Assume G(x)=H’(x)+1 and that H(0)=3 and H(0)=4. Compute the integral from 0 to 2 of g(x)dx

Answer 1

∫[0,2] g(x) dx
= ∫[0,2] (H'(x)+1) dx
= ∫[0,2] H'(x) dx + ∫[0,2] 1 dx
= ∫[0,2] dH/dx + ∫[0,2] 1 dx
= H(2)-H(0) + (2-0)
Now you decide what H is at 0 and 2, since you seem to have mangled that