Mathematics

Session

Definite Integrals - 3

Session Objectives

Definite Integral as the Limit of a Sum

Areas of Bounded Regions

Class Exercise

Definite Integral as the Limit of a Sum

b

h 0a

ƒ x dx = lim h ƒ a + ƒ a+h + ƒ a+2h +…+ ƒ a+ n- 1 h

b- awhere h=

n

b

h 0a

ƒ x dx = lim h ƒ a+h + ƒ a+2h +…+ ƒ a+nh

b- awhere h=

n

OR

Example - 1

2

0

Evaluate: x+3 dx as the limit of a sum.

b

h 0a

Solution: We have


b- a

where h=n

2- 0 2here a=0, b=2, f x = x+3 and h= =

n n

2

h 00

x+3 dx = limh f 0 +f 0+h +f 0+2h + ... +f 0+ n- 1 h

h 0

= limh f 0 +f h +f 2h + ... +f n- 1 h

Solution Cont.

h 0

= limh 3n+h 1+2+3+...+ n- 1

h 0

n n- 1= limh 3n+h×

2

n

n n- 12 2= lim 3n+ ×

n n 2

h 0

= limh 0+3 + h+3 + 2h+3 +... + n- 1 h+3

n

1= lim 6+2 1-

n

=6+2 1- 0 =8

Example - 2

3

1

Evaluate: 2x+3 dx as the limit of a sum.

b

h 0a

Solution: We have


b- awhere h=

n

3- 1 2here a=1, b=3, f x =2x+3 and h= =

n n

3

h 01

2x+3 dx = limh f 1 +f 1+h +f 1+2h +... +f 1+ n- 1 h

Solution Cont.

h 0

= limh 2 1 +3 + 1+h +3 + 1+2h +3 +... + n- 1 h +32 2 2 1

h 0

= limh 5n+2h 1+2+3+...+ n- 1

h 0

n n- 1= limh 5n+2h×

2

n

n n- 12 2= lim 5n+2 ×

n n 2

n

1= lim 10+4× 1-

n

=10+4× 1- 0 =14

Example - 3

2

2

1

Evaluate: x - 1 dx as the limit of a sum.

b

h 0a

Solution: We have


b- awhere h=

n

2 2- 1 1here a=1, b=2, f x = x - 1 and h= =

n n

2

2

h 01

x - 1 dx = lim h f 1 +f 1+h +f 1+2h +... + f 1+ n- 1 h

Solution Cont.

22 22

h 0= limh 1 - 1 + 1+h - 1 + 1+2h - 1 +... + 1+ n- 1 h - 1

22 2 2 2

h 0= limh 0+2h 1+2+3+...+ n- 1 +h 1 +2 +3 +... n- 1

2

h 0

n n- 1 n n- 1 2n- 1= limh 2h× +h ×

2 6

2n

n n- 1 n n- 1 2n- 11 1 1 1= lim 2× × + × h= , h 0 n

n n 2 6 nn

n

1 11- 2-

1 n n= lim 1- +

n 6

1- 0 2- 0 1 4= 1- 0 + =1+ =

6 3 3

Example - 4

3

2

1

Evaluate: 2x +3x+5 dx as the limit of asum.

b

h 0a

Solution: We have

ƒ x dx = lim h ƒ a +ƒ a+h +ƒ a+2h +…+ ƒ a+ n- 1 h

b- a

where h=n

2 3- 1 2here a=1, b=3, f x =2x +3x+5 and h= =

n n

3

2

h 01

2x +3x+5 dx = limh f 1 +f 1+h +f 1+2h +... +f 1+ n- 1 h

Solution Cont.

2 22

2h 0

2.1 +3.1+5 + 2 1+h +3 1+h +5 + 2 1+2h +3 1+2h +5 += lim h

....+ 2 1+ n- 1 h +3 1+ n- 1 h +5

22 2 2 2

h 0= limh 10n+7h 1+2+...+ n- 1 +2h 1 +2 +3 +....+ n- 1

2n

2 7 2 1 4= lim 10n+ × n- 1 n+ × ×n n- 1 2n- 1

n 2 n 3 n

n

1 8 1 1= lim 20+14× 1- + × 1- 2-

n 3 n n

8=20+14× 1- 0 + × 1- 0 2- 0

3

16 118=20+14+ =

3 3

Example - 51

2x

-1

Evaluate: e dx as the limit of a sum.

b

h 0a

Solution: We have


b- a

whereh=n

2x 1- -1 2Here a=-1, b=1, f x =e and h= =

n n

1

2x

h 0-1

e dx = lim f -1 +f -1+h +f -1+2h +...+f -1+ n- 1 hh

Solution Cont.

-2+2 n-1 h-2 -2+2h -2+4h

h 0= lim e +e +e +...+eh

n2h-2

2hh 0

e - 1= lim e

e - 1h

2nh-2

2hh 0

e - 1=e lim

e - 12h

h

2h

4-2 4 -2 2 -2

2h

h 0

e - 1e e - 1 e e - e= = × =

2 2 1 2e - 1lim

2h

4

-22hh 0

1 e - 1=e lim nh=2

2 e - 12h

Areas of Bounded Regions

1. Let f(x) be a continuous function defined on the interval [a, b]. Then, the area bounded by the curve y = f(x), x-axis and the ordinates x = a, x = b is

b

a

= ydx

2. The area bounded by the curve x = f(y), y-axis and the abscissae y = c, y = d is

d

c

= xdy

Areas of Bounded Regions Cont.

1 2

b

1 2

a

3. The area bounded by two intersecting curves y = f x , y =g x

and the ordinates x = a, x = b is

= y - y dx

Example - 6

2Find the area the region bounded by the curve x = 4y and

the straight line x = 4y - 2.

Solution: The given curves are

2

2 xx = 4y y = ... i

4

x +2and x = 4y - 2 y = ... ii

4

22x x +2

= x - x - 2 = 04 4

x 1 x 2 0 x 1 or x 2

1The points of intersection are -1, and 2,1 .

4

x' x

y

y'

x = 4y – 2

x = 4y2

(2, 1)

O

–1, 1 4

Solution Cont.

The required area = Area of the shaded region

2

1 2

-1

= y - y dx

2 2

-1

x+2 x= - dx

4 4

22 3

-1

x x x= + -

8 2 12

4 2 8 1 1 1 9sq. units

8 2 12 8 2 12 8

Example - 7

2 2

Find the area of the smaller region bounded by

x y x ythe ellipse + =1 and the straight line + =1.

16 9 4 3

Solution : We have

2 2 2x y x

+ = 1 y 3 1 ... i16 9 16

x y x+ =1 y = 3 1- ... ii

4 3 4

(i) and (ii) intersect at (4, 0) and (0, 3).

Solution Cont.

x y= 1

4 3+

2 2x y1

16 9+ =

X’ X

Y’

Y

O

(0, 3)

(4, 0)


4

1 2

0

= y - y dx4 42

0 0

x x=3 1- dx - 3 1- dx

16 4

Solution Cont.

4 4

2 2

0 0

3 3= 4 - x dx - 4- x dx

4 4

42

2 2 -1

0

4- x3 x 16 x= 4 - x + sin +

4 2 2 4 2

22 2 -1 2 2 -14- 4 4- 03 4 16 4 0 16 0= 4 - 4 + sin + - 4 - 0 + sin +

4 2 2 4 2 2 2 4 2

-1 -13= 0+8sin 1+0 - 0+8sin 0+8

4

3= 8× - 8 = 3 - 2 sq. units.

4 2

Example - 8

2 2Find the area of the region x, y : x + y 1 x + y .


2 2 2x + y =1 y = 1- x ... i

x + y =1 y =1- x ... ii

(i) and (ii) intersect at (1, 0) and (0, 1).

X’ X

Y’

Y

O

(0, 1)

(1, 0)

x+y =1x+y =1

2 2


1

1 2

0

= y - y dx

Solution Cont.

1

2

0

= 1- x - 1- x dx

122 -1

0

x 1 x= 1- x + sin x - x+

2 2 2

2 22 -1 2 -11 1 1 1 1 0

= 1- 1 + sin 1- 1+ - 1- 0 + sin 0- 0+2 2 2 2 2 2

1 1 1= 0+ × - 1+ - 0+0- 0+0 = - sq. units.

2 2 2 4 2

Example - 9

2 2

Find the area included between the parabolas

y = 4ax and x = 4ay.

Solution : The given parabolas are

2y = 4ax y = 4ax ... i

2

2 xx = 4ay y = ... ii

4a

x = 0 y = 0 and x = 4a y = 4a

224 3x

= 4ax x = 64a x4a

3 3x x - 64a = 0 x = 0 or x = 4a

The points of intersection are (0, 0) and (4a, 4a).

X’ X

Y

Y’

y = 4ax2

x = 4ay2

(4a,4a)

O (0,0)

Solution Cont.

4a

1 2

0

= y - y dx

The required area = Area of shaded region

4a 2

0

x= 4ax - dx

4a

4a332

0

x 1 x= 2 a× - ×

3 4a 32

3

3 2 2 22

4 1 32 16 16= a 4a - 4a - 0+0 = a - a = a sq. units.

3 12a 3 3 3

Example - 10

2 2

2

Sketch the region common to the curves x + y =16

and x = 6y. Also, find the areaboundedby the twocurves.


2 2 2x + y =16 y = 16 - x ... i

2

2 xand x = 6y y = ... ii

6

2Substituting x = 6y in i , we get

2 2y +6y =16 y +6y - 16 = 0

y - 2 y +8 = 0 y = 2 or y = -8

Solution Cont.

2y = 2 x =12 x = ±2 3

The points of intersection are 2 3, 2 and -2 3, 2 .

X’ X

Y’

Y

O

(0, 4)

x+y =162 2

x 2 = 6y

(2 3, 2)( 2 3, 2)-

Solution Cont.

2 3

1 2

0

=2 y - y dx

The required area = Area of shaded region

2 3 22

0

x=2 16- x - dx

6

2 3 22 2

0

x=2 4 - x - dx

6

2 332 2 -1

0

x 16 x x=2 4 - x + sin -

2 2 4 18

Solution Cont.

32 -1 -12 32 3 16 2 3

=2 16- 2 3 + sin - - 0+8sin 0- 02 2 4 18

8 4 3 4 3 16=2 2 3 + - = + sq. units.

3 3 3 3

Example - 11

22 2 2

Find the area of the region enclosed between

the two circles x + y = 1 and x - 1 + y = 1.

Solution : The euation of the given circles are

1 3 1 3i and ii intersect at , and , - .

2 2 2 2

2 2 2x + y = 1 y = 1 - x ... i

2 22and x - 1 + y = 1 y = 1 - x - 1 ... ii

Solution Cont.

x' x

y

y'

O (1, 0)

x2 + y2 = 1

(x - 1)2 + y2 = 11 3,

2 2

1 3,

2 2

1,0

2

Solution Cont.

Required area = area of the shaded region

112

2 2

102

= 2 1- x - 1 dx + 1- x dx

1

12 2-1 2 -1

10

2

1 1 x 1= 2 x - 1 1- x - 1 + sin x - 1 + 1- x + sin x

2 2 2 2

2

2-1 -1

2-1 -1

1 1 1 1 1 1 1= 2 - 1 1- - 1 + sin - 1 - 0 - 1 1- 0 - 1 - sin 0 - 1

2 2 2 2 2 2 2

1 1 1 1 1 1 + 1- 1 + sin 1- 1- - sin

2 2 4 2 2 2

Solution Cont.

-1 -1 -1 -13 1 3 1= - + sin - - sin -1 + sin 1 - - sin

4 2 4 2

3 3 2 3= - - + + - - = - sq.units.

4 6 2 2 4 6 3 2

Thank you

Mathematics

Documents

Transcript of Mathematics