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Find the area enclosed between the curve y=x^3, the x-axis and the line y=-3x+4
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The area we have to find is the small triangle like area above x-axis.
Let's find the intersection points.
`y_1 = x^3` intersect the x-axis at `x = 0`
`y_2 = -3x+4` intersects x-axis at `x = 4/3`
The two functions intersect each other at,
`x^3+3x-4 = 0`
`x^3-x+4x-4 = 0`
`x(x^2-1) +4(x-1) = 0`
`x(x-1)(x+1)+4(x-1) = 0`
`(x-1)(x(x+1)+4) = 0`
`(x-1)(x^2+x+4) = 0`
Therefore two graphs interesect at x =1.
To find the required area, we can integrate `y_1` from 0 to 1 and then integrate `y_2` from 1 to `4/3` .
`A = int_0^1x^3dx + int_1^(4/3)(-3x+4)dx`
`A = [x^4/4]_0^1 + [-3/2x^2+4x]_1^(4/3)`
`A = 1/4 + [8/3-5/2]`
`A = 1/4 + 1/6`
`A = 5/12`
Therefore the area enclosed by the curves is `5/12.`
Posted by thilina-g on May 24, 2012 at 12:27 PM (Answer #1)
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