`g(r) = int_0^r(sqrt(x^2 + 4))dx` Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.

Hello!

Part 1 of the Fundamental Theorem of Calculus states that for a continuous function `f` `F'_a(x)=f(x),` where `F_a(x)=int_a^xf(t)dt.`

Here `f(t)=sqrt(t^2+4)` and `g(x)=F_0(x).`

Therefore

`g'(x)=F'_0(x)=f(x)=sqrt(x^2+4).`

or `g'(r)=sqrt(r^2+4)`  (identical variable replacement).

Hello!

Part 1 of the Fundamental Theorem of Calculus states that for a continuous function `f`
`F'_a(x)=f(x),` where `F_a(x)=int_a^xf(t)dt.`

Here `f(t)=sqrt(t^2+4)` and `g(x)=F_0(x).`

Therefore

`g'(x)=F'_0(x)=f(x)=sqrt(x^2+4).`

or `g'(r)=sqrt(r^2+4)`  (identical variable replacement).