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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, weâ€™ll highlight the mathematical definition of work and learn how we can define work in terms of definite integrals. Weâ€™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*Weâ€™ll show you how to calculate for an objectâ€™s work by evaluating definite integrals, so keep your notes on integral properties and antiderivative formulas handy. For now, letâ€™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, letâ€™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, weâ€™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 weâ€™re observing an object that is moving along the positive horizontal direction from $x=a$ to $x =b$.

Letâ€™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: $\{x_0, x_1, x_2, â€¦, x_n\}$ 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, $F(x)$, letâ€™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} â€“ [x]_{2}^{8}\\&= [(8^2 -2^2) â€“ (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

**Hookeâ€™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 $F(x) = 120x$. Letâ€™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 $27 \phantom{x}\text{J}$