Question

a cable that weighs 6 lb/ft is used to lift 700 lb of coal up a mine shaft 400 ft deep. find the work done. show how to approximate the required work by a riemann sum. (let x be the distance in feet below the top of the shaft. enter $x_i^*$ as $x_i$.) $lim_{n
ightarrowinfty}sum_{i = 1}^{n}(square\times)delta x$ express the work as an integral. $int_{0}^{400}(square\times)dx$ evaluate the integral. $square$ ft - lb

Explanation:

Step1: Analyze the force - work relationship

The work done in lifting an object is $W = Fd$, where $F$ is the force and $d$ is the distance. The force required to lift the coal is a constant $F_{coal}=700$ lb. The force required to lift a small section of the cable of length $\Delta x$ at a distance $x$ from the top of the shaft is $F_{cable}=6\Delta x$ lb.

Step2: Set up the Riemann sum

The work done in lifting a small section of the cable of length $\Delta x$ at a distance $x_i$ from the top of the shaft is $dW_i=(700 + 6x_i)\Delta x$. The total work $W$ is approximated by the Riemann sum $\lim_{n
ightarrow\infty}\sum_{i = 1}^{n}(700+6x_i)\Delta x$.

Step3: Set up the integral

As $n
ightarrow\infty$, the Riemann sum becomes the definite - integral $\int_{0}^{400}(700 + 6x)dx$.

Step4: Evaluate the integral

We use the power rule for integration $\int x^n dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)$ and $\int kdx=kx + C$ (where $k$ is a constant).
\[

$$\begin{align*} \int_{0}^{400}(700 + 6x)dx&=\int_{0}^{400}700dx+\int_{0}^{400}6xdx\\ &=700x\big|_{0}^{400}+6\times\frac{x^{2}}{2}\big|_{0}^{400}\\ &=700\times400+3\times(400^{2}-0^{2})\\ &=280000+3\times160000\\ &=280000 + 480000\\ &=760000 \end{align*}$$

\]

Answer:

1. Riemann sum: $\lim_{n

ightarrow\infty}\sum_{i = 1}^{n}(700 + 6x_i)\Delta x$

1. Integral: $\int_{0}^{400}(700 + 6x)dx$
2. Value of the integral: $760000$ ft - lb