Area Within a Curve

Quote:

Originally Posted by Jughead5267

Hey everybody, newbie here, looking for some quick help. I promised a good friend of mine help with her calculus homework, but I've gotten rusty, and only have a fleeting glimpse of what I should be doing. PLEASE HELP
Here's the functions, of which I've got to find the area within the curve.

y=x^3 - 5x + 2
y=Sqrt(2-x)
y=-Sqrt(2-x)

The graph kinda looks like a horizontal parabola & a vertical snake...

The relevant points of intersection I've found are:
(.12666695, 2.0567), (2,0), & (.69576957, -1.1420)

Of course the x-values are the only ones needed.
Thus
the upper limits for Int1 are .12666695, and .69576957

The upper limts for Int2 are .69576957, and 2

This is where I am stuck.
Do I plug them into Sqrt(2-x) - ((x^3) - 5x + 2), or have I missed a step?

It is beyond late right now, so if I could get some assistance while I sleep, I'll be eternally grateful.

This is an irritating problem. Who came up with this anyway?

Notice that there are actually 3 integrals you need to do:
Integrate from x = 0.126666947 to x = 0.695769572
Integrate from x = 0.695769572 to x = 1.978833423
Integrate from x = 1.978833423 to x = 2

(See the two graphs below. The second is a blown up view of the region around x = 2.)

When you find the area you take the difference between the two functions that are creating the area. So
$A = \int_{0.127}^{0.696}( (\sqrt{2 - x}) - (x^3 - 5x + 2) )~dx + \int_{0.696}^{1.979}( (\sqrt{2 - x}) - (-\sqrt{2 - x}) )~dx$ $+ \int_{1.979}^2 ( (x^3 - 5x + 2) - (-\sqrt{2 - x}) )~dx$

-Dan