How do you use the second fundamental theorem of Calculus to find the derivative of given #int arctan(3t) dt# from #[4,1/x]#?

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Answer 1 · 2016-07-02T19:41:32

#- 1/x^2 arctan(3/x) #

FTC part 2 says that

if #F(x) = int_a^x dt qquad f(t) #

then #(dF)/dx = f(x)#

here you want

#d/dx int_4^{1/x} dt qquad arctan(3t) #

we can start by separating out the integration limits so they match the theorem

#int_4^{1/x} = int_a^{1/x} - int_a^4#

so

#d/dx int_4^{1/x} \ arctan(3t) \ dt#

#= d/dx ( color{green}{int_a^{1/x} dt \ arctan(3t)} - color{red}{ int_a^4 dt \ arctan(3t)} )#

the red term is constant and can be ignored. in fact we could have specified that a = 4 and continued with the basic formulation of FTC pt 2

the green term is not exactly in the form required but the FTC pt 2 can be expressed in terms of a new variable u as

if #F(u) = int_a^u dt \ f(t) #

then #(dF)/(du) = f(u)#

if then #u = u(x) = 1/x# then

#(dF)/(dx) = (dF)/(du) * (du)/dx#

# = f(u) * (du)/dx#

So we can say that

#= d/(du) ( int_a^{u} dt \ arctan(3t) )#

#= arctan(3u) = arctan(3/x)#

and

#= d/dx (int_a^{1/x} dt \ arctan(3t) )#

#arctan(3/x) * d/dx (1/x)#

#- 1/x^2 arctan(3/x) #