Since the area between these 2 curves is bounded by where the 2 curves intersect. We can find the intersection points by setting the 2 functions equal to one another and solving for x.
x3 + x2 + 2x + 6 = 7x2 – x – 4
Solve by setting equal to zero.
x3 – 6x2 + 3x + 10 = 0
This cannot be factored, so we need to find at least one factor using p/q (from back in Precalculus) or by trial and error. Typically these problems use ±1 or ±2 so these are good x values to try to see if one of them sets the equation = 0.
Trying values I found x= –1 to work.
I can then do synthetic division to find the next using x= –1 as my factor.
–1 ¦ 1 –6 3 10
¦ ↓ –1 7 –10
________________
1 –7 10 0
This means after factoring out the (x+1) term (x= –1) we are left with:
x2 – 7x +10
This factors to (x-5)(x-2)
So the three intersection points are x= –1, 2 and 5 (These could have also been found on the calculator if it is a calculator active problem)
To see these points better on the graph, on the calculator I used a window of x: [–2,6] and y: [–10,200] or we could have sketched by hand.
From the graph we see that there are 2 distinct regions we need to find the area of.
The graph shows us which function is on top and which is on bottom.
Now for the calculus....
Using our x-intersection points as bounds we will integrate (top function – bottom function)
2 5
∫ (x3 + x2 + 2x + 6 ) – (7x2 – x –4) dx + ∫ (7x2 – x–4) – (x3 + x2 + 2x + 6)dx
–1 2
Combine like terms before integrating to make the math easier:
2 5
∫ (x3 – 6x2 + 3x + 10 )dx + ∫ (–x3 + 6x2 – 3x–10)dx
–1 2
Integrate:
2 5
1/4 x4 – 2 x3 + 3/2 x2 + 10x | + (–1/4 )x4 + 2x3 – 3/2 x2 – 10x |
–1 2
evaluating we get:
14 – (–25/4) + 25/4 – (–14) = 40.5 = 81/2 units2
Hope this helps!
