Given the functions f(x) = -x^2 + 2x + 3 and g(x) = -x + 3

Question

Given the functions f(x) = 􀀀x2 + 2x + 3 and g(x) = 􀀀x + 3:

(a) Find where the functions f and g intersect.

(b) Express area enclosed by the two functions between their intersection points

as an integral.

(c) Calculate the area enclosed by the two functions between their intersection

points.

Answer 1

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

x^2 -5x = 0

x(x-5) = 0

x=0 or 5

y = 3 or -12

intersection points are (0,3) and (5,-12)

integral of x^2-5x = x^3/3 -5x^2/2

normally evaluated from 0 to 5 =

5^3/3 -5(25)/2 =

125/3 -125/2 =

0, but that treats below axis area as negative, canceling the positive area above the x axis,

giving a false answer

you need to break the area down and add, not subtract the area below the x axis, so

evaluate the integral from 0 to 3

then add the absolute value of the integral from 3 to 5

to get the total area between the line and curve

=9 + 6= 14 = the relevant area

if there's no arithmetic mistake above,(no guarantees), but it's the basic method of calculus, more or less, to find the relevant area

find not just where the lines intersect, but where the lines intersects the x axis, (0,5)