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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