gateway arch, st. louis, missouri 6.1a areas between curves
TRANSCRIPT
Gateway Arch, St. Louis, Missouri
6.1a Areas Between Curves
21 2y x
2y x
How can we find the area between these two curves?
We could split the area into several sections, use subtraction and figure it out, but there is an easier way.
21 2y x
2y x
Consider a very thin vertical strip.
The length of the strip is:
1 2y y or 22 x x
Since the width of the strip is a very small change in x, we
could call it dx.
21 2y x
2y x
1y
2y
1 2y ydx
Since the strip is a long thin rectangle, the area of the strip is:
2length width 2 x x dx
If we add all the strips, we get:2 2
12 x x dx
21 2y x
2y x
2 2
12 x x dx
23 2
1
1 12
3 2x x x
8 1 14 2 2
3 3 2
8 1 16 2
3 3 2
36 16 12 2 3
6
27
6
9
2
The formula for the area between curves is:
1 2Areab
af x f x dx
We will use this so much, that you won’t need to “memorize” the formula!
4
0
1x
2x dx
4322
0
2 1
3 4x x
164 0 0
3
16 12
3 3
4
3
x
y
(0,0)
(1,1)
Sketch the region bounded by the graphs of the function and find the area of the region
1
2
y x
y x
1
42
0
1
2x x dx
5 2
1( 5 ) ( 5) x x x dx
532
1
3 53
xx x
125 175 25 3 5
3 3
25 7
3 3 32
3
x
y
Sketch the region bounded by the graphs of the function and find the area of the region
2 5
5
y x x
y x
5 2
16 5x x dx
6.1a Homework
Pg. 442 1, 2, 5 – 7, 11, 13, 17, 23,
and 25
y x
2y x
y x
2y x
If we try vertical strips, we have to integrate in two parts:
dx
dx
2 4
0 2 2x dx x x dx
We can find the same area using a horizontal strip.
dySince the width of the strip is dy, we find the length of
the strip by solving for x in
terms of y.y x2y x
2y x
2y x
y x
2y x
We can find the same area using a horizontal strip.
dySince the width of the strip is dy, we find the length of
the strip by solving for x in
terms of y.y x2y x
2y x
2y x
2 2
02 y y dy
length of strip
width of strip
22 3
0
1 12
2 3y y y
82 4
3
10
3
General Strategy for Area Between Curves:
1
Decide on vertical or horizontal strips. (Pick whichever is easier to write formulas for the length of the strip, and/or whichever will let you integrate fewer times.)
Sketch the curves.
2
3 Write an expression for the area of the strip.(If the width is dx, the length must be in terms of x. If the width is dy, the length must be in terms of y.
4 Find the limits of integration. (If using dx, the limits
are x values; if using dy, the limits are y values.)
5 Integrate to find area.
Sketch the region bounded by the graphs of the function and find the area of the region
38
81
102
2, 8
y x x
y x
x x
8
2
1 310 8
2 8x x x dx
8
2
2
3 710
8 2x x dx
83 2
2
710
8 4
x xx
18
Sketch the region bounded by the graphs of the function and find the area of the region
2 4 2
2
f x x x
g x x
3
2
0
3 2 2x x x dx
3
2
0
3x x dx 33
2
0
3
3 2
xx
9
2
Sketch the region bounded by the graphs of the function and find the area of the region
3 1
1
f x x
g x x
1 2
3 3
0 1
1 1 1 1x x dx x x dx
1
3
0
2 1 1x x dx
1
2
12 4
3
0
32 1
2 4
xx x
1 32 1 0
2 4
Sketch the region bounded by the graphs of the function and find the area of the region
2f y y y
g y y
3
2
0
2y y y dy
3
2
0
3y y dy
9
2
32 3
0
3 1
2 3y y
Sketch the region bounded by the graphs of the functions and find the area of the region then use a calculator to check your work
4 2
2
2
2
y x x
y x
128
15
2
2 4 2
2
2 2x x x dx
2
2 4 2
0
2 2 2x x x dx 23 5
0
42
3 5
x x
8.533
Sketch the region bounded by the graphs of the functions and find the area of the region then use a calculator to check your work
2
6
1
0, 0 3
xf x
x
y x
3ln10
3
20
60
1
xdx
x
32
03ln 1x
6.908
Sketch the region bounded by the graphs of the function and find the area of the region
sec tan
4 4
2 4 4
0
x xf x
g x x
x
1
0
2 4 4 sec tan4 4
x xx dx
2 42 1 2 2.1797
2
1
2
0
2 4 44 sec
2 4
xx x
2 4 4 44 2
2
Sketch the region bounded by the graphs of the functions and find the area of the region then use a calculator to check your work
4
0
2sin cos 2 0x x dx
0
12cos sin 2
2x x
2sin cos 2
0, 0
f x x x
y x
Sketch the region bounded by the graphs of the algebraic function and find the area of the region
0
0, 1
xf x xe y
x x
1
0
0xxe dx can not be evaluated by hand
you need to use a calculator
1.2556
Find the value of the accumulation function. Then evaluate it for each value of the variable.
2
1
4y x
F y e dx
2
1
8
yx
e
1
2 28 8y
e e
for 1y
1 1
2 21 8 8F e e
0
for 0y
1
20 8 8F
3.1478for 4y
1
2 24 8 8F e
54.2602
Use integration to find the area of the triangle with the given vertices
2, 3
4,6
6,1
A
B
C
Eq for A to B is
912
2y x
Eq for A to C is
5y x
Eq for B to C is
516
2y x
4 6
2 4
9 512 5 16 5
2 2x x dx x x dx
4 6
2 4
7 77 21
2 2x dx x dx
4 6
2 2
2 4
7 77 21
4 4x x x x
14