The Definite Integral: Limits of Riemann Sums and Signed Area

Quick answer

The definite integral ∫ₐᵇ f(x) dx is the limit of Riemann sums as the pieces shrink to zero width, which makes the approximation of the last lesson exact. It measures signed area: regions above the x-axis count positive and regions below count negative, so ∫ of sin x over a full period is 0 even though the shaded area is 4. Integrals add across adjacent intervals, reverse sign when the limits swap, and pass through sums and constant multiples. Many can be found with geometry alone.

What you'll learn

Define the definite integral as a limit of Riemann sums
Interpret a definite integral as signed area
Use the properties of definite integrals
Evaluate integrals of lines, triangles and circles with geometry

From approximation to exact value

Riemann sums approximate the area under a curve, and thinner pieces approximate it better. The definite integral is where they are heading:

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

For a continuous $f$ this limit exists, and every choice of sample points — left, right, midpoint — gives the same limit. The numbers $a$ and $b$ are the limits of integration, and $f$ is the integrand.

Signed area

Each term $f(x_i^*)\,\Delta x$ is negative when $f$ is negative. So the integral counts area above the $x$ -axis as positive and area below as negative:

\int_a^b f(x)\,dx = (\text{area above}) - (\text{area below})

y = sin x

Signed area of sine over a full period

The two regions have the same area, $2$ each. So $\int_0^{2\pi} \sin x\,dx = 0$ , while the total shaded area is $4$ .

Why the limit gives the exact area

For an increasing function, the last lesson showed the left sums sit below the area and the right sums above it. The gap between them is a thin staircase, and as $n$ grows its total width stays $b - a$ while its height, the rise of $f$ across one piece, goes to $0$ . The gap closes.

So the left and right sums squeeze together onto one number, and the area is caught between them. The integral is the one number that every fine enough Riemann sum approaches, which is why it can be computed from any of them.

Properties

For integrable $f$ and $g$ and a constant $c$ :

Property	Statement
zero width	$\int_a^a f(x)\,dx = 0$
reversed limits	$\int_b^a f(x)\,dx = -\int_a^b f(x)\,dx$
adjacent intervals	$\int_a^b f + \int_b^c f = \int_a^c f$
sums and multiples	$\int_a^b (c\,f + g) = c\int_a^b f + \int_a^b g$
comparison	if $f \le g$ on $[a, b]$ , then $\int_a^b f \le \int_a^b g$

Worked examples

Common mistakes

Practice problems

Evaluate $\int_0^5 3\,dx$ with geometry.

Answer

$15$

Full solution

The region is a rectangle $5$ wide and $3$ tall.
Evaluate $\int_{-2}^{2} x\,dx$ .

Answer

$0$

Full solution

Two triangles of area $2$ : one below the axis on $[-2, 0]$ , one above on $[0, 2]$ . They cancel.
Evaluate $\int_0^2 \sqrt{4 - x^2}\,dx$ .

Answer

$\pi$

Full solution

It is a quarter of the circle of radius $2$ : $\tfrac{1}{4}\pi(2)^2 = \pi$ .
Given $\int_1^4 f(x)\,dx = 6$ , find $\int_4^1 f(x)\,dx$ and $\int_1^4 3f(x)\,dx$ .

Answer

$-6$ and $18$

Full solution

Reversing the limits changes the sign; a constant multiple comes out of the integral.
Given $\int_0^2 g = 4$ and $\int_2^5 g = -1$ , find $\int_0^5 g$ .

Answer

$3$

Full solution

Adjacent intervals add: $4 + (-1) = 3$ .
Evaluate $\int_1^3 (x - 2)\,dx$ .

Answer

$0$

Full solution

The line crosses the axis at $2$ . The triangles on $[1, 2]$ and $[2, 3]$ each have area $\tfrac{1}{2}$ , one below and one above.
Find the total area between $y = x - 2$ and the $x$ -axis on $[1, 3]$ .

Answer

$1$

Full solution

Both triangles count as positive for total area: $\tfrac{1}{2} + \tfrac{1}{2} = 1$ .
Evaluate $\int_0^6 |x - 2|\,dx$ .

Answer

$10$

Full solution

Two triangles above the axis: base $2$ , height $2$ (area $2$ ), and base $4$ , height $4$ (area $8$ ).
Is $\int_0^1 x^3\,dx$ larger or smaller than $\int_0^1 x^2\,dx$ ? Why?

Answer

Smaller

Full solution

On $[0, 1]$ , $x^3 \le x^2$ . By the comparison property, the integral of $x^3$ is at most the integral of $x^2$ , and it is strictly smaller because they differ on most of the interval.
A student says $\int_0^{\pi} \cos x\,dx$ must be positive because the curve starts at height $1$ . Evaluate it and explain the error.

Hint

Where is $\cos x$ negative on $[0, \pi]$ ?

Answer

It is $0$ : the area above the axis on $\left[0, \tfrac{\pi}{2}\right]$ cancels the equal area below on $\left[\tfrac{\pi}{2}, \pi\right]$ .

Full solution

$\cos x$ is symmetric about $\left(\tfrac{\pi}{2}, 0\right)$ on $[0, \pi]$ , so the region below the axis is a mirror image of the region above.

The integral is signed area, so the two cancel. The starting height does not decide the sign.

Frequently asked questions

What is a definite integral?

The limit of Riemann sums of f on [a, b] as the width of the pieces goes to 0. For a continuous function the limit exists and does not depend on the sample points.

What does it mean that the integral is signed area?

Area above the x-axis counts as positive and area below counts as negative, so the integral is the area above minus the area below.

What happens if I swap the limits of integration?

The integral changes sign: ∫ from b to a of f equals −∫ from a to b of f.

How is the definite integral different from total area?

Total area counts every region as positive. To find it, integrate |f| or split at the zeros and add the absolute values of the pieces.

What does dx mean in an integral?

It marks the variable of integration and descends from the Δx widths in the Riemann sums the integral is the limit of.

What to learn next

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