Sketch the region bounded by the graphs of the equations. Sketch the region bounded by the graphs of the equations. Find the area of the region.

Question

f(x)

=

x2 + 6x

g(x) = x + 6

Answer 1

f(x) = x^2 +6x is an upward opening parabola

= x^2 +6x +9 -9

= (x+3)^2 -9 in vertex form with vertex = (-3,-9) = minimum point

x intercepts = 0 and -6, the points (0,0) and (-6,0)

y intercept = (0,0)

axis of symmetry is x=-3

g(x)= x+6 is an upward sloping line, with slope = 1 or 45 degrees,

y intercept = (0,6)

x intercept = (-6,0)

the line and parabola intersect at two points

where

x+6 = x^2+6x

x^2 +5x -6=0

(x+6)(x-1) = 0

x =-6, 1

y = 0, 7

the points (-6,0) and (1,7)

evaluate the integral from x =-6 to x=0 then from 0 to 1

the integral of g(x)-f(x) is the integral of x+6 -(x^2+6x) = -x^2-5x+6

= -x^3/3 -5x^2/2 +6x

from -6 to 0

= 72-90 -36

= -54,

from 0 to 1

= -1/3-5/2 +6

= -2/6 -15/6 +36/6

= 19/6 = area above the x axis

total area between the line and parabola

=54 + 19/6

= 54+ 3 1/6

= 57 1/6