Solve the following initial valued problems:
a) g′(x) = sin(x), g(π) = 2.
b) g′′(x) = 18x, g′(0) = 1 and g(0) = 5.
b) g′′(x)=12x2 +2,g′(1)=7andg(1)=3.
I'll do one, and you can follow the method for the others.
g"(x) = 12x^2
g'(x) = 4x^3+c
g'(1)=7, so 4+c=7, so c=3
g'(x) = 4x^3+3
g(x) = x^4+3x+c
g(1)=3, so 1+3+c=3, so c=-1
g(x) = x^4+3x-1