Science:Math Exam Resources/Courses/MATH101 A/April 2025/Question 07 (a)/Solution 1

We know that the position at $t=4$ is given by $$x(4)=\underset{0}{\overset{4}{\int }}v(t)dt.$$ Hence we want to approximate $\underset{0}{\overset{4}{\int }}v(t)dt$ with $4$ trapezoids. $\mathrm{\Delta }x=1$ and $x_i=i$. Thus, $$\begin{array}{rl}\underset{0}{\overset{4}{\int }}v(t)dt& \approx \frac{1}{2}(v(0)+2v(1)+2v(2)+2v(3)+v(4))\\ & =\frac{1}{2}(100+\frac{200}{2}+\frac{200}{9}+\frac{200}{28}+\frac{100}{65})\\ & =\frac{1}{2}(200+\frac{200}{9}+\frac{200}{28}+\frac{100}{68})\\ & =100+\frac{100}{9}+\frac{25}{7}+\frac{10}{13}.\end{array}$$