mathwonk
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TD: that approach in your book is the easiest of the three at elast if one propves the integrability of 1/x and the fundamental theorem of calculus.
now integrability is often not actually rpoved in most books, so to be honest that also is a gap in this so called easy approach.
what I sometime do in my course, instead of assuming integrability of continuous functions, which requires the concept of uniform continuity, is to prove integrability for monotone functions, which is much easier and covers the case of 1/x.
i.e. it is very easy to prove that the upper and lower Riemann sums for a monotone function on [a,b] converge to the same limit, as they differ by the product |f(b)-f(a)| times deltax, which obviously goes to zero as deltax does.
this is Newtons proof. then after knowing your monotone function is integrable, if it is also continuous it is very easy to prove the integral is differentiable, since the difference quotient [F(x+h)-F(x)]/h is bounded above by the area of the rectangle with base h and height f(x+h)-f(x) which approaches zero as h does by the definition of continuity.
If you look in stewart you will probably find that he assumes integrability of continuous functions or maybe proves it in an appendix that most courses skip.
now integrability is often not actually rpoved in most books, so to be honest that also is a gap in this so called easy approach.
what I sometime do in my course, instead of assuming integrability of continuous functions, which requires the concept of uniform continuity, is to prove integrability for monotone functions, which is much easier and covers the case of 1/x.
i.e. it is very easy to prove that the upper and lower Riemann sums for a monotone function on [a,b] converge to the same limit, as they differ by the product |f(b)-f(a)| times deltax, which obviously goes to zero as deltax does.
this is Newtons proof. then after knowing your monotone function is integrable, if it is also continuous it is very easy to prove the integral is differentiable, since the difference quotient [F(x+h)-F(x)]/h is bounded above by the area of the rectangle with base h and height f(x+h)-f(x) which approaches zero as h does by the definition of continuity.
If you look in stewart you will probably find that he assumes integrability of continuous functions or maybe proves it in an appendix that most courses skip.
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