Question

Integral calculus question below! :)

Answer 1

Recall that a differentiable function F(x) is said to be increasing at a point a F'(a)>=0. Show that this function is always increasing:

\[\int\limits_{0}^{e^x}e^{e^{t}}\]

Answer 2

i can tell you this i have trouble with these types of problems sorry i couldnt help ;-;

Answer 3

\[\Large\rm F(x)=\int\limits\limits_{0}^{e^x}e^{e^{\color{orangered}{t}}}dt\]By the Fundamental Theorem of Calculus, Part 1,
we can calculate our derivative,\[\Large\rm F'(x)=e^{e^{\color{orangered}{e^x}}}(e^x)'\]Fundamental Theorem + Chain rule applied above.\[\large\rm F'(x)=e^{e^{e^x}+x}\]So you have an exponential function...
which is never negative, ya? :)

Answer 4

Ok lemme know if you confused when you get back on salt lady :)

Answer 5

Hahaha had a total brain fart. Thanks! @zepdrix

Answer 6

e^x = t and solve