This document discusses using integration to find the area between two curves by considering it as an accumulation process. It explains that we select a representative element, such as a rectangle, and use geometry formulas to relate the area of that element to the functions that define the curves. The area under each representative rectangle is summed to find the total area via integration. Two examples are provided, one finding the area between a parabola and the x-axis using vertical rectangles, and another using horizontal rectangles between two other curves.
1. 7.1b Area of a Region7.1b Area of a Region
Between Two CurvesBetween Two Curves
Integration as an Accumulation Process
2. So far this chapter we have used a representative element – in this case a
representative rectangle - to adapt integration to the area between curves. In
future sections we will continue to do the same thing. We will come up with a
representative element, then using precalculus and geometry formulas, adapt
integration to find an accumulation of those elements.
[ ( ) ( )]
b
a
f x g x dx−∫
For example, in this section we used the idea that Area = (height)(width) and
we found the area of a representative rectangle to show that
∆Area = [f(x) - g(x)]Δx. Then we summed up all those representative
rectangles with integration to get
3. Ex 6 p.451 Describing Integration as an Accumulation Process
Find the area of the region bounded by the graph of y = 4 – x2
and the x-
axis. Describe the integration as an accumulation process.
( )
2
2
2
[ 4 0]x dx
−
− −∫The parabola will intersect the x-axis at x = 2
( )
2
2
2
4 0x dx
−
−
− =∫
As we let the representative rectangle move from left
to right, we can see area accumulating.
( )
1
2
2
5
4
3
x dx
−
−
− =∫
( )
0
2
2
16
4
3
x dx
−
− =∫
( )
1
2
2
27
4
3
x dx
−
− =∫
( )
2
2
2
32
4
3
x dx
−
− =∫
4. Another example. Find area when horizontal rectangle makes sense.
( ) (2 )f y y y= − ( )g y y= −
Graph. Find intersections by setting f(y)=g(y). 2y – y2
= -y 3y – y2
= 0
y = 0, y = 3
(2 )x y y= − 2
2y y= − +
2
( 2 1 1)y y= − − + −
2
(( 1) 1)y= − − −
2
( 1) 1y= − − +
Standard form for sideways
parabola: x = a(y-k)2
+h
a = -1 (opens left), vertex (h, k): (1, 1)
Same size and shape as x=y2
3
2
0
[(2 ) ( )]y y y dy− − −∫
9
2
=
5. As you are creating these integrals, you need to be asking, “is this formula going
to work for every representative rectangle as Δy moves from 0 to 3?”