How do you use the second fundamental theorem of Calculus to find the derivative of given #int dx/lnx# from #[x^2, x^3]#?

1 Answer
Sep 2, 2016

Answer:

#= (x ( x - 1 ) )/( ln x)#

Explanation:

you have #d/dx int_(x^2)^(x^3) dx/lnx# which is pretty ugly as the variable x is overloaded, so we can first re-write as

#d/dx int_(x^2)^(x^3) 1/lnt dt qquad square#

in its simplest form, the Second FTC states that

#d/dx int_a^x f(t) dt = f(x)#

BUT, in this case, we need to apply the Chain Rule as the interval is a function of x, so:

#color(blue)( d/dx int_a^(u(x)) f(t) dt = f(u(x)) u'(x) )qquad triangle#

Finally, to match #square# to #triangle# we do the following

#int_(x^2)^(x^3) 1/lnt dt #

#= int_(x^2)^(a) 1/lnt dt + int_(a)^(x^3) 1/lnt dt #

#=- int_(a)^(x^2) 1/lnt dt + int_(a)^(x^3) 1/lnt dt #

#= int_(a)^(x^3) 1/lnt dt - int_(a)^(x^2) 1/lnt dt #

and now applying #triangle#!!

#= 1/ln(x^3) d/dx(x^3) - 1/ln(x^2) d/dx(x^2) #

#= (3x^2)/ln(x^3) - (2x)/ln(x^2) #

#= (x ( x - 1 ) )/( ln x)#