# 7-5 Applications of Integrals

## Learning Targets

You will be able to

## Concepts / Definitions

**Mass** is a property of an object. This property happens to be a measurement of its *resistance to acceleration*.

**Force** is an interaction between objects equivalent to *mass times acceleration*. In other words, $\vec{F} = m\vec{a}$.

### Work, Revisited

**Work** is the total amount of effort required to perform a task, equivalent to *force times displacement*. $W = \vec{F}*s = m\vec{a}*s$

#### The Metric System

Measurement |
Label |
Units |

Mass |
Kilograms |
$kg$ |

Force |
Newtons |
$kg * \frac{m}{sec^2}$ |

Work |
Joules |
$kg * \frac{m}{sec^2} * m$ |

#### Freedom Units

Measurement |
Label |
Units |

Mass |
Slugs |
$32 * lb$ |

Force |
Pounds |
$lb$ |

Work |
Foot-Pounds |
$ft*lb$ |

Imperial units make no sense.

As far as comparisons go, a Newton is about a quarter of a pound $(1\ N = 0.225\ lbs)$.

A foot-pound is about 1.36 Joules $(1\ ft\ lb = 1.36\ J)$.

### Work Done by a Force

If the force, $F(x)$, is not constant, then we can calculate the work over a tiny integral an infinite amount of times to get the total work.

#### Hooke’s Law and the Spring Constant

Hooke’s Law states that the force required to maintain a spring stretched $x$ units *beyond its natural length* is proportional to $x$, provided $x$ is not too large.

\[F(x) = kx\]
### Work Done Lifting

Consider a stone block attached to the end of a heavy rope. How much work is done in pulling this block to the top of a building?

This problem can be considered as two separate functions: the work done by pulling the stone block to the top, and the work done pulling the heavy rope up.

The stone block is straightforward - you need to exert force equal to its *weight* throughout the interval.

\[w = \int_a^b (weight)(distance\ travelled)\ dx\]
The rope is a little more complicated.

\[w = \int_a^b \frac{weight\ of\ rope\ /\ unit}{distance\ travelled}\ dx\]
Put our two equations together, and we get **Work Done Lifting as an Integral**.

\[w = \int_a^b \frac{m_r\vec{a}}{b-a} + (m_s\vec{a})(b-a)\ dx =\]
Note that $b-a$ is both the length of the rope and the distance travelled. $m_r$ is the mass of the rope, and $m_s$ is the mass of the stone block.

### Work Done Pumping

In a typical pumping problem, you will encounter a tank, a distance needed to travel, and a weight of the fluid.

\[w = \int_a^b (density)\ v(x)\Delta x\ dx\]
Note that $r$ must be adapted for the proper