g(x)= integral 2 to 1 (1+x^2) dx, find d/dx and d^2/dx^2?

Answer 1

∫(2..1) (1+x²) dx is a constant so both these
derivatives are 0.

Answer 2

g(x)
= [x+(x^3)/3] from 1 to 2
= [2+(2^3)/3] - [1+(1^3)/3]
= 4 2/3 - 1 1/3
= 3 1/3

let y=3 1/3 which means that there is no gradient therefore,
dy/dx= 0
and,
d^2y/dx^2= 0

Answer 3

If you're intergrating:
d/dx= x + (1/3)x^3

(d^2)/(dx^2)= (1/2)x^2 + (1/12)X^4