# 4 $h(t)=t^{2}-t + 1$, $h(0)=-\\frac{1}{2}$ to advance in the circuit, locate $h(1)$. © virge cornelius 2016

Explanation:

Step1: Integrate ( h'(t) )

$$ \begin{align*} h(t)&=\int (t^{2}-t + 1)dt\ &=\frac{t^{3}}{3}-\frac{t^{2}}{2}+t + C \end{align*} $$

Step2: Find ( C ) using ( h(0)=-\frac{1}{2} )

Substitute ( t = 0 ) and ( h(0)=-\frac{1}{2} ) into ( h(t)=\frac{t^{3}}{3}-\frac{t^{2}}{2}+t + C ). ( h(0)=\frac{0^{3}}{3}-\frac{0^{2}}{2}+0 + C=-\frac{1}{2} ), so ( C =-\frac{1}{2} ). Then ( h(t)=\frac{t^{3}}{3}-\frac{t^{2}}{2}+t-\frac{1}{2} ).

Step3: Calculate ( h(1) )

Substitute ( t = 1 ) into ( h(t)=\frac{t^{3}}{3}-\frac{t^{2}}{2}+t-\frac{1}{2} ). $$ \begin{align*} h(1)&=\frac{1^{3}}{3}-\frac{1^{2}}{2}+1-\frac{1}{2}\ &=\frac{1}{3}-\frac{1}{2}+1-\frac{1}{2}\ &=\frac{1}{3}+(1 - 1)\ &=\frac{1}{3} \end{align*} $$

Answer:

( h(1)=\frac{1}{3} )