Show integrable is uniformly continuous

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The discussion focuses on proving that the function F(y) = ∫g(x)f(x,y)dx is uniformly continuous given that f is continuous on a compact set H and g is integrable. The initial approach of separating g(x) and f(x,y) to show continuity was unsuccessful due to g(x) not being continuous. A hint suggests leveraging the compactness of H to establish that f is uniformly continuous. This uniform continuity can then be used to demonstrate that F(y) is continuous for all y in the compact interval [c,d]. Once continuity is established, uniform continuity follows from the properties of compact sets.
HF08
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H = [a,b]\times[c,d] . f:H\rightarrowR is continuous, and
g:[a,b]\rightarrowR is integrable.

Prove that
F(y) = \intg(x)f(x,y)dx from a to b is uniformly continuous.


I initially ripped g(x) and f(x,y) apart and tried to show each was continuous. This failed.
In short, I am completely stuck. Please help me.
 
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g(x) is only given as integrable, not necessarily continuous so that couldn't work.
 
Hint: since H is compact, f is, in fact, uniformly continuous. Use that to show for all epsilon>0, there exists a delta>0 such that |f(x,y0)-f(x,y)|<delta for all y such that |y-y0|<delta (for all x). Use that to show that the integral F(y) is continuous at y0. Once you know it's continuous, you don't have to worry about the uniform part, since y is in [c,d], which is also compact.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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