Given x| 5 9 13 17 21 25 29 33 37

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f(x)| 5 5.5 6.8 8.8 9.8 8.1 6.7 5.4 4.2

If n=4, the width of the intervals is `Delta=(37-5)/4=8`

We can estimate `int_5^(37)f(x)dx` by `sum_(i=1)^4f(m_i)Delta` where `m_i` is the midpoint of the `i^(th)` interval.

Thus `sum_(i=1)^4f(m_i)Delta=8[5.5+8.8+8.1+5.4]=222.4`

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**The amount of work done is approximately 222.4J**

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