Area under y=x'squared' - 12

Area under y=x'squared' - 12 - 3 6 1 5 1 2 3 4 9 -2 1 f...

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Unformatted text preview: 3 6 1 5 1 2 3 4 9 -2 1 f First let’s calculate the integral from 0-4 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 x-axis in RED and the 2 4 portion ABOVE the x-axis 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. Press <space bar> end show ...
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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.

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Area under y=x'squared' - 12 - 3 6 1 5 1 2 3 4 9 -2 1 f...

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