Area from a definite integral

**Frostking** · November 25th 2008, 12:32 PM

I am suppose to obtain an area bound by 3x + y = 6 above and below by y = x^2 - 4 and the y axis. I think I should take the integral of the first and subtract the integral of the second one from 0 to 2

In terms of x: 6x - 3x^2/2 - (x^3/3 + 4x) from x = 0 to 2. Am I on the right track on this one????? Please let me know. I asked someone else in my class and they added the two integrals and used |x^2 - 4| for second integral?? Thanks again for this wonderful site!!!! Frost King

**skeeter** · November 25th 2008, 01:00 PM

to find the area between the line and the curve, first find where the line and the curve intersect.

$y = 6 - 3x$

$y = x^2 - 4$

$x^2 - 4 = 6 - 3x<br />$

$x^2 + 3x - 10 = 0$

$(x + 5)(x - 2) = 0$

$x = -5$ , $x = 2$

$A = \int_{-5}^2 (6-3x) - (x^2 - 4) \, dx$

simplify the integrand, integrate, and use the FTC.

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