integration: part 3 application: area under and between curve(s) volume generated

INTEGRATION: PART 3Application: Area under and between Curve(s)

Volume Generated

Integration: Application

AreaVolume Work Pressure

Definite Integration: Refresh

2

ln

1)(

2tan)(

)42()(

)3)(sin()(

6

4

1

4

13

1

8

5

e

e

dxxx

d

dc

dxxxb

dxxxa

Use your calculator to double check your answers

Application : Area Under a Curve

x=bx=a

y=

b

a

dxxfA )(

Shaded area that is bordered by y=f(x), x=a, x=b and x-axis is

Area above x-axis is +ve

Area below x-axis is –ve and would need to |-ve|

A

Application: Area Under a Curve

b

c

c

a

dxxfdxxfA )()(

Application: Area Under a Curve

2unitsDCBAA

Application: Area Under a Curve - Steps

Step-by-Step1. Identify the function.2. Sketch the graph to visualise (if needed)3. Visualise and shade the area in question4. Identify the border(s) for the area5. Perform definite Integration, accordingly.6. If –ve prediction, absolute the value using

the | | sign7. Add together the area (s) (if needed)8. Note in unit2 (An are MUST be +ve)

Area Under a Curve: Example

dxxxxd

dxxxxc

dxxb

dxxa

)6()(

)34()(

)5()(

6)(

0

3

23

3

0

23

2

2

2

2

6

Application: Area Between Curves If f and g are continuous with f(x) =>

g(x) throughout [a , b], then the area of the region between the curves y = f(x) and y = g(x) from a to b is the integral of [f – g] from a to b.

b

a

b

a

dxEBOTTOMCURVTOPCURVEA

dxxgxfA

][

)]()([

Area Between Curves

Area Between Curves - Steps

Step 1: Sketch the curves and note the intersecting points

Step 2: Find the limits of integration by finding the intersecting points (y = y).

Step 3: Write a formula for f(x) – g(x) (depending on the which is the top curve and bottom curve). Simplify it.

Step 4: Integrate f(x) – g(x) of Step3 from a to b. The value obtained is the area (units2).

Intersecting curves example

2

-1 21

1

-2

-1

-2

3

y = 2 - x2

y = - x

y

x

Areas between curvesThe region runs from x = -1 to x = 2.

The limits of integration are a = -1, b =2.

The area between the curves is b

adxxxdxxgxfA

2

1

2 )()2()()(

2

1

32

2

1

2

322

)2(

xx

x

dxxx

2

2

9

3

1

2

12

3

8

2

44 units

Area Between Curves

1. Find area enclosed between y= - x2 + 5x and y=2x

2. Find area between y = x 2 - 2x + 2 and y=-x 2 + 6

3. Find area between y = x 2 – 2x+ 3 and y = 2x3 -12x

Non-intersecting curves Non-intersecting curves exampleexample

/4

1

y = sin x

0

y = sec2x

x-axis

2

(x, g(x))

y-axis

(x, f(x))

x

Area between a curve and a line(trigonometric function)

y =1

y =sin2 x

/2

x-axis

y-axis

0

1

Area Between Curves

1. Find area between y=x2 and y=-x2 [3 , 6]

2. Find area enclosed between y= - x2 + 5x and y=8x+10 [-1 ,0]

3. Find area between y = sin(x) and y=1 [3 , 6]

4. Find area between y = cos(x) and y=-1[-8 , -7]

Integration with respect Integration with respect to y-axisto y-axis

If a region’s bounding curves are described by functions of y, are would be easier calculated horizontal instead of vertical and the basic formula has y in place of x

Area under the curve, you would need to modify the equation to be in terms of y

Formula for Area between curves d

c

dyygyfA )]()([

Integration with respect to y-Integration with respect to y-axisaxis

Find the area that is bounded by x = y + 2 , x = y2 and by the x-axis.

Integration with respect to y-Integration with respect to y-axisaxis

0)2y)(1y(

02yy

y2y2

2

y = -1, y = 2

The region’s right-hand boundary is the line

x = y + 2, so f (y) = y + 2The left-hand boundary is the curve x =

y2, so g (y) = y2. The lower limit of integration is y = 0. We find the upper limit by solving x = y +

2 and x = y2 simultaneously for y:

Y-axis: Examples

1. Find area under the curve for x = 8y – y2 from y = 0 to y = 7

2. Find area between x=y2 and y=2x – 1

3. Find area enclosed between x = (y - 2)2 and x=1

4. Find area between x = 4y – 2y2, y=2x - 1

Application: Volume Generated (Disc Method)

360o

Volume Generated - Steps

STEP 1: Square the equation i.e. (3x)2

STEP2: Perform steps like Area under/between curves

STEP 2.5: Definite Integral as such

STEP 3: State in units3

b

a

dxxfV 2)]([

Application: Volume Generated (2 Curves)

b

a

dxxhxfV 22 )]([)]([

Volume Generated: Examples1. Find volume generated from the curve y = x3

+ x2 – 6x [-3 , 0] rotated along the x-axis.

2. Find volume generated between y = 2x2 - 4x + 6 and y = x2 + 2x + 1 for in respect to the x-axis

3. Find volume generated between y=x2 - 2x +1 and y=2x – 1 being rotated along the x-axis

4. Find volume generated between y = - x2 + 5x and y = 2x rotated along the x-axis