What is the area between the 4 curves: y=x^2-x-6, y=x^2-x-4, x=1, x=6?

Question

This is a calculus question... If you could show me your steps as to how to get the answer I'd really appreciate it!

Answer 1

y = int from 1 to 6 [f(x) - g(x)] dx
y= int 1,6 [x^2-x-4] dx - int 1,6 [x^2-x-6] dx

int 1,6 [x^2-x-4] dx
[(x^3)/3] - [(x^2)/2] - 4x [1,6
inputing 6 and 1 for x you get
[216/3] -18-24 -([1/3] - [1/2] - 4) = 34 1/6 or 205/6

int 1,6 [x^2-x-6] dx
[(x^3)/3] - [(x^2)/2] - 6x
inputing 6 and 1 for x you get
[216/3] -18-36 -([1/3] - [1/2] - 6) = 24 1/6 or 145/6

so you get 34 1/6 - 24 1/6 = 10

Answer 2

integral of h dx between
lower limit 1 and upper limit 6

where h = big y - small y =
(x^2 - x -4) - (x^2 - x -6) = 2


so answer is 10.


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Answer 3

1) find the points of intersection
-solve a system
2) See which points are to the further to the left and right
and top and bottom
3) Integrate the area with respect to those limits