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6036: Area of a Plane Region. AB Calculus. Accumulation vs. Area. Accumulation can be positive, negative, and zero. Area is defined as positive . The base and the height must be positive. h = always Top minus Bottom (Right minus Left). - PowerPoint PPT Presentation
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6036: Area of a Plane Region
AB Calculus
Accumulation vs. Area
Area is defined as positive.
The base and the height must be positive.
Accumulation can be positive, negative, and zero.
h = always Top minus Bottom (Right minus Left)
𝑓 −0=h
h=0 − 𝑓
AreaDEFN: If f is continuous and non-negative on [ a, b ], the
region R, bounded by f and the x-axis on [ a,b ] is
Remember the 7 step method.
b = Perpendicular to the axis!
h = Height is always Top minus Bottom!
( )b
aTA f x dx
a b
( ) 0
lim ( )
b x
h f x
TA f x dx
Area of rectangle
[𝑎 ,𝑏 ]
Ex:
Find the Area of the region bounded by the curve,
and the x-axis bounded by [ 0, ]
siny x x
𝑏=∆ 𝑥 [ 0 ,𝜋 ]h=(𝑥+sin 𝑥 ) −0𝐴= (𝑥+sin 𝑥 ) ∆𝑥
lim𝑛→ ∞
∑ (𝑥+sin 𝑥 ) ∆ 𝑥
𝐴=0
𝜋
(𝑥+sin 𝑥 )𝑑𝑥
𝐴= 𝑥2
2−cos 𝑥|𝜋0
𝐴= 𝜋 2
2− (−1 ) − ( 0−1 )
𝐴= 𝜋 2
2+2
Ex:
Find the Area of the region bounded by the curve,
and the x-axis bounded by [ -1, 1 ]
3 2y x
𝑏=∆ 𝑥 [− 1,1 ]h=0 −(− 3√𝑥−2)
lim𝑛→ ∞
∑ ( 3√𝑥+2 )
−1
1
( 3√𝑥+2 )𝑑𝑥
−1
1 ( (𝑥 )13 +2)𝑑𝑥
34
(𝑥 )43 +2 𝑥| 1
− 1
( 34∗1+2)−( 3
4∗ (1 ) −2)=( 3
4+2)−( 3
4−2)
34
−34+2+2=4
Area between curves
REPEAT: Height is always Top minus Bottom!
( ) ( )b
aTA f x g x dx
a b
f (x)
g (x)1 ( )b
R aA f x dx
2 ( )b
R aA g x dx
Height of rectangle
Area between curves
The location of the functions does not affect the formula.
( ) ( )b
aTA f x g x dx
a b
Both aboveh=f-g
One above one belowh=(f-0)+(0-g)h=f-g
Both belowh=(0-g)-(0-f)h=f-g
<Always Top-bottom>
Area : Method:
Find the area bounded by the curves and
on the interval x = -1 to x = 2
2 1y x
2y x
𝑏=∆ 𝑥 [− 1,2 ]h=(𝑥2+1 ) − (𝑥− 2 )
h=𝑥2−𝑥+3
lim𝑛→ ∞
∑ (𝑥2−𝑥+3 ) ∆ 𝑥
−1
2
(𝑥2−𝑥+3 )𝑑𝑥
𝑥3
3−𝑥2
2+3 𝑥| 2
−1
( 83
−42+3 (2 ))−(−1
3−
12+3 (− 1 ))
93
−32+9=3+9 −1.5=10.5
Area : Example (x-axis):
Find the area bounded by the curves and2( ) 4f x x 2( ) 2g x x
𝑏=∆ 𝑥 [−√3 ,√3 ]
4 −𝑥2=𝑥2− 2
6=2 𝑥2
3=𝑥2
±√3=𝑥
h=( 4 −𝑥2 ) − (𝑥2 −2 )h=6 − 2𝑥2
lim𝑛→ ∞
∑ (6− 2𝑥2 ) ∆ 𝑥
− √3
√3
(6 − 2𝑥2 )𝑑𝑥
6 𝑥− 2( 𝑥3
3 )| √3−√3
6 (√3 ) − 23
(√3 )3 −(− 6√3−( 23 ) (−√3
3 ))6 √3 − 2√3+6√3 − 2√3=8√3
Area: Working with y-axis
Area between two curves.
The location of the functions does not affect the formula.
When working with y-axis, height is always Right minus Left.
( ( ) ( ))
lim ( ( ) ( ))
b y
h h y k y
TA h y k y y
h (y)
k (y)
a
b
( ( ) ( ))b
aTA h y k y dy
Perpendicular to y-axis!
Area : Example (y-axis):Find the area bounded by the curves
and
2 2y x2 2y x
𝑥= 𝑦2
2
𝑥=𝑦+2
2
Perpendicular to y-axis
𝑦2
2= 𝑦+2
2
𝑦 2− 𝑦−2=0
(𝑦− 2 ) ( 𝑦+1 )𝑦=−1𝑎𝑛𝑑 2
𝑏=∆ 𝑦 [− 1,2 ]
h=( 𝑦+22 )−( 𝑦
2
2 )h=
12
( 𝑦+2 − 𝑦2 )
lim𝑛→ ∞
∑ 12
(𝑦 +2 −𝑦 2) ∆ 𝑦
𝐴= 𝑦=−1
𝑦=212
( 𝑦+2 −𝑦 2 )𝑑𝑦
𝐴=12 ( 𝑦
2
2+2 𝑦−
𝑦3
3 )| 2−1
𝐴=12 ( 22
2+2 (2 )− 23
3 )− 12 (−12
2+2 (−1 ) − −13
3 )𝐴=1+2−
86
−14+1 −
16
𝐴=3−2112
=1512
Multiple Regions
1) Find the points of intersections to determine the intervals.
2) Find the heights (Top minus Bottom) for each region.
3) Use the Area Addition Property.
a b c
b =
h = h =
f (x)
g (x)
x
Area : Example (x-axis - two regions):
Find the area bounded by the curve
and the x-axis.
2(1 )y x x
NOTE: The region(s) must be fully enclosed!
Area : Example ( two regions):
Find the area bounded by the curve
and . 3
1y x
NOTE: The region(s) must be fully enclosed!
1y x
Area : Example (Absolute Value):
Find the area bounded by the curve and the
x-axis on the interval x = -2 and x = 3
( ) 2 3f x x
PROBLEM 21
Velocity and Speed: Working with Absolute Value
DEFN: Speed is the Absolute Value of Velocity.
The Definite Integral of velocity is NET distance (DISPLACEMENT).
The Definite Integral of Speed is TOTAL distance. (ODOMETER).
Total Distance Traveled vs. Displacement
The velocity of a particle on the x-axis is modeled by the function, .
Find the Displacement and Total Distance Traveled of the particle on the interval, t [ 0 , 6 ]
3( ) 6x t t t
Updated:
• 01/29/12
• Text p 395 # 1 – 13 odd
• P. 396 # 15- 33 odd