Question

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:

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