Question

Sorry, the last post was mistyped.

Consider the functions f(x) = (x^3/x^4+1) and g(x) = (x/x^4+1). Let R denote the region in the first quadrant bounded by the curves y = f(x) and y = g(x). Find the exact volume of the solid that has R as its base if every cross section by a plane perpendicular to the x-axis is a rectangle of height 3. ("Exact volume" means no calculator numbers.)

Answer 1

assuming the usual carelessness with parentheses, I assume we're talking about
f(x) = x^3/(x^4+1)
g(x) = x/(x^4+1)
The curves intersect at (0,0) and (1,1)
So, each thin rectangle of thickness dx has its base the distance between the two curves: (x-x^3)/(x^4+1)
That makes the volume
v = ∫[0,1] 3(x-x^3)/(x^4+1) dx = 3/8 (π-ln4)
Now, you may be wondering how to get this tidy little result.
∫x^3/(x^4+1) dx is easy, since if u=x^4+1, that's just ∫du/u
But what about ∫x/(x^4+1) dx ? That's almost as easy, since then if u=x^2, it's almost exactly ∫ du/(u^2+1)