Question

If `g(x)` is the antiderivative of `f(x) = root3(x^2+4x)`, what is the value of `g(1)`?

`g(5) = 7`.

Answer 1

`g(1) ~~ -3.882`

Explanation:

The Fundamental Theorem of Calculus tells us that

`int_1^5 root3(x^2+4x) dx = g(5)-g(1)`

Approximation or technology gets us

`int_1^5 root3(x^2+4x) dx ~`~ 10.882`.

So

`10.882 ~~ 7 - g(1)` and `g(1) ~~ 7-10.882 = -3.882`.

You'll probably need approximation or technology because the general antiderivative involves the hypergeometric function which is not taught in introductory calculus classes.

`int root3(x^2+4x) dx = (3root(3)(x^2+4x)(8((x+4)/x)^(2/3)color(white)(.)_2F_1(1/3,2/3;4/3;-4/x)+x^2+6x+8))/(5(x+4))+C`

where `color(white)(.)_2F_1(a,b;c;x)` is the hypergeometric function. (No, I cannot explain it.)