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# Work Calculus – Definition, Definite Integral, and Applications

**Work**, in physics, indicates the amount of force acting through a given distance. In this article, we’ll highlight the mathematical definition of work and learn how we can define work in terms of definite integrals. We’ll also get to see how essential integral calculus is in physics and engineering in our discussion.

** Work in calculus is simply equal to the area under the curve of **$\boldsymbol{F(x)}$,

**$\boldsymbol{F(x)}$**

*where***$\boldsymbol{x =a}$**

*is simply the function representing the force acted upon from***$\boldsymbol{x = b}$.**

*to*We’ll show you how to calculate for an object’s work by evaluating definite integrals, so keep your notes on integral properties and antiderivative formulas handy. For now, let’s do a quick recap of what work is and learn how definite integrals come into place!

**What is work in calculus?**

Work in calculus reflects the area under the curve of $F(x)$. In our introductory physics class, work is simply the** product of the force acting upon the object over a given distance (or displacement)**.

\begin{aligned}W &= Fd\end{aligned}

Keep in mind that in SI metric system, force is measured in Newtons ($N$), the distance is in meters ($m$), and work is in terms of $N \cdot m$ or in joules ($J$). For example, if a person exerts a force of $10 \phantom{x}N$ to lift an object $40 \phantom{x}m$ off the floor, the amount of work done is as shown below.

\begin{aligned}W &= F\cdot d\\&= (10 \phantom{x}N) \cdot (40 \phantom{x}m)\\&= 400 \phantom{x}N\cdot m\\&= 400 \phantom{x}J\end{aligned}

Now, let’s extend the definition of work by considering the fact that the force applied can be variable.

**Work definition in calculus**

In the earlier discussion, we’ve shown how work can be calculated when given a constant value for the force exerted. However, there are instances, **when force is not constant**. Suppose that we’re observing an object that is moving along the positive horizontal direction from $x=a$ to $x =b$.

Let’s say $F(x)$ is the function representing the amount of force exerted over the interval, $[a, b]$. If $F(x)$ is continuous throughout the interval, we can approximate the amount of work done by partitioning the interval into subintervals with the following endpoints: $\{x_0, x_1, x_2, …, x_n\}$ and a uniform width of $\Delta x$. Hence, we can estimate the total amount of work as shown below:

\begin{aligned}W &\approx \sum_{i =1}^{n} f(c_i)\Delta x_i \end{aligned}

To find a better approximation, as we have learned before, can be acquired by evaluating the quantity as $n\rightarrow \infty$.

\begin{aligned}W &=\lim_{n\rightarrow\infty} \sum_{i =1}^{n} f(c_i)\Delta x_i\\ &= \int_{a}^{b} F(x)\phantom{x}dx \end{aligned} |

This means that **work exerted on a moving object is simply the area under the curve of the function for force**, $F(x)$ from $x=a$ to $x =b$.

**How to solve work problems in calculus?**

Now that we understand how we can define work in terms of the definite integral of the force function, $F(x)$, let’s break down the steps to solve problems involving work:

- Identify the function that represents the force, $F(x)$, exerted from $x=a$ to $x= b$.
- Set up the expression for work in terms of $F(x)$: $W = \int_{a}^{b} F(x)\phantom{x}dx$.
- Evaluate the definite integral by applying key integral properties and formulas.

For example, if we want to calculate the amount of work done given $F(x) = 2x + 1$ Newtons over the interval, $[2, 8]$. Hence, we can calculate the amount of work done as shown below.

\begin{aligned}W &= \int_{2}^{8} (2x +1)\phantom{x}dx\\&= \int_{2}^{8}2x \phantom{x}dx + \int_{2}^{8}1 \phantom{x}dx\\&= 2\left[\dfrac{x^2}{2}\right]_{2}^{8} – [x]_{2}^{8}\\&= [(8^2 -2^2) – (8 -2)]\\&= 54 \end{aligned}

This means that the work done to move the object is equal to $54$ Newton-meters.

*Example 1*

A particle is moving along the $x$-axis and located $x$ feet from the origin. If a force of $x^2 + x + 4$ pounds is being acted upon the article, calculate the total amount of work done to move the particle from $x = 2$ to $x = 6$.

__Solution__

We can see that the force acting upon the particle is equal to $F(x) = x^2 + x +4$. Hence, use this expression to calculate the amount of work done to move the particle over the interval, $[2, 6]$.

\begin{aligned} W&= \int_{2}^{6} (x^2 + x + 4) \phantom{x}dx\end{aligned}

Evaluate the definite integral to find $W$.

\begin{aligned} W&= \int_{2}^{6} x^2\phantom{x}dx +\int_{2}^{6} x\phantom{x}dx +\int_{2}^{6} 4 \phantom{x}dx\\ &= \left[\dfrac{x^3}{3} \right ]_{2}^{6} + \left[\dfrac{x^2}{2} \right ]_{2}^{6} + [4x]_{2}^{6}\\&= \left[\left(\dfrac{6^3}{3} – \dfrac{2^3}{3} \right ) +\left(\dfrac{6^2}{2} – \dfrac{2^2}{2} \right )+ (4\cdot 6 – 4\cdot 2) \right ]\\&= \dfrac{208}{3} + 16 + 16\\&= \dfrac{304}{3}\end{aligned}

This means that the total amount of work done upon the particle is $\dfrac{304}{3}\phantom{x}\text{ft-lb}$.

*Example 2*

We need a force of $30 \phantom{x}\text{N}$ to hold a spring that has been stretched from its natural length of $0.20 \phantom{x}\text{m}$ to a length of $0.45 \phantom{x}\text{m}$. Calculate the amount of work done if we stretch the spring from $0.25 \phantom{x}\text{m}$ to $0.48 \phantom{x}\text{m}$.

** Hint**: Recall that

**Hooke’s Law**states that the force needed to maintain the stretch of $x$ units is $F(x) = kx$, where $k$ is the spring constant.

__Solution__

Calculate the sprint constant by using the given values: $F(x) = 30 \phantom{x}\text{N}$ and $x = 0.45 -0.20 = 0.25 \phantom{x}\text{m}$.

\begin{aligned} k&= \dfrac{F(x)}{x}\\&= \dfrac{30}{0.25}\\&= 120\end{aligned}

Using $k = 120$, we have $F(x) = 120x$. Let’s use this expression for the force needed to maintain a stretch from $0.25 \phantom{x}\text{m}$ to $0.48 \phantom{x}\text{m}$. Hence, we have the following expression for the amount of work needed:

\begin{aligned}W &= \int_{0.25}^{0.48} 120x \phantom{x}dx \end{aligned}

Evaluate the definite integral to find the exact value of work in $\text{N}\cdot \text{m}$.

\begin{aligned}W &= 120\int_{0.25}^{0.48} x \phantom{x}dx\\&= 120\left[\dfrac{x^2}{2} \right ]_{0.25}^{0.48}\\&= 60\left[x^2 \right ]_{0.25}^{0.48}\\&= 60[(0.48)^2 – (0.25)^2]\\&= 10.074 \phantom{x}\text{N}\cdot \text{m}\\&= 10.074 \phantom{x}\text{J} \end{aligned}

This means that the total amount of work done is $10.074 \phantom{x}\text{N}\cdot \text{m}$ or $10.074 \phantom{x}\text{J}$.

**Practice Questions**

1. A particle is moving along the $x$-axis and located $x$ feet from the origin. If a force of $2x^2 + 4x$ pounds is being acted upon the article, calculate the total amount of work done to move the particle from $x = 1$ to $x = 5$.

2. We need a force of $40 \phantom{x}\text{N}$ to hold a spring that has been stretched from its natural length of $0.10 \phantom{x}\text{m}$ to a length of $0.35 \phantom{x}\text{m}$. Calculate the amount of work done if we stretch the spring from $0.15 \phantom{x}\text{m}$ to $0.60 \phantom{x}\text{m}$.

**Answer Key**

1. $\dfrac{392}{3} \phantom{x}\text{ft-lb}$

2. $27 \phantom{x}\text{N}\cdot \text{m}$ or $27 \phantom{x}\text{J}$