Integration and Area Foundation- Definite Integrals

Integration and Area

Foundation- Definite Integrals

Today’s Goals–To understand why Integration can find the area under a Curve

–To Introduce definite Integrals

Recap

We have already Introduced Integration as the inverse of Differentiation

We know that we add a constant for INDEFINITEIntegrals and that

Cn

xdxx

nn

1.

1

Remember from differentiation


Remember from differentiation

dx

dy

x

yofLimit

0x

What is integration


Integration is summing things up

It means to piece things together , to add things up !!

Consider y=x2 If we wanted to find the area under this Curve between x=0 and x=1 we could use strips like this :

8 Strips dx=1/8

10 Strips dx=1/10

30 Strips dx=1/30

50 Strips dx=1/50

Summing these strips gives an underestimate of the area that decreases as the number of strips increases


Summing these strips gives an overestimate of the area that decreases as the number of strips increases


Finding the Area

With increasing strips, ∆x ∆y tend to Zero and

we can write and

And So

If write the area of each of the individual rectangle as ∆A

Then the area of each individual rectangle is between

ΔxΔyyΔAxyΔ

Δyyx

ΔAy

0xdx

dA

x

ΔALim

y

dx

dA dxyA

∆y

y

∆x

δy

y

Δx


ba

The area between a and b under the curve is

x

y

xyArea ~

As the number of rectangles increases the approximation to the area improves

0xn

Limit xy0xArea =

This Limit is written as

b

aydx

The Area under the curve from x=b to x=a , This is the Definite Integral and is written as

The following slides explore this

a) The area of shaded rectangle 7 is y6.Δx ; (8 Strips here)

The approximation of the total required Area is

∆x

b) The area of shaded rectangle 7 is (y6+dy6).Δx and of the total required area is

7

0iy x

7

0

xyy ii < Area <

As n increases the approximation of the area gets better

Underestimate Overestimate

y0 y1

y2 y3

8

1

xyi

7

0iy x

7

0

xyy ii

Δx


Δx y7

0i < Area <

8

1

xyi

7

0

8

1

Axyxy ii

where ∆A is difference in areas

7

0

8

1 x

Ayy ii

1

01

n

i

n

i x

A

xyy

let n →∞ and so ∆x, ∆y, ∆A → 0

Limit

Taking an infinite amount of rectangles

1

01

n

i

n

i dx

dAyy

dx

dA

x

A

Limit

∆X→0

This RHS is the definition of the 1st derivative

And So And so this leaves

dx

dAyyn 0

Change in AreaWith respect to (w.r.t)Change in x


Using our knowledge of taking antiderivatives

dxdx

dAdxyy 0n

1

01

n

i

n

i dx

dAyy

dx

dAyyyyyy nn 11021 ..............

dx

dA

x

A

RHS becauseLimit ∆X→0

n

00n dxydxyyA

This is the value of x at the far side of the curve

This is the value of x at the near side curve

In our specific example we have

1x

0x

2dxx1

0

3

3

x

=

1

0

33

3

0

3

1

=

3

1

x

y=x2

10

y


Splitting Areas for Integration

Where a curve is below the x-axis the integral is negative

Therefore if the curve crosses the x axis we need to split the integration into seperate parts

dxxfdxxfArea

0

1

2

0)()(

where |x| (the ‘modulus of x’) means the positive value of x

ExampleFind the area enclosed by the x axis and the curve

dxxxxdxxxxArea

0

1

2

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

)1)(2( xxxy

y = 0 when x = 0x = 2 and x = -1

The curve is below the axis for 0 < x < 2

and above the axis for -1 < x < 0

Example (needs checking!!)Find the area enclosed by the x axis and )1)(2( xxxy

dxxxxdxxxx 2)1)(2( 22

0

32

0

3

804

3

84

34

2

0

234

xxx

dxxxxdxxxx 2)1)(2( 20

1

30

1

)112

7()1

3

1

4

1(0

34

0

1

234

xxx

unitssqareaTotal .12

13

12

5

3

22

3

22

12

5

Area between 2 curves f(x) and g(x)


dxxgxfAreab

a ))()((

provided f(x) and g(x) don’t cross between a and b


dxxgxfAreab

a ))()((

Example

First find the points of intersection of curvey=x2 and line y=x+2

At the points of intersection

22 xx

022 xx

0)1)(2( xx

12 xorxThese will be our limits of integration


dxxgxfAreab

a ))()((

dxxxA 2

1

2)2(

2

1

32

32

2

x

xx

3

222

2

2 32

A

3

)1()1(2

2

)1( 32

3

842

3

12

2

12

143

2

18

Area = 4½ square units

Example

Area between the curve and the y axis

To find this we rearrange the integral as

dyxAready

cy

yxA x

δy

c

d

Area between the curve and the y axis

dyy 23

1

2

Rearrange as x=y2+2 dyyfArea

y

y

3

1)(

3

1

3

23

y

y

2

3

16

3

33

2

3

169

3

212

Area = 12 2/3 square units

Splitting Areas for Integration

We need to split integrals like this into 2 partsBecause the Integral under the X-axes gives a negative result

dx)x(fdx)x(fArea0x

1x

2x

0x

Finding Integrals between the Y axes and f(x) we rearrange the Integral as follows

dy)y(fArea3y

1y

Rearrange as x=y2+2

Documents

Integration and Area Foundation- Definite Integrals