Definite Integrals

students, imagine you need to find the total amount of water that flows through a pipe in one hour, even though the flow rate changes every minute 💧. Or suppose you want the exact area under a curve on a graph, where the height keeps changing. That is where definite integrals become powerful. In this lesson, you will learn what definite integrals mean, how they connect to accumulation, and how they are used in calculus to solve real problems.

By the end of this lesson, you should be able to:

explain the main ideas and terminology behind definite integrals,
use the notation $\int_a^b f(x)\,dx$ correctly,
connect definite integrals to area, accumulation, and the Fundamental Theorem of Calculus,
apply IB-style reasoning to evaluate and interpret definite integrals,
describe how definite integrals fit into the wider study of calculus.

What a definite integral means

A definite integral measures the total accumulation of a quantity over an interval. The expression $\int_a^b f(x)\,dx$ means that the function $f(x)$ is being added up from $x=a$ to $x=b$. The numbers $a$ and $b$ are the limits of integration, and $dx$ shows that the accumulation happens with respect to $x$.

A helpful way to think about this is as a limit of many small rectangles. If a curve is split into many narrow strips, then each strip has width $\Delta x$ and height $f(x)$ for some point in the strip. The area of one strip is approximately $f(x)\Delta x$. Adding all the strips gives an estimate of the total area. As the strips become thinner and more numerous, the estimate becomes exact. This limit process is the idea behind the definite integral.

The notation is important. In $\int_a^b f(x)\,dx$:

$f(x)$ is the integrand,
$x$ is the variable of integration,
$a$ is the lower limit,
$b$ is the upper limit.

If $f(x)\ge 0$ on $[a,b]$, then $\int_a^b f(x)\,dx$ is the area under the curve and above the $x$-axis. If $f(x)$ is negative in part of the interval, then the integral gives signed area, not just geometric area. That means regions below the axis contribute negatively.

For example, if $f(x)=x^2$ on $[0,2]$, then $\int_0^2 x^2\,dx$ gives the total area under the parabola from $x=0$ to $x=2$. Since $x^2\ge 0$ there, this is also the geometric area.

From sum to exact value

Before calculus, people often approximate totals using rectangles or trapezoids. Definite integrals give an exact value when the function is nice enough to integrate. This makes them useful in science and economics.

Suppose the velocity of a car is $v(t)$ metres per second. Then $\int_0^5 v(t)\,dt$ gives the displacement of the car over the first $5$ seconds. Why? Velocity tells you how fast position changes, so adding small velocity changes over time gives total change in position.

This same idea appears in many situations:

distance travelled from a velocity function,
total revenue from a marginal revenue function,
mass from a density function,
total charge from a current function.

A definite integral is therefore an accumulation tool. It combines tiny changes into a final result.

Here is a simple example.

If $f(x)=3$, then

$\int_1^4 3\,dx = 3(4-1)=9.$

This makes sense because a constant function creates a rectangle with height $3$ and width $3$. The area is $9$.

Now consider $f(x)=x$. Then

$\int_0^2 x\,dx = \left[\frac{x^2}{2}\right]_0^2 = \frac{4}{2}-0=2.$

The graph of $y=x$ from $0$ to $2$ forms a right triangle with base $2$ and height $2$, so the area is $\frac{1}{2}\cdot 2\cdot 2=2$.

The Fundamental Theorem of Calculus

One of the most important ideas in calculus is the connection between differentiation and integration. This is called the Fundamental Theorem of Calculus.

If $F'(x)=f(x)$, then

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

This rule turns a difficult accumulation problem into an easier evaluation problem.

That means you do not usually need to compute a limit of rectangles directly in IB Mathematics Analysis and Approaches SL. Instead, you find an antiderivative and subtract its values at the limits.

Example:

$\int_1^3 (2x+1)\,dx.$

An antiderivative of $2x+1$ is $x^2+x$. So

$\int_1^3 (2x+1)\,dx = \left[x^2+x\right]_1^3 = (9+3)-(1+1)=10.$

This result can also be checked geometrically. The graph of $y=2x+1$ is above the $x$-axis on $[1,3]$, so the integral is a positive area.

The Fundamental Theorem of Calculus is powerful because it shows that differentiation and integration are inverse processes. Differentiation finds rate of change. Integration finds total change.

Interpreting signs and area

A common mistake is to think every definite integral equals area. That is not always true. The value of a definite integral is signed area.

If a function crosses the $x$-axis, then parts above the axis count positively and parts below count negatively. For example, if $f(x)$ is positive on $[a,c]$ and negative on $[c,b]$, then

$\int$_a^b f(x)\,dx = $\int$_a^c f(x)\,dx + $\int$_c^b f(x)\,dx.

The second part may reduce the total.

To find total geometric area, you may need to split the integral and use absolute values of each region separately. For instance, if a graph is below the axis on part of the interval, the geometric area there is positive, even though the definite integral over that section is negative.

Example:

If $f(x)=-2$ on $[0,3]$, then

$\int_0^3 -2\,dx = -6.$

But the geometric area between the graph and the $x$-axis is $6$ square units.

This distinction is very important in exam questions. Always read the wording carefully: does the question ask for area, or for the value of the integral?

Using definite integrals in IB-style problems

IB questions often ask you to evaluate, interpret, or apply definite integrals in context. Good reasoning matters as much as calculation.

Example 1: area under a curve

Find the area under $f(x)=x^2$ from $x=1$ to $x=3$.

Since $x^2\ge 0$ on this interval,

$\int_1$^3 x^2\,dx = $\left[$$\frac{x^3}{3}$$\right]_1$^3 = $\frac{27}{3}$-$\frac{1}{3}$=$\frac{26}{3}$.

So the area is $\frac{26}{3}$ square units.

Example 2: displacement from velocity

A particle has velocity $v(t)=4-t$ for $0\le t\le 5$. Find the displacement.

Compute

$\int_0$^5 (4-t)\,dt = $\left[4$t-$\frac{t^2}{2}$$\right]_0$^5 = 20-$\frac{25}{2}$ = $\frac{15}{2}$.

So the displacement is $\frac{15}{2}$ metres. Notice that the velocity is positive at first and negative later, so the particle changes direction. The definite integral still gives net displacement.

Example 3: interpreting context

If $r(t)$ is the rate at which water enters a tank in litres per minute, then

$\int_0^{10} r(t)\,dt$

represents the total volume of water added in the first $10$ minutes. Units matter! The units of the integral are the units of the integrand multiplied by the units of the variable of integration.

That is why if $r(t)$ is in litres per minute and $t$ is in minutes, the result is in litres.

Connection to the rest of calculus

Definite integrals sit at the centre of calculus because they connect several major ideas:

differentiation tells how a quantity changes,
integration tells how much accumulates,
antiderivatives link the two.

In IB Mathematics Analysis and Approaches SL, definite integrals are used in topics such as:

area between a curve and the axes,
area between two curves,
kinematics problems involving displacement and distance,
interpreting rates of change in context.

They also support deeper reasoning. For example, if $f(x)$ is increasing and positive on an interval, then its integral on that interval will reflect both its size and the length of the interval. If the interval is larger, the accumulation is often larger too.

A definite integral can also be estimated numerically when an exact antiderivative is not available. In IB, you may meet methods such as the trapezium rule, which approximates area using trapezia. This connects the exact idea of integration with practical approximation.

Conclusion

Definite integrals are a fundamental part of calculus because they measure total accumulation across an interval. The notation $\int_a^b f(x)\,dx$ represents the exact sum of infinitely many small contributions. Through the Fundamental Theorem of Calculus, you can evaluate definite integrals using antiderivatives, making them much easier to compute.

students, the key ideas to remember are area, accumulation, signed values, and real-world interpretation. Once you understand these ideas, definite integrals become a flexible tool for solving problems in mathematics, physics, and other applied settings.

Study Notes

A definite integral is written as $\int_a^b f(x)\,dx$.
$f(x)$ is the integrand, $a$ is the lower limit, and $b$ is the upper limit.
A definite integral measures total accumulation over an interval.
If $f(x)\ge 0$ on $[a,b]$, then $\int_a^b f(x)\,dx$ equals the area under the curve.
If $f(x)$ is partly below the $x$-axis, the integral gives signed area.
The Fundamental Theorem of Calculus states that if $F'(x)=f(x)$, then $\int_a^b f(x)\,dx=F(b)-F(a)$.
Definite integrals connect directly to differentiation, since integration and differentiation are inverse processes.
In applications, definite integrals can represent displacement, total distance, mass, volume, charge, or revenue.
Units matter: the result of an integral has units from the integrand multiplied by the variable of integration.
For total geometric area, split the interval where the graph crosses the axis and use positive areas.
In IB questions, always check whether the task asks for the value of the integral or the geometric area.
Numerical methods like the trapezium rule estimate definite integrals when exact methods are difficult.