Question

Find the area of the curve between y=3-x^2 and y=2x.

Answer 1

\[2x = 3 - x ^{2}\]
\[x^{2} + 2x -3 = 0\]
(x-1) (x+3) = 0
x = 1 , y = 3 -1 =2
x = -3 , y =3 - 9 = -6
(1,2) , (-3, -6)

Answer 2

Now that you have the points of intersection and that
3 - x^2 < 2x for all x in the open interval (-3, 1) you can set up the integral:
\[A = \int\limits_{-3}^{1}(3 - x ^{2}-2x)) dx.\]

Answer 3

That works out to
\[(3x - x ^{3}/3 - x ^{2})|_{x=1} - (3x - x ^{3}/3 - x ^{2})|_{x=-3},\]
and the rest of the work is arithmetic.

Answer 4

thanks