A plot of the two functions is shown in:
To find where they intersect, you set them equal:
3|x| = x2-10
So we can work without the absolute value operation, lets consider the two cases x<0 and x≥0 separately.
3(-x) = x2-10 (-x b/c we know x<0 and only want nonnegative values)
x2+3x-10 = 0 (bring -3x over to right hand side, then rewrite eqn)
(x+5)(x-2) = 0
x=-5 or x=2 (discard x>0 answer b/c we're considering negative x)
So x=-5 at one intersection
To get the (x,y) coord's: plug in x=-5 to either of your functions
3x = x2-10
x2-3x-10 = 0
(x+2)(x-5) = 0
x=-2 or x=5 (discard x<0 answer b/c we're considering x≥0 here)
x=5 is at the other intersection
To get (x,y) coord's plug in x=5 to either function.
About integrating to find the area:
As you look at the two functions, notice that they are both symmetrical about the y-axis. So, you can define h(x) to be:
h(x) = 3|x| - (x2-10). Now, due to the symmetry, you can say the total area between the two curves is the same as twice the area under the part that goes to the right of the y-axis (i.e. the part where x≥0). So,
Area = 2 ∫ (3x - x2 + 10) dx evaluated from x=0 to x=5.
We were able to just write 3x because we're integrating over the x ≥ 0 portion only.
Does that make sense?