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Университет имени Сулеймана Демиреля

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EXERCISES 1.1

A.Antiderivative and Indefinite Integral

1.Evaluate the integrals.

a.	dw	b. dz
c.	d cos x	d. d(x3 +3x2	1)

2.Given f(x) = d(x2 1) and f(1) = 2, find f(5).

3.x f (x) dx = x2 +5x+1 is given. Find f(x).

4.x3 f (x) dx = x5 4x3 x+1 is given. Find f(2).

5.f (x) = 5x2 – 4x + 1 and f(1) = 3 are given. Find f(3).

6.Evaluate the integrals.

a.(cos x+4x3 2e2x ) dx

b.(5x3 3x2 +5) dx

c.		5dx		d.	8
		1+ x	2		7x dx

e.	cos4x dx			f.	7sin x dx

7.Evaluate the integrals, using the basic formulas for integration.

a.	x5	dx	b.	4 dx
c.	x 3	dx	d. 15 dx
				x
e. 3x7 dx			f.	( 13 +	16	19 ) dx
				x	x	x

8.Evaluate the integrals, using the basic formulas for integration.

sin4x dx

cos5x dx

cos

sin

4sec2 4x dx

tan2 x dx

(cot2 x+2) dx

1 x2

x2 +5

x2 +1 dx

5cos(8x 4) dx

1 4x2

sin

m. 9x2 +1 dx

x dx

cot2 x dx

(tan2 x 1) dx

B.Integration Methods

9.Evaluate the integrals using the substitution method.

a.sin(4x+1) dx

b.(1+ x2 + x3 )8 (2x+3x2 ) dx

c.	(1 x2 )7			x dx	d.	x cos(x2				5) dx
e.			1	2 dx	f.		cos x			dx
		1	16x				1+sin	2	x

g.x 1+ x2 dx

h.(x4 + x2 ) (2x3 + x) dx

Integrals

10.Evaluate the integrals using the substitution method.

a.	x cos x2				dx	b.	x sin(5x2 +7) dx
c.		(1	1	4	dx	d.	5sin x	dx
		(1	3x)				cos x
e.			x		dx
e.		5x 1			dx
		5x 1

11.Evaluate the integrals using the method of integration by parts.

a. ex x dx			b. x2 ex dx
c.	x3	ex dx	d. x sin x dx
	2		f. arccos x dx
e.	xx	dx	f. arccos x dx
	e
g.	ln(x+5) dx		h. log x dx
i.	arccot x dx		j. cos(ln x) dx
k.	sin2 x e2x dx

12. Evaluate the integrals by using partial fractions.

a.		9	4 dx	g.			1	dx
		(3x+1)				2	+4x+4
						x
h.		1	dx	t.		xdx
	(x+1)3				3+ x4

13. Evaluate the integral of each radical function.

a.			5x 1 dx		b.	1 x dx
c.	x 1+ x2 dx				d. 3 x+1 dx
e.			x	dx	f.	5x	dx
			1 x2			5x2 +3

g.		5	1+ x dx		h.	x2
						3 1+ x3 dx
i.			x 2 +3 dx
			x 2

14. Evaluate the integral of each radical function.

a.	1 4x2			dx	b.	dx
						1 x2
c.	x			dx	d.	16x2	9	dx
c.	16x	2		dx	d.	x		dx
	16x		+1			x

e. 16 9x2 dx

15.Evaluate the integral of each trigonometric function.

a.	sin2	x cos x dx		b.	sin x cos x dx
c.	cos3	x sin5	x dx	d.	cos2	x sin x dx
e.	sin3	x cos5	x dx	f.	sin4	x cos3 x dx
g.	cos4	x sin3 x dx		h.	cos4x cos3x dx
i.	sin5	x cos7	x dx	j.	sin3x cos4x dx
k.	sin7x sin5x dx			l.	sin3x cos8x dx
m.	cos2x sin4x dx			n. cos5x sin x dx
o.	cos x cos4x dx			 p. sin6		x cos6 x dx

16.Evaluate the integral of each function by using the substitution t = tan 2x .

1+sin x

1 sin x dx

40	Algebra 11

A. EVALUATING DEFINITE INTEGRALS

		1. Definition of the Definite Integral

Definition
Definition		definite integral
		Let f(x) be a continuous function defined on an interval [a, b]. Then the area between the
		graph of f(x) and the x-axis is called the definite integral of f(x) betwen a and b.

For example, in the figure opposite, the shaded area A

shows the definite integral of f(x) on the interval [a, b].

We can write this expression as A = f (x) dx .

y
	y = f(x)
	A
a	b	x
a	b

THE DEFINITE INTEGRAL

Note

If the graph is below the x-axis then its integral will

be negative. However, area is a positive quantity so

we reverse the sign: A = – f(x) dx.

We say that the definite integral can be negative but

the area is always positive.

Note

If part of the graph is below the x-axis and part of the graph is above the x-axis then the integral will be the algebraic sum of the areas.

In the figure, all of the areas A, B, C are positive

numbers, so f (x) dx = A – B+C.

Integrals

y
a		b		x
				x
		A
y
A		C		x
			b	x
	B		b
	B
				41

Note

For linear functions we can use geometric methods (by finding the area of a triangle) to calculate the area under a curve.

EXAMPLE 57 Find the area of the region between the graph of y = 3x and the x-axis on the interval [0, 4].


Solution The shaded area is				y
Solution The shaded area is
4	1			12
A = 3x dx =	1	4 12 = 24	square units, by the
A = 3x dx =	2	4 12 = 24	square units, by the
0	2

formula for the area of a triangle.

	A
0	4	x
0	4

EXAMPLE 58 In the figure, the areas of the shaded parts A, B and C are given as A = 7 cm2, B = 9 cm2 and C = 8 cm2. Find the total area of the shaded region in the figure and evaluate the integral on the interval [a, b].

Solution Total area = A + B + C = 7 + 9 + 8 = 24 cm2

The integral on [a, b] = f (x) dx = A – B + C

= 7 – 9 + 8 = 6.

y
	A		C
	A
a		B	b	x
a			b

42	Algebra 11

2. The Fundamental Theorem of Calculus

Theorem Fundamental Theorem of Calculus

Let f(x) be a function such that f : [a, b] R. If F (x) = f(x) and f(x) dx = F(x) + c, then

b b

f (x) dx =(F(x)+ c)\|= F(b) F(a)
a	a

Here we use the notation | to show the boundaries of the integral.

Note

The Fundamental Theorem of Calculus shows us that we do not need a constant of integration c when we evaluate a definite integral. For example, suppose we write F(a) + c instead of F(a), and F(b) + c instead of F(b). Then by the Fundamental Theorem of Calculus,

f (x) dx =(F(b)+ c) (F(a)+ c) = F(b) – F(a) + c – c = F(b) – F(a).

		FUNDAMENTAL THEOREM OF CALCULUS

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

	59	1
EXAMPLE		x dx =?
		x dx =?
		0
Solution		1		2	1	2			2		1.
Solution		x dx = x			\|=	1	–	0	=		1.
		0	2		0	2		2			2
	60	5
EXAMPLE		x2	dx =?
		x2
		3
		5		3	5		3		3	= 125 27			98 .
Solution		x2	dx = x		\|= 5			–	3	= 125 27		=	98 .
		3		3	3	3			3		3		3
		3
	61
EXAMPLE		sin x dx =?
		sin x dx =?
		0

Solution		sin x dx = cos x\| = –cos – (–cos 0) = 1 + 1 = 2.
		0					0
		0

Integrals

Proof

Check Yourself 10

Evaluate the definite integrals.

5			e
a. x3	dx	b. cos x dx	c. 1		dx
1		0	1	x
Answers
a. 156	b. 0	c. 1

3. Properties of the Definite Integral

Let f: [a, b] R and g: [a, b] R be two integrable functions. Then the following properties hold:

	a
1.	f(x) dx =0
	a
a f (x) dx F(a) – F(a) 0
a
	b	a
2. f(x) dx = – f(x) dx
	a	b
b				a
f (x) dx F(b) – F(a) –(F(a) – F(b)) – f( x) dx
a				b
		b		b
3.	For any c R,		c f(x) dx = c	f(x) dx.
		a		a
	b				b	b
c f (x) dx = c F(b) F(a) = c F(b) c F(a)= c F(x)\|= c f (x) dx
	a				a	a
	a					a
	b		b	b
4. [ f(x) g(x)] dx = f(x) dx g(x) dx
	a		a	a
b	b
f (x) dx g(x) dx =[F(b) – F(a)] [G(b) – G(a)]					[F(b) G(b)] – [ F( a) G( a)]
a	a
			= [ F G](b) – [ F G](a)
			b
			= [ f (x) g(x)] dx
			a
				c	b	c
5.	For any a, b, c R with a b c, then f(x) dx =					f(x) dx + f(x).
				a	a	b
c
f (x) dx = F(c) – F(a)= F(c) – F(a)+ F(b) – F(b) = [F(b) – F(a)]+[F( c) – F( b)]
a
	b		c
	= f (x) dx+ f (x) dx
	a		b

44	Algebra 11

						b	b
	6.	If f(x) g(x) for every x [a, b], then					f(x) dx g(x) dx.
						a	a
Proof	If f(x) g(x) then f(x) – g(x) 0.
	Let us write h(x) = f(x) – g(x), so h(x) 0 for every x on [a, b].
			b
	We know that h(x) dx is the area between the graph of h(x) and the x-axis.
			a
				b
	If h(x) 0 then h(x) dx 0.
				a
		b		b
	7.	f (x) dx		f (x) dx
		a		a
			b
Proof	We know that f (x) dx is the area between the graph of f(x) and the x-axis on the interval
			a
	[a, b]. So we have two possible cases:
	Case 1: f(x) 0, for every x [a, b] or f(x) 0, for every x [a, b].
		y					y
								y = f(x)
			a	b	x
					x
				A			A
					y = f(x)		a	b	x
					y = f(x)		a	b

b	b		b
In the first figure, A = – f(x) dx or	f( x) dx= – A and \| – A\|=\| A\|=			f( x)	dx.
In the first figure, A = – f(x) dx or	f( x) dx= – A and \| – A\|=\| A\|=			f( x)	dx.
a	a		a
b	b
In the second figure, A = f (x) dx =		f (x)	dx. This concludes half of the proof.
In the second figure, A = f (x) dx =		f (x)	dx. This concludes half of the proof.
a	a

Case 2: Let c [a, b]. If f(x) 0 for x [a, c] and			y
		b			y = f(x)
f(x) 0 for x [c, b] then f (x) dx = A2 – A1					y = f(x)
f(x) 0 for x [c, b] then f (x) dx = A2 – A1
		a		A2
			a	A2
	b		a		x
and so	f (x) dx = A2 – A1 .		c	b	x
and so	f (x) dx = A2 – A1 .		c	b
	a		A1
	a
	b
However, f (x)		dx =\| A2 \|+\| A1 \| and
	a
A2 – A1	A2 + A1	by the triangle inequality.
b	b
So f (x) dx f (x) dx.
a	a

Integrals

Note

1.All continuous functions have integrals on a closed interval [a, b].

2.A function with a countable number of points of discontinuity has an integral on the

closed interval [a, b]. For example, if the points c1, c2, …, cn [a, b] are the points of

b	c1	c2	b
discontinuity of f(x) then f (x) dx = f (x) dx+			f (x) dx+...+ f(x) dx.
a	a	c1	cn

	62	3
	62	3
EXAMPLE		x dx =?
		a	3
Solution		We know f (x) dx = 0, so x dx = 0.
		a	3

	63	4
EXAMPLE		5x2 sin3 4x dx=?
		5x2 sin3 4x dx=?
		4
Solution		In a similar way to the previous example we can write 4	5x2	sin3 4x dx = 0.
		4

	64 3
EXAMPLE		(2x2	+3x 4) dx =?
		(2x2	+3x 4) dx =?
1
3			+3x 4) dx = 2	3		3		3	3	3		2	3	3
Solution (2x2			+3x 4) dx = 2	x2	dx+3 x dx 4			dx = 2 x		\|+3	x		\| 4 x\|
1				1		1		1	3	1		2	1	1
			= 2	(9	1)+3	( 9	1) 4	(3 1)=	52+12			8 =		64.
					3	2	2		3					3

	65	/ 2
EXAMPLE		(sin x+cos x) dx =?
		(sin x+cos x) dx =?
		0
		/ 2	\|/ 2=( –cos	+sin	)
Solution		(sin x+cos x) dx =( cos x+sin x)	\|/ 2=( –cos	+sin	)	– (– cos 0 + sin 0)
		0	0	2	2

= (–0 + 1) – (–1 + 0) = 2.

46	Algebra 11

	66	1
EXAMPLE		3x+1 dx = ?
		3x+1 dx = ?
		0

Solution 1 We can calculate the integral using substitution: Let u = 3x + 1 so du = 3 dx.

Now we need to calculate the boundaries in terms of u. The lower bound is 3 0 + 1 = 1.

The upper bound is 3 1 + 1 = 4. So we can write

If we calculate a definite integral using the substitution method, we must always remember to calculate the boundaries in terms of the substitution.

1	3x+1 dx =	1	4	1 u32 4			2u32 4		2 432	2 13	2	16 2	=	14	.
	3x+1 dx =	3	u du =	3	3	\|=	9	\|=	9	9	=	9	=	9	.
0		3	1	3	3	1	9	1	9	9		9		9
					2

Solution 2 Alternatively, in this example we can calculate the integral directly:

1		2(3x+1)	3	2 \|= 2 43 / 2		2 13 / 2	= 16 2	14.
3x+1 dx =		2(3x+1)		2 \|= 2 43 / 2		2 13 / 2	= 16 2	14.
				1
0		9		0	9	9	9	9
Check Yourself 11
Evaluate the definite integrals.
2					4				3
a. x5	cos4x dx				b. (x3	+4x2 3x 2) dx		c.	3
a. x5	cos4x dx				b. (x3	+4x2 3x 2) dx		c.	(x2 + x 2) dx
2					1				2
									2
					3
d. (2cos x sin2x) dx					e. 2x+3 dx
/ 2					1
Answers
a. 0	b.	477		c.	25	d. –1	e. 26
		4			6		3

4. Leibniz’s Rule

Rule
	Leibniz’s Rule
					x
	Let f : [a, b] R be a continuous function such that F(x)= f (t) dt. Then
		d	x		a
	F (x)=
				f (t) dt	= f (x).

		dx a

Leibniz’s Rule gives us two important corollaries:

Integrals

EXAMPLE 67

Solution

EXAMPLE 68

Solution

COROLLARY TO LEIBNIZ'S RULE

Let u(x) and v(x) be two differentiable functions. Then

	v(x)
1.	F(x)=	f(t) dt
1.	F(x)=	f(t) dt	F (x)= f v(x) v (x)
	a
	v(x)
2.	F(x)=	f (t) dt
	F(x)=	f (t) dt	F (x)= f v(x) v (x) f u(x) u (x).

u(x)

	x
F(x)= cos t2			dt	is given. Find F (						).
F(x)= cos t2			dt	is given. Find F (						).
	1							2
										v(x)
We know from corollary 1 to Leibniz’s Rule that F(x)= f(t) dt											F (x) = f(v(x)) v (x).
										a
	x
F(x)= cos t2			dt and f(t) = cos t2, then F (x) = cos x2 (x) = cos x2.
	1
So F (		)= cos(			)2 = cos		=	2	.
So F (	2	)= cos(		2	)2 = cos	4	=	2	.
	2			2		4		2

2
F(x)= x (t2 4t+1) dt is given. Find F (2).
x
We know from corollary 2 to Leibniz’s Rule that,
v(x)
F(x)= f (t) dt	F (x)= F(v(x)) v (x) – f(u(x)) u ( x).

u(x)

So if we have F(x)= (t2 4t+1) dt then

F (x) = f(x2) (x2) – f(x) (x) where f(t) = t2 – 4t + 1, F (x) = (x4 – 4x2 +1 ) (2x) – (x2 – 4x + 1 ) 1,

F (2) = (24 – 4 22 + 1) (2 2) – ( 22 – 4 2 + 1) = 7.

48	Algebra 11

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