How can I find $h(y)$?

Question

Given the following function $$ f(t) = \begin{cases} t &\text{for }t\leq2\,,\\ 2 &\text{for } t\geq 2\,,\\ \end{cases} $$ and $$h'(y)=f(y)\,,$$ how can I find $h(y)$ ?

Answer 1

An antiderivative is not unique. So $c=h(0)$ gives $$h(y)-h(0)=\int_0^yf(t)dt=\int_0^ytdt =y^2/2$$ for $y\le 2$, so $$h(y)=y^2/2+c$$ when $y\le 2$. But if $y>2$, $$h(y)-h(2)=\int_2^yf(t)dt=\int_2^y2dt=2(y-2).$$ Id est, $$h(y)=h(2)+2t-4=(2^2/2+c)+2y-4=2y-2+c$$ when $y>2$. (You should check btw that $h$ is differentiable, and it is.)