(Sections 3.9, 4.1 - WordPress.com · 2020. 4. 7. · (f) f(x)=x5 sec(x)tan(x)+ 1 2 p x (g) f(x)= 2 x2 +sec(x)tan(x) 2. Solve the following initial value problems. (a) y0 =sinx, y(0)

MTH132 - SS20 - Section 24

Integrals - Work Book

(Sections 3.9, 4.1 - 4.3)

Name:

1

3.9 Antiderivatives

• Finding the antiderivative of a function f(x) on the interval I means to find a

function F (x) such that when you di↵erentiate F (x) you must get f(x) for all x in

I.i.e Find F (x) such that F 0

(x) = f(x) for all x in I.

• If F is an antiderivative of f , then the most general antiderivative of f is F (x)+cfor some constant c.

• Use derivative rules we have learned to find the antiderivatives.

Exercises:

1. Find the most general antiderivative of the given functions.

(a) f(x) = x5

(b) f(x) = 2x2 � 3x4 � 5

(c) f(x) = 4 cos (x) + 8

(d) f(x) = 2 sec2(x)� x5

(e) f(x) = x+ 2 sin (x)

2

Since ( x6)'

= 6×5

F Cx ) = txt C

F Cx ) = zzx3 - 3¥15 X + c

F Cx ) = 4 Sin Xt 8 X t C

FCXS = 2 tan x - Ift c

F CX ) = IzXZ - 2 Cos X t c

(f) f(x) = x5 � sec (x) tan (x) +1

2px

(g) f(x) =2

x2+ sec (x) tan (x)

2. Solve the following initial value problems.

(a) y0 = sin x, y(0) = 2

(b)dy

dx= �6 sin x, y(0) = 4

(c)dy

dx=

1

x2, y(1) = 0

3

g II't

- It I

FCX ) = II

- Sec x t II + c

42

F Cx ) = IXG

- Sec X t I k+ c

T 2×-2

F ex ) -

- 21ft't secxtc =

-2¥ + see Xtc

Ycxs= - Cos x t C

X=0, Y ( O ) = 2 =

- Cos Otc ⇒ 2 = - I t C ⇒ C = 3

-

y CX) = - Cosx -13

YCX ) = 6 Cos Xtc

YCO ) = 4 = 6 cos Otc ⇒ 4=6 t C ⇒ c = - 2

b CX ) = 6 Cos x. 2

- 2

←X

- Iycxs-

- ¥ t c =

-

Ftc

ya ) = o =-

lyte → O = - Itc ⇒ ⇐ I

ycx ) =-

¥+1

3. The acceleration of an object moving along the x-axis is a(t) = 3 sin (t). Find its velocity

and position functions, v(t) and s(t) if v(0) = 1 and s(0) = 3.

4.1 Area and Distance

Estimating area between a given function f(x) and the x-axis onthe interval [a, b] using rectangles:

• First, identify the number of rectangles that will be used to estimate the area. Call

it n.

• If the rectangles are of the same width, then the width of a rectangle is given by,

�x =b� a

n.

• Now, divide the interval [a, b] into n number of intervals with a length of �x. Let’sdenote the ith interval by [xi, xi+1].

• The height of each rectangle, i, will be determined by the function value at a ’suitable’

x value, denoted by f(x⇤).

• Then, the area will be given by the sum of the areas of the rectangles.

• How to choose the x⇤value:

– To get a right-hand sum, let x⇤be the right end point of each interval.

– To get a left-hand sum, let x⇤be the left end point of each interval.

– To get an upper sum (over-estimate), let x⇤be the largest value of each

interval.

– To get a lower sum (under-estimate), let x⇤be the smallest value of each

interval.

4

VIT) -

- act ) ⇒ Vct ) = - 3 Cost t C,

f- O ⇒ VCO ) = I =-3 Cos of C,

⇒ I =-3 t C,

⇒ 4=4

Vct ) =- 3 Cost t 4

SET , = Vct ) ⇒ sets =-3 Sint t att Cz

t -

- O ⇒ SCO ) =3 = -3 Sin of 4 Costs ⇒ 3 = Cz

SCH = - 3 Sint t 4 t t 3

Exercises:

1. Using three equally-spaced rectangles of equal width, find the upper sum approximation

of the area between the curve y =1

xand the x-axis from x = 2 to x = 8.

2. Estimate the net area under the curve y = x(x � 2), between x = 0 and x = 4, using 4

rectangles with equal width, with heights of the rectangles determined by the height of the

curve at left endpoints and right endpoints.

5

Ab

A- 2. YC 2) t 2. yet ) -12 . y( 6) iii.= 2 [ It -4 -1¥ ] I I I

18×24 68=2[ Etf -1¥ ]

Ax=b=8=I⇒=2[ ¥ ]

A = I6

Axe 4-41=14--1For left endpoints : X o I 2 3 4

y O - I 0 3 8

AT 1. ycojt 1. y (1) t to y (2) t 1.913 )

= I [0+43+0-13]= 2

For right end Points :

Art I [ DCD -19123T yes ) -1943 ]

= I [-1+01-3+8]= 10

3. Estimate the area under the graph of f(x) = �x2+ 8x+ 9 from x = 0 to x = 8 using the

areas of 4 rectangles of equal width.

(a) Estimate the area using left endpoints.

(b) Estimate the area using an upper sum (over-estimate).

6

- ( x - 45+25I

yA

(a) Are 2 [ fast fast to 4) tf " ] !

!

!

,

!!

I:{II,

" +25+213 [¥. ×

=152

(b) A = 2 [ f (2) tf (4) tf (4) t fig , ) 1*84--0=2= 2 [ 2 It 251-25+21 ]

= 2 [ 92 ]= 184

4.2 Definite Integrals

• Sigma Notation:

*

nPi=1

ai = a1 + a2 + ...+ an

*

nPi=1

1 = n

*

nPi=1

i =n(n+ 1)

2

*

nPi=1

i2 =n(n+ 1)(2n+ 1)

6

• Definite Integral

* To get a better estimate for the area, we use infinitely many rectangles to

approximate the area.

* The definite integral,

Z b

a

f(x) dx, gives the net area between the curve f

and the x-axis.

* If f is continuous on [a, b], thenZ b

a

f(x) dx = limn!1

nX

i=1

f(xi)�x, where �x =b� a

nand xi = a+ i�x.

* If the function is below the x-axis, then the area is counted negatively.

• Properties of definite integrals:

(i)

Z b

a

[f(x) + g(x)] dx =

Z b

a

f(x) dx+

Z b

a

g(x) dx

(ii)

Z b

a

kf(x) dx = k

Z b

a

f(x) dx

(iii)

Z b

a

k dx = k(b� a)

(iv)

Z b

a

f(x) dx = �Z a

b

f(x) dx

(v)

Z c

a

f(x) dx =

Z b

a

f(x) dx+

Z c

b

f(x) dx

Exercises:

1. Compute the following.

(a)

3Pi=1

2i

7

= 21+22+23 = 2+4+8 = 14

(b)

30Pi=1

(3 + 2i)

(c) limn!1

nX

i=1

2i

n2

2. Suppose

Z 2

�1

f(x) dx = 6 and

Z 3

�1

f(x) dx = 8. Find

Z 2

3

5f(x) dx.

3. Evaluate the following integrals. (Hint: Geometric shape of the graph could be useful)

(a)

Z 3

0

|x� 1| dx

(b)

Z 5

0

|x� 2| dx

8

¥2,1 = h, Erie hCh

= 773 t.IE,

zi =3 :{ it 27€,

i

=3 ( 30 ) +2 . 30£31 )

2

= Got 930= 990

-

- FI . E Eh.

i = fig . E- hc

= thing,

h4÷= fin, Kz

= thing, I°

In

= I

3I I I

§ fax , ax t § FIX )d× = f CX ) DX⇒ ffcxgdx = 8 - 6 - 2

--

-

6

②8 O

2

3%5 fix )dx= 5 fgfcxsdx = -5 §fc×, ax =- 5 (2) = - IO

O O

z- . -

-0I• I

I { f } 2

= I (1) G) tz (2) (2) o IJ

= Iz t 2

= IZ

,

{ its= Iz ( 2) (2) tz (3) (3) 0 2 5

= 2+92=21

(c)

Z 0

�5

|x+ 3| dx

(d)

Z 3

0

p9� x2 dx

(e)

Z 4

�4

p16� x2 dx

4. Express the following limit of the sums in the form of a definite integral.

limn!1

nX

i=1

3

r2 +

4i

n.4

n

5. The graph of a function f(x) is given below. What is the value of

Z 3

0

f(x) dx?

9

4! Is= I C 2) (2) t Iz (3) Cz )

- s -3 °

= Zt Iz = 132y

①b

y=

^

92=9 - XZ Ex

= ¥ test, gyp,

* 5=-9-3

0 3

ab

5- IF

42=16×2 Ex-

40

4

= I IT (4)2 X2ty2= 16

= 16-2-11= SIT

( from 9=2 to b - 6)6

I = f FF dx

2

XiAx=b-nI =

- - Ixi-ati.snE In h

3

eyugetcancelled ftcx )dx="Green area

"

O

= I (2)= 2

4.3 Fundamental Theorem of Calculus

• Theorem 1: Let f be a continuous function on [a, b]. If F (x) =

Z x

a

f(t) dt for

a x b, then F 0(x) = f(x).

• If F (x) =

Z b(x)

a(x)

f(t) dt, then F 0(x) = f(b(x))b0(x)� f(a(x))a0(x).

• Theorem 2: Let f be a continuous function on [a, b] and F be any antiderivative

of f . Then

Z b

a

f(x) dx = F (b)� F (a).

Exercises:

1. Find the derivative of the functions given below.

(a) F (x) =

Z 1

cosx

p5� t2 dt

(b) F (x) =

Z 4

x3

1

t2 + 1dt

(c) F (x) =

Z 2x2

1

sin t

1 + t2dt

(d) F (x) =

Z x2

x

1

t2 + 1dt

10

Cos X

= - f IF dt

I

E' ex , = - 4¥55. Ceosx )

'

=- I × . sin x

3x

=- f at

4

F' cxs -

-- ¥4 ,

.

'

= -×¥ .

3×2

F' as =

s.in#I3.ex5--sinfIYf.4x=*TaiEkx4ii--

1¥-

¥

2. Evaluate the following integrals.

(a)

Z 2

0

(3x2+ 5x� 3) dx

(b)

Z 2

1

3� 2x2

x4dx

(c)

Z 4

1

t3 +pt

t2dt

(d)

Z 9

1

x� 1px

dx

11

= §. X

's

+5¥ - 3×12

= tea 5-3×102

°

= ( 23T ( 212-3 Cz ) ] - [ 03+2.02 - 3. o )2

=8 t Ia - 4 - 6 = 8 t to -6 I 12

= § -

2×x⇒dx=

§(3.

II 2 .

52) DX

=

a I ? =-

Ist It != t I ] - fifty ] = fifty - I .

-

Is - I =

-

is

= §¥zt t) at =

§t + t

'

%) at

= It"

I ! it -¥1,4= -

⇐ ] - CI -

E) = C ' I-

I ] - [ ¥ D= s - I - C - Z )

= 7. LZ =

= §¥yz

- xtyzdx = § x"

-

Ik dx

= x÷'

- XIII ! = zx% . ax"

1,9=3 xrx - arx 1,9

= [23959-259] - [ I if - 2K ) = 23.27- 6 -

23+2= 18--6 - Zz +2 = 14-23

= 403

3. Determine the value(s) of a such that

Z a+1

a

(2x+ 3) dx = 10

12

Atl

12×-13) dx = to

a

-

att

2122+3×1= co

a

XZ -13×19+1 = 10

a

[Latif -13cal - is ] - [ a' +3 a ] = to

¥2at I -1/39+3-¥34 = to

2 a +4.

- 10

2 a = 6

a =3