Question

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]`?

Answer 1

`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`

Answer 2

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)`