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Answer 1 · 2016-11-19T22:26:42

#F(x)# is a strictly increasing function

If #F(x)=int_1^(x^2)e^(-sin(k))dk# then

#d/dxF(x)=e^(-sin(x)) > 0# so

#F(x)# is a strictly increasing function

Answer 2 · 2016-12-04T02:28:50

#F(x)# is decreasing for #x in (-oo, 0)#
#F(x)# has a minimum at #x=0#
#F(x)# is increasing for #x in (0, oo)#

Suppose #G(x)# is the antiderivative of #e^(-sin(x))#, i.e. #G'(x) = e^(-sin(x))#. Then

#F'(x) = d/dxint_1^(x^2)e^(-sin(k))dk#

#=d/dx(G(x^2)-G(1))#

#=d/dxG(x^2)-d/dxG(1)#

#=2xG'(x^2)#

#=2xe^(-sin(x^2))#

Then, as #e^(-sin(x^2))>0# for all #x in RR#, the sign of #F'(x)# will match the sign of #x#.

Because #F(x)# is decreasing where #F'(x)<0# and increasing where #F'(x) > 0#, this gives us our final result:

#F(x)# is decreasing for #x in (-oo, 0)#
#F(x)# has a minimum at #x=0#
#F(x)# is increasing for #x in (0, oo)#

This is the graph of #F(x)#:

enter image source here