Use midpoint rule to estimate the average velocity?

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SUMMARY

The discussion focuses on using the midpoint rule to estimate the average velocity of a car over the first 12 seconds. The midpoint rule is applied with 3 sub-intervals, calculated as $$\frac{12}{4} = 3$$, leading to $$n = 3$$. The average velocity is determined using the formula $$\overline{v}(t)=\frac{1}{12}\int_0^{12}v(t)\,dt$$, approximated by evaluating the velocity function at midpoints of the intervals. The final expression for the integral approximation is $$\int_0^{12}v(t)\,dt\approx4\left(v(2)+v(6)+v(10) \right)$$, requiring values from the provided graph.

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shamieh
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Use the midpoint rule to estimate the average velocity of the car during the first 12 seconds.

Click here to see the graph from my book.

i understand the midpoint rule is
$$\frac{b - a}{n}$$

so $$\frac{12}{4} = 3$$
so $$n = 3$$

I also know that

$$\frac{1}{12} \int^{12}_{0} v(t)dt $$

But now I'm stuck... any guidance anyone can offer would be great.
 
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The average velocity would be given by (as you stated):

$$\overline{v}(t)=\frac{1}{12}\int_0^{12}v(t)\,dt$$

Using the Midpoint rule to approximate the integral in this expression, with 3 sub-intervals of equal width ($n=3$), we could state:

$$\int_0^{12}v(t)\,dt\approx\frac{12-0}{3}\sum_{k=1}^3\left(v\left(2(2k-1) \right) \right)=4\left(v(2)+v(6)+v(10) \right)$$

Do you see that we evaluate the velocity function at the midpoint of each sub-interval?

Now you just need to read the needed values from the given graph.
 

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