Do you know what the antiderivative of 1 is?
yes, but you have to add C as a constant of integration because both y = x and y = x + 2 have 1 as a derivative. So when you take the antiderivative of h"(t) = 1, you get h'(t) = x + C Then you use the fact that h'(2) = 3 to solve for C
so.. 2+C=3 then, solve for C
right. so the first derivative is h'(t) = x + 1. now take the antiderivative of that
h(t) = 1/2x^2+x+C Use h(4)=5 to find C
Right. h(t) = ½t² + t - 7. All that's left is to find h(6)
thank you so much :)