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f (x)dx f (c)(b a) .

A number f(c) is called a mean-value of f(x) in [a, b].

11. а) f (x)dx 2 f (x)dx , if f (x) is an even function;

a	0
a

b)f (x)dx 0 , if f (x) is an odd function.

12. b f x dx b f t dt.

The variable of integration is a “dummy variable” in the sense that any other variable produces the same result, that is, the same number.

13. If	f (x)		has integral in [a, b] and x [a; b] , then
			x
а) Ф(x) f (t)dt exists and is continuous in [a, b] ;
			a
	d			d	x
b)	d	Ф(x)		d	f (t)dt f (x) .
b)		Ф(x)		dx	f (t)dt f (x) .
	dx			dx	a
					a

6.3.Evaluating definite integrals

6.3.1.The fundamental theorem of integral calculus by Newton-

Leibniz

This section shows that there is an intimate connection between the definite integral and the derivative. This relationship, expressed in the fundamental theorem of integral calculus by Newton-Leibniz, provides a tool for computing many, but not all, definite integrals without having to form a single approximating sum. The argument will be intuitive.

Fundamental theorem. If f (x) is continuous on interval [a,b] and F(x) is any antiderivative of f (x) there, then

f (x)dx F(b) F(a) .

Usually, F(b) – F(a) is abbreviated by writing

b	b
f (x)dx F(x)	b	F(b) F(a).
a	a
a

152

The substitution method and integration by parts are the most general techniques.

6. 3.2. Substitution technique for a definite integral

Let f(x) be a continuous function on the interval [a, b], x (t) be a differentiable function on the interval [ ; ] , and ( ) a , ( ) b , then

f (x)dx f ( (t)) (t)dt .

6.3.3.Integration by parts

The integration by parts of definite integral is made according to the formula:

b	b	b
udv uv	a	vdu,
a	a	a
a		a

where u and v are the differentiable functions of x on the interval [a, b].

T 6. Typical problems

Evaluate the given definite integrals.

1. 1 3x2 2x 3 dx.

Solution. Here f(x) = 3x2 2x + 3, a = 2; b = 1. Since an antiderivative of 3x2 2x + 3 is F(x) = x3 x2 + 3x, then:

1 3x2 2x 3 dx F 1 F 2 1 1 3 8 4 6 21.

If we had chosen F(x) to be x3	x2 + 3x + C, then F(1) F( 2) =
1 1 3 C 8 4 6 C 21,	as before. Since the choice of the

value of C is immaterial, for convenience we shall always choose it to be 0, as originally done. Usually F(b) F(a) is abbreviated by writing: F x ba .

Hence we have 1 3x2 2x 3 dx x3 x2 3x 1 2

= (1 1 + 3) ( 8 4 6) = 21.

153

1								1																							3										1						3					3
1								1
2. 3			1 xdx 3									1 xd (1 x)																					3 (1 x)4															(0 1)					.
2. 3			1 xdx 3									1 xd (1 x)																		4			3 (1 x)4															(0 1)					.
0								0																						4											0			4							4
0								0

3		1 tg2 x										3						1									dx								3					1
3.							dx																																							d(1 tg x)
3.		(1 tg x)			2		dx																2		cos				2			x					(1 tg x)							2		d(1 tg x)
		(1 tg x)											(1 tg x)												cos							x					(1 tg x)
	4												4																						4
				1											1										1												1						1					2 3		.

						3
			1 tg x									1 tg										1 tg														1				3		2						2
			1 tg x																																	1				3		2						2
						4
						4											3													4
3													3																									3
4. 2			1 cos 2x dx											2																								2
															sin x						dx					sin xdx													( sin x)dx
															sin x						dx					sin xdx													( sin x)dx
				2
				2

2												2													2
				cos x						cos x										3
																				3
																					2	( 1 0) (0 ( 1)) 2 .

											2
5. Find the mean-value of																	f (x)													1						in [2; 3].
5. Find the mean-value of																	f (x)								x 2				x 1							in [2; 3].
Solution. By property 10																															b
																									1						b
																f (c)									1						f (x)dx ,
																f (c)								b	a						f (x)dx ,
																								b	a						a
																															a
where f (c) is the mean-value of																					f (x)					in [a; b] , c [a; b] .
We have
																																1
			1	3				dx								3								d	x							2								2									2x 1		3
				3												3								d	x
f (c)																																											arctg
f (c)			3			2																															2						arctg
			3	2 2 x				x 1								2								1	2									3						3								3			2
				2 2 x				x 1								2								1	2									3						3								3
																			x
																			x
																								2										2
																								2										2
						2			(arctg							5			arctg									3)							2			arctg				3			.
						3			(arctg							3			arctg									3)							3			arctg				9			.
						3										3																			3							9
Used a formula arctg arctg arctg																																						.
Used a formula arctg arctg arctg																																						.
																																	1

154



6. Show that 4		(x4 x2 ) tg xdx 0 . Do not evaluate the definite

integral.	4
integral.
Solution. Limits		of integration	are symmetrical to zero, namely if
f (x) (x4 x2 ) tg x		is odd, by property 11 the given integral is equal to 0.
We have:
	f ( x) (( x)4 ( x)2 ) tg( x) (x4 x2 ) tg x f (x) ,
consequently, f (x) is the odd function.
7. Show that the substitution is not used x sin t				to obtain the integral
		2
		ò x 3 1-x2 dx .
		0
Solution. Domain of		t is ( ; ) correspondent x		[ 1; 1] , however the
interval	of the integral is [0; 2] .		Therefore, do not use the substitution
x sin t	to obtain the integral

In Problems 8 to 10 use the substitutions to evaluate the given definite integrals.

4 x2dx

8. I

x3 1

Solution. Case 1. I

t x3 1;

1 x 4

x 4

dx;

3 x 2

x 2

1 ln

ln 63 ln 7

1 ln 9.

Case 2. Since t = x3 1, when x = 2 we have t = 7 when x = 4

t = 63. Thus				I	1	63 dt		1 ln	t	63	1	ln 63 ln 7	1 ln 9.
										63

2		dx			3	7	t	3		7	3		3
					3		t	3			3		3


9.			.
9.	2	cos x	.
0	2	cos x
0

155

Solution. The substitution tg 2x t transforms the integral to

	2									tg		x	t,		x 2 arctg t,
	2				dx					tg		2	t,		x 2 arctg t,								x
	0				dx								2dt					1 t		2					0			2
			2 cos x					dx							, cos x					2	,				0			2	=

																		1 t2					t		0		1
											1 t2							1 t2					t		0		1
1					2dt								1		dt		2					t		1		2				1
1

													2						arctg									arctg				.
							1 t		2				2	3	t	2	3		arctg			3		0		3		arctg		3	3 3	.
0	(1	t		2			1 t						0	3	t		3					3		0		3				3	3 3
	(1	t			) 2				2
							1 t		2
							1 t
			ln 5		e x	xe x 1 dx.
10.					e x	xe x 1 dx.
		0			e		3
		0

Solution. Let e x 1 t , the limits of integration with respect to t are: when x = 0 we have e0 1 0 and when x = ln5 t =

eln 5 1 2 . Evaluate

						e x		t 2					1 , e x dx 2tdt ,									dx				2tdt	.
						e x		t 2					1 , e x dx 2tdt ,									dx					.
Then																									t 2 1
Then
ln 5	e x		e x		1		2		(t 2				1)	t	2tdt			2				t 2				2		t 2 4		4
						dx											2									dt 2					dt
	e	x			3	dx				t		2			2	1	2				t	2	4			dt 2		t	2	4	dt
0	e				3		0			t			4 t			1		0			t		4			0		t		4
0							0											0								0
						2					4												t			2 4 .
						2
					2 (1									)dt		2(t		2 arctg						)
					2 (1					t	2			)dt		2(t		2 arctg					2	)
						0				t			4										2			0
						0
In Problems 11 to 13 apply integration by parts.
11. Evaluate I 4 ln x dx.
							1				x
Solution. We try							u ln x							and dv				dx	. Then							du dx		and
Solution. We try							u ln x							and dv				x	. Then							du dx		and
																		x									x
				1		1												4		4			x dxx 4 ln 4 4								4
v = x				2 dx 2x 2 . Thus I = 2												x ln x		1		2										x	1
v = x																		1		2											1
																				1

156

=4ln4 8 + 4 = 4(ln4 1).

12. 2 x 2 sin xdx .

Solution. Apply integration by parts twice:


													dv sin xdx,
						2	x 2 sin xdx					u x 2 ,
						2	x 2 sin xdx					u x 2 ,
					0							du 2xdx,	v	cos x
					0

							2				2			u x,				dv cos xdx,
							2				2
x 2	cos x		2		2 x cos xdx						2 x cos xdx
		0					0				0			du dx, v sin x


										2
				2x sin x					2	2 sin xdx			2 cos x			2	2 .
								0		0					0

13. 2 sin n xdx ( n is a natural number).

Solution.



J n	2 sin n	xdx	u sinn 1 x,	dv	sin xdx,
			du (n 1)sinn 2 xcos xdx,	v	cos x
	0



				2
cos x sin n 1	x		2	(n 1) cos2	x sin n 2	xdx
		0		0

(n 1) 2 (1 sin 2 x) sin n 2 xdx (n 1) 2 sin n 2 xdx (n 1) 2 sin n xdx .

Thereby

J n (n 1)J n 2 (n 1)J n .

Thereafter

157

J n nn 1 J n 2 .

If n 2k 1 is an odd number, we get a recursion formula. Recall that

J1 sin xdx cos x

1 .

Then

(2k 2)!!

(2k 2)(2k 4) 4 2

J 2k 1

( k N ).

(2k 1)!!

(2k 1)(2k 3) 5 3

If n 2k is an even number and recalling that J 0

we get a recursion formula

J 2k

(2k 1)!!

(2k 1)(2k 3) 3 1

( k N ) .

(2k)(2k 2) 4 2

(2k)!!

T 6. Exercises for class and homework

Evaluate the given definite integrals.

	16	x 4
1.		x 4				dx .
1.						dx .
	1	x 2
	2			dx
4.				dx					.
4.		2 cos x 3							.
	0	2 cos x 3
	0
	4		x
7.			x		dx .
7.		x 1			dx .
	1	x 1
	ln14		e	x	e		x	5
10.			e		e			5		dx .
10.				e	x					dx .
	ln 5			e			1
	ln 5

13. a 2 x 2 dx .

2	2
2.			x			x 2 1dx .
	0
	2	sin				1
		sin				1
5.					x			dx .
5.				x	2			dx .
	1			x

	1				x
8.					x		dx .
8.			1 x				dx .
	0		1 x
	0
	ln 2
11.							1 e2x dx .
		0
		2						dx
14.								dx	.
14.									.
			2 x5					x 2 1

	e	dx
3.		dx		.
3.				.
	1	x 1 ln 2 x
	2	dx
6.		dx	.
6.		x(x 1)	.
	1	x(x 1)
	1
	14	x
9.		x	dx .
9.		2 x	dx .
	2	2 x

12. x 2 1 x 2 dx .

15. 2 dx .

1 x x3

158

	3		dx
16.			dx	.
16.		x	x 2 5x 1	.
	1	x	x 2 5x 1
	e 2
19.	ln(x 2)dx .
	0

17. xe x dx .


20. 3		xdx		.
20. 3		2		.
	sin		x
	4

Answers

18. 4 x cos 2xdx .

21. ln 2 xdx ;

1. 12.				2.		26	. 3.		.	4.			2	arctg		1		.		5. 1.		6. ln					4	.		7.	3.	8. 2				. 9. 88 .
						3		2					5			5										3									2				3
11.	3		ln(2				3) .	12.					.		13.		a	2	.	14.			1		(				7	3	8) .			15.			1	ln	8	.
11.	2		ln(2				3) .	12.			16		.		13.	4			.	14.	32				(					2	8) .			15.		2		ln	5	.
	2										16					4					32									2						2			5
16. ln	7 2			7	.	17.1 2/ e .				18.				1	. 19. 2 2ln 2 .						20.					(9 4					3)	1	ln	3	. 21.			e 2 .
16. ln		9			.	17.1 2/ e .				18.		8		4	. 19. 2 2ln 2 .						20.			36		(9 4					3)	2	ln	2	. 21.			e 2 .
		9										8		4										36								2		2

T 6. Individual test problems

Use the substitution to evaluate the integrals.

293 (x 2)2

1.3 3 3 (x 2)2 dx .

8	x 1 1
4.		dx .

3	x 1 1

ln 5	e x	xe x 1 dx .
7.	e x	xe x 1 dx .
0	e		3
0
4		dx
10.		dx					.
10.							.
0 1		2x 1
0.5 ln 2			e x dx
13.			e x dx
								.
		e	x	e		x		.
	0	e		e
	0
2				dx
16.				dx					.
16.									.
0	x 1				(x 1)3

ln 2							dx
2.							dx					.
2.			e	x	(3 e				x		)	.
	0		e		(3 e						)
	0
	5						dx
5.							dx						.
5.													.
	0	2x					3x 1
ln 2
8.				e x			1dx .
	0
	7 / 3					xdx
11.						xdx				.
11.										.
	2 / 3					3x 2
	0					dx
14.						dx					.
14.						3					.
	11						x 1
	5				xdx
17.					xdx			.
17.								.
	0			3x 1

8	xdx
3.		.

3	x 1

	2 ln 2				dx
6.									.
6.				e	x				.
	ln 2			e			1
	ln 2
	5		xdx
9.			xdx			.
9.						.
	0		x 4
	ln 3				dx
12.					dx						.
12.			e	x	e		x				.
	ln 2		e		e
	ln 2
	1		x2 dx
15.			x2 dx					.
15.			4					.
	0	(1 x)
	0
	3 7			x2 dx
18.				x2 dx						.
18.										.
	0			9 x3

159

		5			x	2	dx								/ 2		cos xdx							0	x
19.					x		dx					.		20.			cos xdx					.		21.	1 e x dx .
19.				(x 1)								.		20.								.		21.	1 e x dx .
		2		(x 1)				x 1							0	4 sin x								ln 3	1 e
	ln 2					dx									e3		dx							ln 3	dx
22.						dx							.	23.			dx				.			24.	dx			.
22.													.	23.							.			24.				.
		0			e x 1 e 2x										1 x 1 ln x									ln 2	1 e x
	e3				ln xdx										26		x3dx							9	xdx
25.					ln xdx						.			26.			x3dx						.	27.	xdx	.
25.								2			.			26.			(1 x	2	)	2 / 3			.	27.	x 1	.
	e	2 x(1 ln								x)					7		(1 x		)					4	x 1
	e														7									4
	13		(x 1)dx												ln12		dx							1	xdx
28.			(x 1)dx						.					29.			dx			.				30.	xdx		.
28.			3						.					29.						.				30.			.
	0				2x 1										ln 5		4 e x							1	5 4x

Topic 7. Improper integrals

Improper integrals: integral of integration unbounded and integrand unbounded. Definitions and evaluation. Convergent and divergent improper integrals.

Literature: [1, section 8], [3, term 7, §2], [6, section 9, п. 9.4], [7, section 11, §7], [9, §40].

T 7. Main concepts

7.1. Improper integrals: integral of integration unbounded

There is the first type of improper integral. Suppose f(x) is continuous and nonnegative for a x ∞ (see fig. 2.2).

у=f(x)

О а
	b х

Fig. 2.2

We know that integral b f x dx is the area

of the region between the curve y = f(x) and the x-axis from x = a to x = b. As b ∞, we may

think of lim b f x dx as the area of the

b a

unbounded region that is shaded in fig.2.2. This limit is abbreviated by

f x dx ,

160

called the first type of improper integral. If this limit exists, f x dx is said

to be convergent or to converge to that limit. In this case the unbounded region is considered to have a finite area, and this area is represented by

f x dx . If the limit does not exist, the improper integral is said to be

divergent and the region does not have a finite area.

We can remove the restriction that f(x) 0. In general, the improper


integral f x dx is defined by
0
	f x dx	lim A	f x dx .
		A
a		a
			b
Other forms of improper integrals are			f x dx and	f x dx . In each

of three forms of improper integrals, the interval over which the integral is evaluated has infinite length.

The improper integral b f x dx is defined by

If this limit exists,

divergent.

b	f x dx	lim	b	f x dx .
		B
			B

b f x dx is said to be convergent. Otherwise, it is


The improper integral				f x dx is defined in terms:

			c	A
	f (x)dx	lim		f (x)dx lim	f (x)dx ,
		B		A
			B	c

where c is the real number.

If both integrals on the right side are convergent, then f x dx is said to

be convergent. Otherwise, it is divergent.

161

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