Applications of Integration

The Most Direct Application: Area

The geometric interpretation of the definite integral is the area under a curve. We can extend this idea to find not just the area between a curve and the x-axis, but also the more general case of the area enclosed between two different curves.

Area Between Two Curves

To find the area of the region bounded by two functions, $f(x)$ and $g(x)$, between $x=a$ and $x=b$, we use a simple idea: the area between the curves is the area under the top curve minus the area under the bottom curve.

If $f(x) \ge g(x)$ on the interval $[a, b]$, then the area is:

$$\text{Area} = \int_{a}^{b} [f(x) - g(x)] \,dx$$

In words: Integrate the 'top function' minus the 'bottom function' over the interval.

Example: Area Between a Parabola and a Line

Find the area of the region enclosed by $y=x^2$ and $y=x+2$.

1. Sketch: The line $y=x+2$ is above the parabola $y=x^2$ in the enclosed region.

2. Find Limits: Set the functions equal to find the intersection points.
$x^2 = x+2 \implies x^2 - x - 2 = 0 \implies (x-2)(x+1) = 0$.
The curves intersect at $x=-1$ and $x=2$. So, $a=-1$ and $b=2$.

3. Set Up Integral:
Top function = $x+2$. Bottom function = $x^2$.
$\text{Area} = \int_{-1}^{2} [(x+2) - x^2] \,dx$

4. Integrate:
First, find the antiderivative of $x+2-x^2$: $\frac{x^2}{2} + 2x - \frac{x^3}{3}$.
Now, evaluate at the limits:
$\left[\frac{2^2}{2} + 2(2) - \frac{2^3}{3}\right] - \left[\frac{(-1)^2}{2} + 2(-1) - \frac{(-1)^3}{3}\right]$
$= \left[2 + 4 - \frac{8}{3}\right] - \left[\frac{1}{2} - 2 + \frac{1}{3}\right] = \left[\frac{10}{3}\right] - \left[-\frac{7}{6}\right] = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = 4.5$

The Balance Point of an Object

The center of mass is the unique point where the weighted relative position of the distributed mass sums to zero. You can think of it as the 'average' position of all the mass in an object, or the point where the object would be perfectly balanced if you placed it on a single support.

For a simple system of discrete particles, the center of mass is a weighted average:

$$X_{cm} = \frac{m_1x_1 + m_2x_2 + ...}{m_1 + m_2 + ...} = \frac{\sum m_ix_i}{\sum m_i}$$

But what about a continuous object, like a metal rod or a plate, where the mass is spread out? For this, we need integration.

From Summation to Integration

To find the center of mass of a continuous body, we use the same logic as before, but we replace the discrete sum with a continuous sum (an integral). We divide the object into an infinite number of tiny pieces of mass, $dm$, each at a position $x$.

The formula for the x-coordinate of the center of mass becomes:

$$X_{cm} = \frac{\int x \,dm}{\int dm}$$

The denominator, $\int dm$, is simply the total mass of the object, $M$. So, the formula is often written as:

$$X_{cm} = \frac{1}{M} \int x \,dm$$

Introducing Linear Density (λ)

The term $dm$ must be expressed in terms of a spatial variable, such as $dx$, using the concept of density. For a one-dimensional object, we use linear mass density, $\lambda$, which is the mass per unit length.

$$\lambda = \frac{\text{mass}}{\text{length}} \implies dm = \lambda \,dx$$

If the density is uniform, $\lambda = M/L$. If it is non-uniform, $\lambda$ will be a function of position, $\lambda(x)$.

Example: Center of Mass of a Uniform Rod

Problem: Find the center of mass of a thin, uniform rod of length $L$ and mass $M$, placed along the x-axis from $x=0$ to $x=L$.

1. Set Up the Integral:
We use the formula $X_{cm} = \frac{1}{M} \int x \,dm$.
Since the rod is uniform, its linear density is constant: $\lambda = M/L$.
We can write $dm = \lambda \,dx = (M/L)dx$.
The integral will be from one end of the rod (0) to the other (L).

2. Substitute and Integrate:
$X_{cm} = \frac{1}{M} \int_{0}^{L} x \left(\frac{M}{L}dx\right)$
We can pull the constants $M/L$ out of the integral.
$X_{cm} = \frac{1}{M} \cdot \frac{M}{L} \int_{0}^{L} x \,dx = \frac{1}{L} \int_{0}^{L} x \,dx$
Now, integrate $x$:
$X_{cm} = \frac{1}{L} \left[\frac{x^2}{2}\right]_{0}^{L}$
$X_{cm} = \frac{1}{L} \left(\frac{L^2}{2} - 0\right) = \frac{L^2}{2L} = \frac{L}{2}$

This is the expected result: the center of mass of a uniform rod is at its geometric center.

For a fully worked example using a non-uniform, position-dependent density $\mu(x)=\lambda x$ — including a dimensional-analysis check on $\lambda$ — see Example 49 in the Examples section of this chapter.

Chapter 6 Summary: Applications of Integration

Here is a summary of the core physical ideas and formulas developed in this chapter. Notice that every result follows the same pattern: identify a thin element, write its small contribution, and integrate to add up all the elements.

• Area under a curve / between two curves: $$\text{Area} = \int_{a}^{b} f(x)\,dx \qquad \text{Area} = \int_{a}^{b} [f(x) - g(x)]\,dx\ \ (f(x) \ge g(x))$$
• Center of mass of a continuous body: $$X_{cm} = \frac{1}{M}\int x\,dm, \qquad dm = \lambda\,dx$$ where $\lambda$ is the linear mass density (constant for a uniform body, a function $\lambda(x)$ for a non-uniform one).
• Displacement and velocity from integration: $$\Delta v = \int_{t_i}^{t_f} a(t)\,dt, \qquad \Delta x = \int_{t_i}^{t_f} v(t)\,dt$$
• Work done by a variable force: $$W = \int_{x_i}^{x_f} F(x)\,dx$$
• Area of a circle by integration (ring method): slicing the circle into thin concentric rings of radius $x$ and thickness $dx$ gives $dA = 2\pi x\,dx$, so $$A = \int_0^{r} 2\pi x\,dx = \pi r^2$$
• Volume of a sphere by integration (shell method): slicing the sphere into thin concentric shells of radius $x$ and thickness $dx$ gives $dV = 4\pi x^2\,dx$, so $$V = \int_0^{r} 4\pi x^2\,dx = \frac{4}{3}\pi r^3$$
• Mass of a rod with non-uniform density $\mu(x)=\lambda x$: the mass element is $dm = \mu(x)\,dx = \lambda x\,dx$, so the total mass is $$M = \int_0^{L} \lambda x\,dx = \frac{\lambda L^2}{2}$$

Reversing the Direction: From Acceleration to Position

In differentiation, we learned how to find velocity and acceleration from a position function. Integration allows us to do the reverse. If we know an object's acceleration, we can find its velocity. If we know its velocity, we can find its displacement. This is used throughout physics for analyzing motion.

Finding Velocity from Acceleration

Since acceleration is the derivative of velocity ($a = dv/dt$), it follows that velocity is the antiderivative of acceleration.

$$v(t) = \int a(t) \,dt$$

The definite integral gives the change in velocity over a time interval:

$$\Delta v = v_f - v_i = \int_{t_i}^{t_f} a(t) \,dt$$

Finding Displacement from Velocity

Similarly, since velocity is the derivative of position ($v = dx/dt$), displacement is the integral of velocity.

The displacement (change in position, $\Delta x$) of an object between time $t_i$ and $t_f$ is the definite integral of its velocity function over that interval.

$$\Delta x = x_f - x_i = \int_{t_i}^{t_f} v(t) \,dt$$

This means that the displacement is the net area under the velocity-time graph. The total distance traveled, however, would be the integral of the speed, |v(t)|, which requires finding the area and treating any parts below the axis as positive.

Example: Motion with Variable Acceleration

An object starts from rest and has an acceleration given by $a(t) = 6t$ m/s². Find its displacement between t=0 and t=3 s.

1. Find the Velocity Function:
First, we must integrate the acceleration to find the velocity.
$v(t) = \int a(t) \,dt = \int 6t \,dt = 3t^2 + C$.
Since the object starts from rest, $v(0)=0$. Plugging this in: $3(0)^2 + C = 0 \implies C=0$.
So, $v(t) = 3t^2$.

2. Find the Displacement:
Now, integrate the velocity function from t=0 to t=3.
$\Delta x = \int_{0}^{3} 3t^2 \,dt$
The antiderivative of $3t^2$ is $t^3$.
$\Delta x = \left[t^3\right]_{0}^{3} = 3^3 - 0^3 = 27$ meters.

For another worked example of this technique, applied to the velocity function $v=2t+4$, see Example 51 in the Examples section of this chapter.

From Abstract Sums to Physical Quantities

Integration is more than a rule for finding antiderivatives — it is a systematic way of adding together an infinite number of infinitesimally small contributions. Whenever a physical quantity is built up continuously, rather than in discrete jumps, that quantity can be expressed as an integral. This chapter collects the most important physical applications of this idea: finding the area under a curve, locating the center of mass of a body whose density is not uniform, recovering displacement and velocity from a known velocity or acceleration function, computing the work done by a force that changes with position, and — as you will see in the worked examples — deriving the area of a circle and the volume of a sphere from first principles.

In every case the strategy is the same: slice the quantity into a large number of thin elements, write down the small contribution made by one representative element, and then let the sum over all elements become an integral as the element size shrinks to zero. Once this pattern is mastered it will reappear throughout your study of physics — from mechanics to electromagnetism.

How This Chapter Is Organized

• Finding Areas Using Integration — the area under a curve and the area enclosed between two curves.
• Finding the Center of Mass by Integration — locating the balance point of a continuous body with non-uniform density.
• Finding Displacement and Velocity by Integration — recovering velocity from acceleration, and displacement from velocity.
• Work Done by a Variable Force — computing work when the force changes with position, such as in a spring.
• Key Concepts — a consolidated summary of the formulas used throughout the chapter.
• Examples — additional worked problems, including the area of a circle, the volume of a sphere, the mass of a rod with position-dependent density, and the displacement from a linear velocity function.

Beyond Constant Forces

In introductory physics, we often calculate work as $W = F \cdot d$, but this simple formula only applies when the force is constant. What if the force changes as the object moves? For example, the force required to stretch a spring increases the more you stretch it. To find the work done by a variable force, we need integration.

Work as the Area Under the Force Curve

Just as displacement is the area under a velocity-time graph, the work done by a force is the area under a force-displacement graph.

To calculate this, we divide the total displacement into many tiny displacements, $dx$. Over each tiny displacement, the force $F(x)$ is almost constant. The small amount of work done is $dW = F(x)dx$. To find the total work, we sum (integrate) all these small pieces of work over the total displacement from $x_i$ to $x_f$.

$$W = \int_{x_i}^{x_f} F(x) \,dx$$

Example: Work Done in Stretching a Spring

A classic example of a variable force is the force exerted by a spring, described by Hooke's Law: $F(x) = kx$, where $k$ is the spring constant and $x$ is the displacement from its equilibrium position.

Problem: How much work is required to stretch a spring with a spring constant $k$ from its equilibrium position ($x=0$) to a displacement $x=L$?

1. Set Up the Integral:
We use the work formula with the force function $F(x)=kx$ and integrate from the initial position $x_i=0$ to the final position $x_f=L$.
$W = \int_{0}^{L} kx \,dx$

2. Integrate:
The antiderivative of $kx$ is $\frac{1}{2}kx^2$.
$W = \left[\frac{1}{2}kx^2\right]_{0}^{L}$
$W = \left(\frac{1}{2}kL^2\right) - \left(\frac{1}{2}k(0)^2\right) = \frac{1}{2}kL^2$

This is the well-known formula for the potential energy stored in a spring.