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__Theorem:__Let f be continuous on [a,b]. The function g defined on [a,b] by http://tutorial.math.lamar.edu/AllBrowsers/2413/DefnofDefiniteIntegral_files/eq0051M.gif [Broken]

is continuous on [a,b], differentiable on (a,b), and has derivative

**g'(x)=f(x) for all x in (a,b)**

**1) Given that g is defined by g(x)=definite integral of sin (pi t) from 0 to x for all real numbers x, determine g'(-2), if it exists.**

[In this case a=0, according to the above theorem g'(x)=f(x) for all x in (a,b), but the theorem doesn't say that g'(x)=f(x) for x<a, and in this case -2<0=a. Does it mean that g'(-2) does not exist? ]

**2) Let g be defined by g(x)=definite integral of 1/ (1+t^2) from 0 to x for all real numbers x.**

2a) Show that if x<0, then g(x)<0 using properties of integrals.

2a) Show that if x<0, then g(x)<0 using properties of integrals.

[Sorry, I really have no clue about part a and I need some help on it...]

**2b) Find the intervals on which g is increasing or decreasing.**

[According to the theorem on the top g'(x)=1/ (1+x^2) is true only for all x in (a,b), in this case a=0, then how can I find g'(x) for the part x<0 ?]

I hope someone would be nice enough to help me out! Thanks a lot!

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