Find the area between curves using integration y=x^2+2x+3,y=X+4,X=-3 and X=3

Answer 1

x^2+2x+3=x+4

x^2+x-1=0

x=-.5+/-.5sqr5= 1/2 +/-sqr5/2= (1+/-sqr5)/2= about 1.6, -.6.

y=(5+/-sqr5)/2= about 3.6, 2.4

points of intersection are about (-.6, 2.4). (1.6, 3.6). x=-3, 3 are irrelevant to the problem, just distracting surplusage

points of intersection are ((1-sqr5)/2, (5-sqr5)/2) and ((1+sqr5)/2, (5+sqr5)/2)

integration limits are x=(1+/-sqr5)/2 = about -.6, 1.6

integral of (x+4- x^2-2x-3 = -x^2-x+1)

= -x^3/3-x^2/2 +x

evaluated from 1/2-sqr5/2 to 1/2+sqr5/2

= -(1/2+sqr5/2)^3- (1/2 +sqr5/2)^2+1/2+sqr5/2 -[(1/2-sqr5/2)^3+(1/2-sqr5/2)^2+1/2-sqr5/2]

= -2-sqr5 -3/2-sq5/2 +1/2+sqr5/2 -[2-sqr5 +3/2 -sqr5/2 +1/2 -sqr5/2]

= -3-sqr5 -[4-2sqr5 ]

= -7 +sqr5

|sqr5-7|=about 4.6

(no guarantees no mistakes above)