How do you use the second fundamental theorem of Calculus to find the derivative of given #int [(ln(t)^(2))/t]dt# from #[3,x]#?

2 Answers
Apr 7, 2016

Answer:

#int_3^xln^2(t)/tdt=(ln^3(x)-ln^3(3))/3#

Explanation:

First, note that using the substitution #u = ln(t) => du = 1/tdt# we have

#intln^2(t)/tdt = intu^2du = u^3/3+C = ln^3(t)/3+C#

The second fundamental theorem of calculus states that
#F'(x) = f(x)=> int_a^bf(x)dx = F(b)-F(a)#

By the above, we have #d/dt ln^3(t)/3 = ln^2(t)/t#, and so

#int_3^xln^2(t)/tdt = ln^3(x)/3-ln^3(3)/3#

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

Apr 7, 2016

Answer:

Instead of actually finding the function and then differentiating, one could reason as follows:

Explanation:

The Second Fundamental Theorem of Calculus says that

If #f# is continuous on #[a,b]#, then

#int_a^b f(x) dx = F(b)-F(a)#
where #F# is a function for which #F'(x) = f(x)# for all #x# in #[a,b]#.

In this case we are using the variable #t# in the integrand and the variable #x# as the upper limit of integration.

We want the derivative of #int_3^x [ln(t)^2/t] dt#.

Note that since we are asked about the interval #[3,x]#, we must have #x > 3# (otherwise the interval is either empty or undefined).

So, #ln(t)^2/t# is continuous on #[3,x]#, and

#int_3^x [ln(t)^2/t] dt= F(x) - F(3)# where #F# is a function such that #F'(x) = ln(x)^2/x#.

And there is our answer. The derivative of #int_3^x [ln(t)^2/t] dt# is #ln(x)^2/x#.

NOTE

The First Fundamental Theorem of Calculus says this directly. It says:

If #f# is continuous on #[a,b]# and #g# is defined for all #x# in #[a,b]# by,

#g(x) = int_a^x f(t) dt#, then #g# is continuous on #[a,b]# and #g'(x) = f(x)# for all #x# in #(a,b)#