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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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ctkh66 | Student, Grade 11 | eNoter

Posted May 24, 2012 at 11:35 AM via web

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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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thilina-g | College Teacher | (Level 1) Educator

Posted May 24, 2012 at 12:27 PM (Answer #1)

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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`

`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.`

 

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