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6 1
5 1 2 3 4 9
2
1 f
First let’s calculate the integral from 04 of the function ( x ) = 3x 2 − 12
4 ∫ 3x
0 2 4 − 12 = x − 12 x = 16 − 0 = 16
3 0 Now, please note that there are essentially TWO pieces to calculate the area on.
One of those areas is in the domain 0<x<2 and the other is in the domain 2<x<4.
Press <space bar> or click mouse to continue 3
6 1
5 1 3 2 4 9
2
1 Notice that I have shaded the portion of the graph BELOW the xaxis in RED and the
2
4
portion ABOVE the xaxis in 3 x 2 − 12 This emphasizes− 12
GREEN. + 3 x 2 the positive and negative aspects
of those respective areas. Let’s look at the calculation as the SUM of these two shaded
0
2
areas. The first integral will be taken on2the interval 0<x<2 4
and the second on the integra
3
3
x − 12 x + x4 − 12 x
2
2<x<4.
4 ∫ ∫ 3x0− 12 2(∫ )x −+ 2 +4∫ 3 x 2 −4)2 2 3 − 12(2 )
∫
2 − 12(2) − 3 − 1 = 0 3 1 3 − 12( 1 − 02 2 3 0 0 2 2 [ 8 − 24 bar> or click mouse 8 continue
Press <space− 0 + 64 − 4to− 8 − 24
ress <space bar> to continue 3
6 1
5 + 32
1
− 16 2 3 4 9
2
1 8 − 24 − [ 0 ] + 64 − 48 − [ 8 − 24]
−16 + 32
= 16
4 My hope is that you now see that there are in essence TWO separate areas of the graph
essentially “in conflict”. You can merely “subtract” the negative area from the positive area
to find the integral, BUT you could also envision this as an area of 16 AND an area of 32
resulting in TOTAL area of 48. In much the same sense that it would take 48 square feet o
Carpet to cover both the red and green areas.
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This note was uploaded on 06/15/2011 for the course MATH 122 taught by Professor Kustin during the Spring '08 term at South Carolina.
 Spring '08
 KUSTIN

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