Can someone please help me?
Find the area between y= 2-3x^2, y=6, x=1, x=5
Nope, because the area under the curve of -3x^2 is negative. And you are trying to find the total absolute area between two curves. Therefore (2-3x^2) becomes (-2+3x^2)
if you get a negative answer, that means your answer is wrong.
$\displaystyle \int\limits_1^5 {(6 - 2+ 3x^2)} dx = 140$ NOT -140
140 is the total area between the two curves, since the area under the x-axis is negative so you have to take into account for that.
Here, I just edited your steps
$\displaystyle \int\limits_1^5 {(6-2-3x^2)}dx$
= $\displaystyle \int\limits_1^5 {(4-3x^2)}dx$
$\displaystyle 4x+(3x^3/3)$
$\displaystyle 4x+x^3$
$\displaystyle [4(5)+5^3]- [4(1)+(1)^3]$
$\displaystyle [145]- [5]$
$\displaystyle =140$