Simplify the following:
d/dx[∫(t/lnt)dt] where the integral is a definite integral with bounds from x to x2
The Fundamental THeorem of calculus says:
Suppose f is continuous on [a,b]
Then ∫abf(x)dx=F(b)-F(a) where F is any antiderivative of f
The Attempt at a Solution
So this is how I thought to solve it. Let f(t)=t/lnt
First notice that f(t) continuous on (1,c] forall c in ℝ => f(t) integrable and there exists an antiderivative F(t) such that F'(t) = f(t)
So let F(t) be such an antiderivative. ***By the fundamental theorem of calculus,
∫f(t)dt (where the integral bounds are from x to x2=F(x2)-F(x)
d/dx[∫f(t)dt] (where the integral bounds are from x to x2=d/dx[F(x2)-F(x)] = d/dx[F(x2)] - d/dx(F(x)] = 2xF'(x2) - F'(x) [by the Chain Rule] = 2x(x2/ln(x2)) - x/lnx = 2x3/2lnx - x/lnx [by rules of logarithm] = (x3-x)/lnx (common denominator and added them together)
So I get that as long as the bounds on the integral are from (0,c), the answer to
d/dx[∫(t/lnt)dt] where the integral is a definite integral with bounds from x to x2 = (x3-x)/lnx
OK here is my problem, the book solves this almost exactly the same way, but they never make any mention of the bounds on the integral or continuity. I guess I am confused. Do I not need to think about where the function is continuous? It's just part of the FTC says "Suppose f is continuous on [a,b]" so it seems to me like I can only apply it when that part holds. Why don't you need to think of continuity in this case? There is no restriction in the problem statement about x. How do you solve this problem for any general x?
Also the other problem, is the line where I put ***. I think I've done something wrong here because this function is only continuous on the open region (1,c] for all c, but the FTC requires f be continuous on a closed bounded region [a,b]. So do I just pick a number great than 1, say [1.1,c]? I mean what's the formal way to solve this with regards to the continuity aspect?