Approximating Definite Integrals — Question 9

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Question 9

A velocity is measured every second: t (s)01234v (m/s)03564\begin{array}{c|ccccc} t\text{ (s)}&0&1&2&3&4\\ \hline v\text{ (m/s)}&0&3&5&6&4 \end{array} Use T4T_4 to estimate displacement.

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Question 9 – Solution

Step 1: Find the time width. h=4−04=1 second.h=\frac{4-0}{4}=1\text{ second}. Step 2: Apply the trapezoidal formula. T4=h2[v0+2v1+2v2+2v3+v4]=12[0+2(3)+2(5)+2(6)+4]=12[0+6+10+12+4]=12(32)=16.\begin{align*} T_4&=\frac h2[v_0+2v_1+2v_2+2v_3+v_4]\\ &=\frac12[0+2(3)+2(5)+2(6)+4]\\ &=\frac12[0+6+10+12+4]\\ &=\frac12(32)=16. \end{align*} Step 3: Include units. Velocity times time gives displacement in meters. 16 m\boxed{16\text{ m}}

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