Fundamental Theorem of Calculus and Uniform Convergence

Suppose f is Reimann integrable on [a,b] and let F(x)= ∫_a^x▒f(t)dt for all x ∈[a,b]. Prove that F is continuous on [a,b] (hint: f must be bounded)
Consider F(x+h)= ∫_a^(x+h)▒f(t)dt= ∫_a^x▒f(t)dt+ ∫_x^(x+h)▒f(t)dt=F(x)+∫_x^(x+h)▒f(t)dt
Given that f is Reimann Integrable in [a, b] and hence is bounded on [a, b].
So, there exists some real number M such that for all x in [a, b] so in [x, x+h] for any h such that the second interval lies totally inside the first interval.
Taking the limit as ,
We have lim┬(h→0)⁡F(x+h)=lim┬(h→0)⁡F(x)+lim┬(h→0)⁡∫_x^(x+h)▒f(t)dt
lim┬(h→0)⁡F(x+h)≤lim┬(h→0)⁡F(x)+lim┬(h→0)⁡∫_x^(x+h)▒Mdt=lim┬(h→0)⁡F(x)+lim┬(h→0)⁡〖Mh=F(x)〗+0
Hence the function F(x) is continuous on [a, b]

Let F(x) = {█(x^2 sin⁡(1/x) if 0<|x|≤1,@0 if x=0)┤ and let f(x) = F'(x)
Prove that F'(x) exists for all x ∈[-1,1]
We have
We know that for all x in [-1, 1]. Applying sandwich theorem, we have
Hence F'(0) exists and is equal to zero. As the point x=0 is the only critical point where the differentiability is at danger and we have F'(0)=0. So, the function F'(x) exists.
Find f(x) for all x ∈[-1,1], and prove that f is reimann integrable on [-1, 1]
So, differentiating the given function, we have

Find ∫_(-1)^1▒f(x)dx
from fundamental theorem of integral calculus.
Hence the value of the integral is

Let f_n(x) = 1/n sin(n^2 x)
Prove f_n converges uniformly on R to a differentiable function, yet 〖f'〗_n(0) diverges.
Point wise convergence of is given by f(x)=0 as for all real x.
Now, as . Hence the convergence is uniform and the function is f(x)=0.
Now, which is obviously divergent.
Let f_n (x)= 〖sin〗^n (x) for all x ∈[0,π]. Prove that 〖f'〗_n is not uniformly convergent on [0,π]. (hint: suppose false and deduce a contradiction)
Suppose the given statement is false, which means that 〖f'〗_n is uniformly convergent on [0,π].
We have
We have
We check the convergence of the real sequence
Point wise convergence for this sequence is f(x)=0 for all x except for s . and for we have that is convergent. Hence there is no point wise convergence hence the convergence of is not uniform. According to the hint given, the convergence of should have been uniform we are getting the result in negative. Hence our assumption, that 〖f'〗_n is uniformly convergent on [0,π] is wrong. Hence 〖f'〗_n is not uniformly convergent on [0,π].

Note: theorem included below to be used as an aid with the hint in #4

Suppose f_n is defined on a finite interval I and 〖f'〗_n is continuous on I. Suppose 〖f'〗_n converges uniformly on I. Suppose moreover that there exists at least one point a ∈ I such that f_n (a) is a convergent sequence of real numbers. Then there exists a differentiable function f such that f_n→f uniformly on I, and f'(x) ≡ lim┬(n→∞)⁡〖〖f^'〗_n (x)〗 on I

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