IntegrationArea under Curve

Stained-glass windows

How much glass is needed?

How much glass is needed?

History

Calculus was historically developed to find a general method for determining the area of geometrical figures.

When these figures are bounded by curves, their areas cannot be determined by elementary geometry.

Integration can be applied to find such areas accurately.

Trapezoidal Rule

Also known as Trapeziod/Trapezium Rule

An approximating technique for calculating area under a curve

Works by approximating the area as a trapezium

How much glass is needed?

1Area of a trapezium height sum of parallel sides

2

2

1 1Shaded Area (1)(4 3.8) (1)(3.8 3)

2 21 1

(1)(3 2) (1)(2 0)2 2

10.8 units

Actual Area = 10.67 units2.

Trapezoidal Rule

1Area of a trapezium height sum of parallel sides

2

1Shaded Area ( ) ( )

2b a f a f b

(2, 4)

(1, 1)2y x

2

1 1Shaded Area (1)(0 1) (1)(1 4)

2 2

3 units

From diagram, clearly, it is an overestimate.

Actual Area = 2.67 units2.

Using Integration to find exact area

1

0

12

0

2

Shaded Area

2

0.5

x dx

x

unit

Using Integration to find exact area

4

0

42

0

42

0

2

Shaded Area 2

2

2

16

x dx

x

x

unit

Why?

Using rectangles to approximate the area under the line

- Show Geogebra

Using rectangles to approximate the area under the line

Dividing the area under the line into 4 strips,

We will start to approximate the area by finding the area of the rectanglesWidth of each rectangle = 0.251 1 31 1

44

1

444 2Area

1 1,

4 4

1 1,

2 2

3 3,

4 4

1

41

4

1

4

3 1

8 2actual area underestimate

1 1 3

4 2

1

44

3

8

What about n strips?

width of each rectangle =1

n

1

n

1

n

1

n

1

n

0

Find the height of each rectangle

Write down the statement for the area of each rectangle and sum them up

1

n

2

n

3

n

4

n

5......n

n

n

1......

n

n

What about n strips?

Dividing the area under the line into n strips,

1 1 11 2 3 1..

1.

n n n

nArea

n n nn n

width of each rectangle =1

n

1

n

1

n

1

n

1

n

01

n

2

n

3

n

4

n

5......n

n

n

1......

n

n

2

221 1

2 2

n n

n

n

5 5,n n

21 2 3 ...

11n

n

2

1 ( 1)

2

n n

n

What about n strips?1

2

1

2Are

na

1

n

1

n

1

n

1

n

01

n

2

n

3

n

4

n

5......n

n

n

1......

n

n

As we increase the no. of rectangles, the white triangles will be filled up by the rectangles and we will get a better approximation of the area.

,1

20As n

n

1

2Area

What about a curve?Similarly, we divide the area under the curve into n strips.

width of each rectangle =1

n

1

n

1

n

1

n

1

n

1

n

2

n

3

n

4

n

5......n

n

n

1......

n

n

Find the height of each rectangle

Write down the statement for the area of each rectangle and sum them up

What about a curve?Similarly, we divide the area under the curve into n strips.

width of each rectangle =1

n

1

n

1

n

1

n

1

n

1

n

2

n

3

n

4

n

5......n

n

n

1......

n

n

2

2,

55

n n

2 2 2 21 11 2 3 11

.1

. .n

Arn n

ean n nn nn

2 2 23

21 2 3 ... ( 11

)n

n

3

1 ( 1)(2 1)

6

n n n

n

What about a curve?

1

n

1

n

1

n

1

n

1

n

2

n

3

n

4

n

5......n

n

n

1......

n

n

2

2,

55

n n

3

1 ( 1)(2 1)

6

n n nArea

n

2

1 1, 0

2 6As n

n n

1

3Area

2

2

2 3 1

6

n n

n

2

1 1 1

3 2 6n n

Area by Integration

12

0

x dx

1

3

13

03

x

Example

Find the area under the curve

y = (x – 3)2

y = (x – 3)2

between x = 3 and x = 6

Find the area under the curve between x = 3 and x = 6

y = (x – 3)2

y = (x – 3)2

y = (x – 3)2

y = (x – 3)2

2( 3)y x

62

3

( 3)x dx63

3

2

( 3)

3

9

x

units

2 2 2 2 2 ( 1)1 2 3 4 ... ( 1)

2

n nn

2 2

2 2

( 1) 2 1

( 1) 2 1

n n n

n n n

2 2

2 2

2 2

2 2

2 2

2 1 2 1 1

3 2 2 2 1

4 3 2 3 1

5 4 2 4 1

( 1) 2 1k k k

1

2

3

4

n

n

n

n

n k

Adding both sides,

2 2

2 2

( 1) 1 2(1 2 3 ... )

( 1) 11 2 3 ...

2

k k k

k kk

2 2

2

( 1) 11 2 3 ...

2

( 1) ( 1)

2( 1)( 1 1)

2( 1)

2

k kk

k k

k k

k k

2 2 2 2 2 ( 1)1 2 3 4 ... ( 1)

2

n nn

3 3 2

3 3 2

( 1) 3 3 1

( 1) 3 3 1

n n n n

n n n n

3 3 2

3 3 2

3 3 2

3 3 2

3 3 2

2 1 3(1) 3(1) 1

3 2 3(2) 3(2) 1

4 3 3(3) 3(3) 1

5 4 3(4) 3(4) 1

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

1

2

3

4

n

n

n

n

n k

2 2 2 2 ( 1)(2 1)1 2 3 ...

6

k k kk

Adding both sides,3 3 2 2 2 2( 1) 1 3(1 2 3 ... ) 3(1 2 3 ... )k k k k

2 2 2 2 3 3

32 2 2 2

32 2 2 2

2

2

2

3(1 2 3 ... ) ( 1) 1

2( 1) 2 2 3 ( 1)3(1 2 3 ... )

2

2( 1) 2( 1) 3 ( 1)(1 2 3 ... )

6

( 1)2( 1) 2 3

6

( 1) 2 4 2 2 3

6

( 1)(2 )

6( 1)(2

3 ( 1

1)

6

)

2k k k

k k k kk

k k k kk

k k k

k k k k

k k k

k k k

k k

( 1)

2

k k