Introduction to the Integral

Have you ever wondered how to find the area of a shape with curved edges? Or how to calculate the total distance traveled when your speed keeps changing? These questions puzzled mathematicians for centuries, and their answers led to one of the most powerful ideas in all of mathematics: the integral.

If you have already studied derivatives, you have learned about rates of change - how fast things are happening at each instant. The integral asks the opposite question: if we know the rate at which something is happening, how much has accumulated over time? This process of “undoing” a derivative, of adding up infinitely many infinitely small pieces, is what integration is all about.

Do not worry if this sounds abstract right now. We will start with something very concrete - finding areas - and build up to the general concept step by step. By the end of this lesson, you will see that integration is a natural and intuitive process, even if the notation takes some getting used to.

Core Concepts

The Area Problem

Let us begin with a simple question: what is the area under a curve?

For a rectangle, you learned the formula $A = \text{base} \times \text{height}$ in elementary school. For a triangle, $A = \frac{1}{2} \times \text{base} \times \text{height}$. But what about the region bounded by a curve like $y = x^2$, the $x$-axis, and two vertical lines?

This is the area problem, and it motivated much of the development of calculus. The key insight is that we can approximate curved regions using shapes we already understand: rectangles.

Consider the region under the curve $y = x^2$ from $x = 0$ to $x = 2$. We cannot directly apply any formula we know, but we can break this region into vertical strips. If we approximate each strip with a rectangle, we get an estimate of the total area. The more rectangles we use, the better our approximation becomes.

Approximating Area with Rectangles

Here is the strategy: divide the interval $[a, b]$ into $n$ equal subintervals, each of width

$$\Delta x = \frac{b - a}{n}$$

For each subinterval, we construct a rectangle. The width of each rectangle is $\Delta x$, but we need to decide how tall to make each one. The height will be the value of the function at some point in that subinterval - but which point?

This choice gives us different types of approximations, each with its own advantages.

Left, Right, and Midpoint Sums

Left Riemann Sum: Use the left endpoint of each subinterval to determine the height.

$$L_n = \sum_{i=0}^{n-1} f(x_i) \Delta x$$

where $x_i = a + i \cdot \Delta x$ for $i = 0, 1, 2, \ldots, n-1$.

Right Riemann Sum: Use the right endpoint of each subinterval to determine the height.

$$R_n = \sum_{i=1}^{n} f(x_i) \Delta x$$

where $x_i = a + i \cdot \Delta x$ for $i = 1, 2, 3, \ldots, n$.

Midpoint Riemann Sum: Use the midpoint of each subinterval to determine the height.

$$M_n = \sum_{i=1}^{n} f\left(\frac{x_{i-1} + x_i}{2}\right) \Delta x$$

Each approach gives a slightly different answer, but as we use more and more rectangles (as $n \to \infty$), all three methods converge to the same value - the exact area under the curve.

For an increasing function, the left sum underestimates the area while the right sum overestimates it. For a decreasing function, the opposite is true. The midpoint sum typically gives a better approximation than either the left or right sum for the same number of rectangles.

The Definite Integral as a Limit

The definite integral is defined as the limit of Riemann sums as the number of rectangles approaches infinity:

$$\int_a^b f(x),dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x$$

Here, $x_i^*$ represents any point in the $i$-th subinterval. The remarkable fact is that for continuous functions, this limit exists and is the same regardless of which points we choose in each subinterval.

The definite integral $\int_a^b f(x),dx$ represents the signed area between the curve $y = f(x)$ and the $x$-axis from $x = a$ to $x = b$. The area is positive where $f(x) > 0$ (above the $x$-axis) and negative where $f(x) < 0$ (below the $x$-axis).

This might seem strange at first - negative area? But it makes perfect sense in applications. If $f(t)$ represents velocity, then positive values mean you are moving forward and negative values mean you are moving backward. The signed area gives you your net displacement, not the total distance traveled.

The Fundamental Theorem of Calculus

Now we come to one of the most beautiful results in all of mathematics. The Fundamental Theorem of Calculus (FTC) connects the two main operations of calculus - differentiation and integration - revealing them to be inverse processes.

Fundamental Theorem of Calculus, Part 1:

If $f$ is continuous on $[a, b]$ and we define

$$F(x) = \int_a^x f(t),dt$$

then $F$ is differentiable on $(a, b)$ and $F’(x) = f(x)$.

In other words, if you integrate a function and then differentiate the result, you get back to where you started.

Fundamental Theorem of Calculus, Part 2:

If $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$ (meaning $F’(x) = f(x)$), then

$$\int_a^b f(x),dx = F(b) - F(a)$$

This is extraordinary. It says that to find the area under a curve, you do not need to compute limits of Riemann sums. You just need to find an antiderivative and evaluate it at the endpoints.

We often write $F(b) - F(a)$ as $\bigl[F(x)\bigr]_a^b$ or $F(x)\Big|_a^b$.

Why the Fundamental Theorem Works

The intuition behind the FTC is beautiful. Imagine you are tracking the total rainfall in a bucket over the course of a day. Let $f(t)$ be the rate of rainfall (inches per hour) at time $t$, and let $F(t)$ be the total accumulated rainfall up to time $t$.

The rate at which the total is changing is exactly the current rate of rainfall: $F’(t) = f(t)$. And the total rainfall from time $a$ to time $b$ is the difference $F(b) - F(a)$.

Integration adds up the rate of change to find the total accumulation. Differentiation finds the rate of change from the total accumulation. They are two sides of the same coin.

Notation and Terminology

Term	Meaning	Example
$\int_a^b f(x),dx$	Definite integral from $a$ to $b$	Signed area under curve
Riemann sum	Approximation using rectangles	$\sum_{i=1}^n f(x_i) \Delta x$
Integrand	The function being integrated	$f(x)$ in $\int f(x),dx$
Limits of integration	The bounds $a$ (lower) and $b$ (upper)	In $\int_0^5$, the limits are 0 and 5
$dx$	Indicates the variable of integration	Tells us to integrate with respect to $x$
$\Delta x$	Width of each subinterval	$\Delta x = \frac{b-a}{n}$
Antiderivative	A function whose derivative is $f$	$F(x)$ where $F’(x) = f(x)$
Fundamental Theorem	$\int_a^b f’(x),dx = f(b) - f(a)$	Derivatives and integrals are inverses
Signed area	Area counting below $x$-axis as negative	$\int_0^{2\pi} \sin x,dx = 0$
Partition	Division of $[a,b]$ into subintervals	${0, 0.5, 1, 1.5, 2}$ partitions $[0,2]$

Examples

Key Properties and Rules

Understanding these properties will help you work with integrals efficiently and recognize patterns that simplify calculations.

Linearity Properties

Constant Multiple Rule: $$\int_a^b c \cdot f(x),dx = c \int_a^b f(x),dx$$

You can factor constants out of an integral. This makes sense: if you double the height of every rectangle in a Riemann sum, you double the total area.

Sum and Difference Rule: $$\int_a^b [f(x) \pm g(x)],dx = \int_a^b f(x),dx \pm \int_a^b g(x),dx$$

The integral of a sum equals the sum of the integrals. This allows you to break complex integrands into simpler pieces.

Properties Involving Limits of Integration

Reversing Limits: $$\int_a^b f(x),dx = -\int_b^a f(x),dx$$

Swapping the limits of integration changes the sign. Intuitively, you are now measuring area in the “backward” direction.

Zero-Width Interval: $$\int_a^a f(x),dx = 0$$

If the lower and upper limits are the same, there is no interval to integrate over, so the integral is zero.

Additivity of Intervals: $$\int_a^b f(x),dx + \int_b^c f(x),dx = \int_a^c f(x),dx$$

You can split an integral at any point and add the pieces back together. This is useful when a function has different formulas on different parts of the interval, or when it crosses the $x$-axis.

Comparison Properties

Non-negativity: If $f(x) \geq 0$ for all $x$ in $[a, b]$, then $\int_a^b f(x),dx \geq 0$.

Monotonicity: If $f(x) \leq g(x)$ for all $x$ in $[a, b]$, then $\int_a^b f(x),dx \leq \int_a^b g(x),dx$.

Bounding: If $m \leq f(x) \leq M$ for all $x$ in $[a, b]$, then $m(b-a) \leq \int_a^b f(x),dx \leq M(b-a)$.

This last property gives you quick bounds on an integral without having to compute it exactly.

Real-World Applications

The integral appears throughout science, engineering, economics, and everyday life. Here are some of the most common applications.

Distance from Velocity

If $v(t)$ is your velocity at time $t$, then the integral

$$\int_{t_1}^{t_2} v(t),dt$$

gives your displacement (net change in position) from time $t_1$ to time $t_2$.

For example, if you drive north at 60 mph for one hour and then south at 30 mph for one hour, your displacement is $60 - 30 = 30$ miles north, even though you traveled a total distance of 90 miles.

If velocity is always positive (you never reverse direction), then displacement equals total distance traveled. Otherwise, you need to account for the signed area.

Total Cost from Marginal Cost

In economics, the marginal cost $MC(x)$ is the cost of producing one additional unit when you are already producing $x$ units. The integral

$$\int_a^b MC(x),dx$$

gives the total cost of increasing production from $a$ units to $b$ units.

Similarly, if $MR(x)$ is marginal revenue, then $\int_a^b MR(x),dx$ gives the additional revenue from selling units $a$ through $b$.

Work Done by a Variable Force

In physics, work equals force times distance - but only when the force is constant. If the force varies with position, you need integration:

$$W = \int_a^b F(x),dx$$

This is essential for problems involving springs (where force depends on displacement), gravity (where force depends on distance from a planet), and many other situations.

Accumulation Over Time

Integration naturally models any process where something accumulates over time:

• Population growth: If $r(t)$ is the rate of population change, then $\int_0^T r(t),dt$ is the total population change over time $T$.

• Water flow: If $f(t)$ is the rate of water flowing into a tank (gallons per minute), then $\int_0^t f(s),ds$ is the total amount of water added by time $t$.

• Pollution: If $p(t)$ is the rate at which pollutants enter a lake, then $\int_0^T p(t),dt$ is the total amount of pollution added over the time period.

• Drug concentration: In pharmacology, integrating the rate of drug absorption gives the total amount of drug in the bloodstream.

Average Value of a Function

The average value of a function $f$ on the interval $[a, b]$ is defined as:

$$f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x),dx$$

This generalizes the idea of an average to continuous functions. For example, if $T(t)$ is the temperature at time $t$, the average temperature over a day is

$$T_{\text{avg}} = \frac{1}{24} \int_0^{24} T(t),dt$$

Self-Test Problems

Test your understanding with these practice problems. Try to solve each one before checking the answer.

Summary

Here are the key ideas from this lesson:

• The definite integral $\int_a^b f(x),dx$ represents the signed area between the curve $y = f(x)$ and the $x$-axis from $x = a$ to $x = b$.

• We can approximate integrals using Riemann sums - adding up the areas of rectangles. Left, right, and midpoint sums use different points to determine rectangle heights.

• As the number of rectangles increases, Riemann sums converge to the exact integral: $$\int_a^b f(x),dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x$$

• The Fundamental Theorem of Calculus connects integration and differentiation: if $F’(x) = f(x)$, then $$\int_a^b f(x),dx = F(b) - F(a)$$

• Key properties of integrals include:

  • Linearity: constants factor out, sums split apart
  • Reversing limits changes the sign
  • Integrals over adjacent intervals can be combined

• Integrals have wide applications: finding areas, computing distances from velocities, total quantities from rates, work from variable forces, and average values of functions.

• For increasing functions, left sums underestimate and right sums overestimate. For decreasing functions, the opposite is true.

• The integral gives signed area: regions below the $x$-axis contribute negatively, which is meaningful in applications like displacement.

You now have the foundational understanding of integration. In future lessons, you will learn techniques for finding antiderivatives of more complex functions and explore further applications of this powerful tool.