x (1 2 6 x )3dx .
	x		x 2 x 4
19.		x 2 dx				.			20.						dx			.
	x	2 2x 5										(2x 3) 4x x 2
22.		x 2 dx						.	23.						dx				.
22.								.	23.										.
	x	2 2x 2								(x			1) 6x x 2 5
25.		x 4 dx					.		26.					dx			.
25.							.		26.				2 x x 2				.
	x	2 4x 5									x		2 x x 2
28.	2x x 2			dx .					29.			1 x 2			dx .
28.		x	2	dx .					29.			2 x		2	dx .
		x										2 x

Answers

15.						dx						.
15.		x(1		x			3	)	1/ 4			.
		x(1		x				)
18.				dx					.
18.									.
			3 1 x3
21.						dx								.
21.			(6x 8 x2 )3											.
			(6x 8 x2 )3
24.					dx						.
24.	x			x	2						.
	x			x		x 1
27.		(2x 2				3x)dx							.
27.													.
			x 2 2x 5
30.				xdx						.
30.										.
	x					x 2 1

			1.				ln		x 3				x2 6x 11									C .						2.			arcsin			x 1				C			. 3.					3		2x2 12x 15
			1.				ln		x 3				x2 6x 11									C .						2.			arcsin							C			. 3.					2		2x2 12x 15
																																		2												2
2			2 ln \| x 3									x2	6x 15 / 2								\| C .4. 6								x x2				4 arcsin(2x 1) C .5. 2																		5 4t t2
3arcsin(						t			2	) C ,			where					t sin t .						6. 2					x 44 x 4ln								1 4 x					C .				7.		3			3		x	1 4
						t			2																																							3					x	1 4


							3																																									16					x	1
							3																																									16					x	1
3			x			1 7							6 6			x5					x 66 x 6arctg6 x C .11.
		3							C .8.									2																					(		x 2)				1 x arcsin										x C .
28		3							C .8.				5					2																					(		x 2)				1 x arcsin										x C .
28			x			1							5
12.		12 3			4		x 1)				7		3		4	x 1)			4		C .13. u					3		3u C ,where									u (1 x						2 / 3 1/ 2				.14.					t	7		3	t	5
		7		(								3		(																															)
				(								3		(																															)						7				5
																																																			7				5
	t3		t C ,							where			t			1 x2						. 17.			1		ln			t 1		8ln			2t 1						15 ln			t		1		5					C ,
																									1																							5
																																																	t
																								2																	2										1
																								2																	2										1
where t								x2 x 4					2			.		18.		1		ln	u2 u 1								1	arctg		2u 1					C , where							u				3 1 x3					.
where t												x				.		18.		6		ln	(u	1)2							3	arctg				3			C , where							u						x			.
												x								6			(u	1)2							3					3																x
19.			x 3					x2 2x 5 ln									x 1					x2 2x 5								C .			20.			1				ln		x 6				60x 15x2									C .
			x 3																																	1						x 6				60x 15x2
			2																																	15									2x 3

144

21.

8 x2

x 2

C .

23. Remark. Use a substitution

t .

2(x 2)

2 6x 8 x2

x 1

24.

x2 x 1

C .

26. Remark. Use a substitution

t . 30. Remark.

x2 1 .

Multiple both numerator and denominator by

T 5. Individual test problems

5.1. Find the integrals.

5.1.1.			dx			.						5.1.2.					dx		.							5.1.3.						dx						.
5.1.1.		8x x2				.						5.1.2.			2 4x 1				.							5.1.3.												.
	4	8x x2												x	2 4x 1													2		3x 2x2
5.1.4.			dx			.						5.1.5.						dx			.						5.1.6.					dx					.
5.1.4.						.						5.1.5.									.						5.1.6.										.
	x	2 6x 8												2	8x 2x2													3			2x x2
5.1.7.			dx				.					5.1.8.					dx		.								5.1.9.					dx							.
5.1.7.		2x x2					.					5.1.8.			6x x2				.								5.1.9.												.
	2	2x x2											1		6x x2														x		2 10x 4
5.1.10.				dx					.			5.1.11.						dx								. 5.1.12.								dx								.
5.1.10.									.			5.1.11.				4x2 8x 3										. 5.1.12.																.
		2x 3 x2														4x2 8x 3															1	2x x2
5.1.13.				dx						.		5.1.14.						dx						.			5.1.15.							dx											.
5.1.13.										.		5.1.14.												.			5.1.15.																		.
		4x2 x 4														2 4x x2															x	2 2x 4
5.1.16.				dx					.			5.1.17.						dx					.				5.1.18.							dx							.
5.1.16.	3x 2 x				2				.			5.1.17.						2					.				5.1.18.				x	2									.
	3x 2 x															2x		8x 10													x		5x 6
5.1.19.				dx							.	5.1.20.					dx			.							5.1.21.					dx				.
5.1.19.				2							.	5.1.20.				x	2			.							5.1.21.								2	.
	16x			8x 3												x		x 1												2 x 2x
5.1.22.				dx								. 5.1.23.						dx									. 5.1.24.							dx						.
5.1.22.		4x2 4x 3										. 5.1.23.					5 6x 9x2										. 5.1.24.													.
		4x2 4x 3															5 6x 9x2															3x 2x2
5.1.25.			dx		.							5.1.26.						dx							.		5.1.27.							dx									.
5.1.25.	1 x x2				.							5.1.26.				1 2x x2									.		5.1.27.					4 3x x2											.
5.1.28.			dx					.				5.1.29.						dx				.					5.1.30.							dx										.
5.1.28.			2					.				5.1.29.										.					5.1.30.						2											.
	3x		6x 9													3 x x2																x		4x 8

145

5.2. Find the integrals.

5.2.1.	(3x 1)dx				.						5.2.2.	(4x 3)dx					.										5.2.3.	(5x 2)dx										.

	x2 4x 5												7 2x x2															x2 6x 10
5.2.4.	(7x 3)dx			.							5.2.5.		(3 x)dx					.									5.2.6.	(2 3x)dx			.

	6 2x x2												x2 2x 2															x2 4x 6
5.2.7.	(x 3)dx					.					5.2.8.		(2x 1)dx						.								5.2.9.	(4x 3)dx				.

	x2 8x 18												x2 2x 3															6 4x x2
5.2.10.	(5x 4)dx									.	5.2.11.		(x 2)dx									.					5.2.12.	(3x 5)dx								.
	x2 8x 17
													4 2x x2															x2 4x 8
5.2.13.	(3x 2)dx						.				5.2.14.		(6x 1)dx								.						5.2.15.	(3x 1)dx					.

	5 4x x2												x2 2x 9															8 4x x2
5.2.16.	(4 x)dx								.		5.2.17.		(7x 4)dx											.			5.2.18.	(9x 1)dx						.

	x2 2x 4												11 4x x2															5 2x x2
5.2.19.	(4x 3)dx						.				5.2.20.		(3x 7)dx										.				5.2.21.	(3x 1)dx												.

	3 2x x2												12 4x x2															x2 6x 11
5.2.22.	(2 x)dx									.	5.2.23.		(x 4)dx													.	5.2.24.	(3 4x)dx											.
	14 4x x2
													x2 8x 20															x2 4x 7
5.2.25.	(5 x)dx										5.2.26.		dx														5.2.27.	( x 5)dx
								.												.															.
	x2 4x 6												1 2x x2															x2 2x 9
5.2.28.	(x 3)dx									.	5.2.29.		(3x 2)dx												.		5.2.30.	(x 4)dx									.
	21 8x x2
													x2 4x 10															13 6x x2
5.3. Use trigonometric substitution to find the integrals.
5.3.1.	1 x2		dx .								5.3.2.		x2 1			dx .											5.3.3.	4 x2		dx .
	x																											x
													x
5.3.4.	1 x2		dx .								5.3.5.		4 x	2	dx .												5.3.6.	x2 9		dx .
	x	4																										x

5.3.7.	x2 4		dx .								5.3.8.		4 x2			dx .											5.3.9.	9 x2	dx .
	x2												x4															x4

146

5.3.10.						x2 4					dx .								5.3.11.				4 x				2 3					dx .				5.3.12.					dx												.
5.3.10.						x		4			dx .								5.3.11.						x	6						dx .				5.3.12.				(x	2					1)				5			.
						x																			x															(x						1)
5.3.13.						x2 9					dx .								5.3.14. x				3		9 x						2	dx .				5.3.15.				x2 1									dx .
5.3.13.							x				dx .								5.3.14. x						9 x							dx .				5.3.15.				x2									dx .
5.3.16.								dx									. 5.3.17.								dx							.				5.3.18.				x2 9								dx .
5.3.16.				x 2				x2		1 3							. 5.3.17.					x 2			x 2			1				.				5.3.18.				x2								dx .
5.3.19.								dx					.						5.3.20.				x2 9						dx .							5.3.21.				dx												.
5.3.19.			x3					x2 1					.						5.3.20.					x4					dx .							5.3.21.		x2			x2 9											.
5.3.22. x						2	1 x				2	dx			.					5.3.23.			x2 4					dx .								5.3.24.				16 x						2			dx .
5.3.22. x							1 x					dx			.					5.3.23.				x	2			dx .								5.3.24.				x	4								dx .
																								x																x
5.3.25.								dx						.					5.3.26.				x2 9							dx .						5.3.27.					dx													.
5.3.25.						(x 2 9)3								.					5.3.26.					x4						dx .						5.3.27.				(x2 4)3														.
5.3.28.						x2 dx					.							5.3.29.				16 x					2			dx .				5.3.30.						16 x 2											dx .
5.3.28.						9	x2				.							5.3.29.						x	4					dx .				5.3.30.						x	2										dx .
						9	x2																	x																x
5.4. Find the integrals.
5.4.1.						1			x 1											dx .								5.4.2.									x 1 1										dx .
5.4.1.	(1					3		x 1)												dx .								5.4.2.					(	3		x 1 1)				x 1							dx .
	(1							x 1)						x 1																			(			x 1 1)				x 1
5.4.3.	3 (x 1)2										6 x						1				dx .							5.4.4.					(3			x 1)( x 1)								dx .
5.4.3.						x	1				3		x		1						dx .							5.4.4.								6 x5								dx .
						x	1						x		1																					6 x5
5.4.5.	x 3 x2								6 x							dx .												5.4.6.							2x 1 3				2x 1							dx .
5.4.5.	3															dx .												5.4.6.																		dx .
						x(		x 1)																													2x 1
5.4.7.								x 1									dx .											5.4.8.							x 1 23 x 1										dx .
5.4.7.	3				x 1				6		x				1		dx .											5.4.8.					3			x 1			x 1						dx .
					x 1						x				1																		2			x 1			x 1
5.4.9.							6 x 1											dx .									5.4.10.								x		x 3 x2						dx .
5.4.9.		3				x 1												dx .									5.4.10.										3						dx .
						x 1					x 1																									x(	x 1)
5.4.11.								6 x 3											dx .									5.4.12.						x 3 x2 6 x								dx .
5.4.11.			3			x	3												dx .									5.4.12.						3								dx .
						x	3						x 3																							x(	x 1)

147

5.4.13.

3x 1 1

dx .

5.4.14.

3x 1 2

dx.

3x 1 3

3x 1

3 3x 1

3x 1

5.4.15.

dx.

5.4.16.

x 3

dx .

3 (2x 1)2

2x 1

x 6 x 1

5.4.17.

x 3

dx .

5.4.18.

3x 1 1

dx .

x 3

3x 1

5.4.19.

x 1

dx .

5.4.20.

dx .

5.4.21.

dx .

x (

x 1)

5.4.22.

5.4.23.

4 x

dx .

5.4.24.

x 3 x

dx .

3x 3 x2

x 1

5.4.25.

x 3

dx .

5.4.26.

dx .

5.4.27.

dx .

x 43

4x 3

x 3

5.4.28.

dx .

5.4.29.

dx .

5.4.30.

dx .

x 3

Topic 6. Definite integrals

Definitions. Existence of the definite integral. Properties of the definite integral. The fundamental theorem of integral calculus by NewtonLeibniz. Basic methods of definite integrals application.

Literature: [1, section 7], [3, section 7, §2], [4, section 7, §23], [5, section 6], [6, section 9, ch. ch. 9.1, 9.2], [7, section 11, §§1–6] [9, §§35–39].

T 6. Main concepts

6.1. Definitions

148

The sums of the form f ( i ) xi (Fig.2.1) were used to estimate

i 1

certain quantities such as area, mass, distance, and volume. The larger n is and the shorter the sections xi xi xi 1 are, the closer we would expect these

approximating sums to be to the quantity we are trying to find. We are really interested in what happens to these approximating sums as all the sections in the partition are chosen smaller and smaller. This leads to the notion of the

y=f(x)

f(ξ2)

f(ξ1)

f(ξn)

О x0=aξ1 x1 ξ2 x2

xn-1 ξn xn= x

Fig.2.1

definite integral of a function over an interval, which will be defined after we introduce a measure of the “fineness” of a partition.

Definition. The mesh of a partition is the length of the longest section (or sections) in the partition.

Definition.	If	f (x) is a function defined	on [a, b]	and	the	sums
n
f ( i ) xi approach a certain number as the mesh of partitions of						[a, b]
i 1
shrinks toward	0	(no matter how the sampling	number i	is	chosen in
[xi-1 , xi ] ), that certain number is called the definite integral of					f (x)	over
[a, b] .

The numbers a and b are called limits of integration; a is the lower limit and b is the upper limit. The symbol x is called the variable of integration and f(x) is the integrand.

The definite integral is also called the definite integral of f (x) from a to

b and integral of f (x) from a to b. The symbol for this number is b	f x dx .
a

The symbol comes from the letter S of Sum; the dx traditionally suggests a

149

f x x f x .

small section of the x axis and will be more meaningful and useful later. It is important to realize that area, mass, distance traveled, and volume are merely applications of the definite integral. (It is a mistake to link the definite integral too closely with one of its applications, just as it narrows our understanding of number 2 to link it always with the idea of two fingers.)

Slope, velocity, magnification, and density are particular interpretations or applications of the derivative, which is a purely mathematical concept defined as limit:

f (x) = lim

x 0

Similarly, area, total distance, mass, and volume are just particular interpretation of the definite integral, which is also defined as limit:

b	f x dx	lim	n	f	x .
		mesh 0		i	i
a			i 1

In advanced calculus it is proved that, if f(x) is continuous, then

lim f i xi

mesh 0 i 1

exists; that is, a continuous function always has a definite integral. For emphasis we record this fact, an important result in advanced calculus, as a theorem.

Theorem. Let f(x) be a continuous function defined on [a, b]. Then the approximating sums

f ( i ) xi

i 1

Approach a single number as the mesh of the partition of [a, b]

approaches 0. Hence b f x dx exists.

Geometrical interpretation of the definite integral. If f(x) is

continuous and f(x) 0 on [a,b] then b f x dx can be interpreted as the

area of the region bounded by the curve y = f(x), the x-axis and the lines x

=a and x = b, as indicated in fig. 2.1.

6.2.Properties of the definite integral

150

Let f(x) and g(x) be continuous functions ant let 1							and 2 be the
constants. Then
1.	If	f (x) 0
			b
			f (x)dx 0 .
			a
2.	If ( a b ) in [a; b] , then
			b	b
			b	b
			f (x)dx		f (x)	dx .


			a	a
3.	If	f (x) g(x) for all x in [a; b]		( a b ), then
			b	b
			f (x)dx g(x)dx .

4.If the limits of integration are equal, we have

f (x)dx 0 .

5.f (x)dx f (x)dx .

	b	b	b
6.	f (x) g(x) dx f (x)dx g(x)dx .
	a	a	a
	b	b	b	b
7.	1 f (x)dx 2 g(x)dx 1 f (x)dx 2 g(x)dx .
	a	a	a	a
8.	If a, b and c are numbers, then
		b	c	b
		f (x)dx f (x)dx f (x)dx .
		a	a	c
9.	If m and M are numbers and m f (x) M for all x in [a; b] (), then
		b
		m(b a) f (x)dx M (b a) if a b ,
		a
		and m(b a) b	f (x)dx M (b a) if a b.
		a

10. If a and b ( a b ) are numbers. Then there is a number c between a and b such that

151

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