The Definite Integral

Riemann Sums

Partition $[a,b]$ into $n$ sub-intervals. For each, erect a rectangle whose height is $f$ evaluated at a chosen sample point. The Riemann sum is

As $n\to\infty$, $S_n\to\int_a^b f(x)\,dx$ for any choice of sample points, provided $f$ is integrable.

Independence of Sample Points

Fix a large $n$. No matter where you pick the sample point within each sub-interval, the sum barely changes - it always converges to the same value.

Drag the red dots below to move sample points within their sub-intervals. The total stays near $\int_0^\pi \sin x\,dx = 2$.

The sample points (red dots) are draggable horizontally within each sub-interval.

Properties of the Definite Integral

We know $\int_0^1 e^{-x^2}\,dx$ exists. Can we pin down its value without finding an antiderivative? Surprisingly, we can squeeze it between two numbers using nothing but common sense.

On $[0,1]$, the exponent $-x^2$ ranges from $0$ to $-1$, so $e^{-1}\le e^{-x^2}\le e^0 = 1$. The function is sandwiched between two constants. Since the interval has length 1:

$e^{-1}\approx 0.368\;\le\;\displaystyle\int_0^1 e^{-x^2}\,dx\;\le\; 1$

This is a special case of the general bounding property:

Linearity. This extends naturally. If we can split integrands and pull out constants, we can decompose hard integrals into easier ones:

$$\int_a^b\bigl[c_1 f + c_2 g\bigr]\,dx = c_1\int_a^b f\,dx + c_2\int_a^b g\,dx$$

Additivity. Integrals respect cutting. If $a < b < c$:

$$\int_a^c f = \int_a^b f + \int_b^c f$$

This is how we handle piecewise functions and singularities.

Mean Value Theorem for Integrals. The bounding property says $m\le\frac{1}{b-a}\int_a^b f\le M$. That middle quantity is the average value of $f$. If $f$ is continuous, the Intermediate Value Theorem says $f$ must hit every value between $m$ and $M$. So there exists $x_1\in[a,b]$ with:

FTC connection. $\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$. Differentiation undoes integration - the rate of change of accumulated area is the height of the curve.

When Integrals Misbehave

Everything so far assumed $f$ is bounded on a finite interval. What happens when the interval stretches to infinity, or $f$ blows up?

Infinite limits. We define

$$\int_a^\infty f\,dx = \lim_{b\to\infty}\int_a^b f\,dx$$

If the limit exists, the integral converges; otherwise it diverges.

Example 1: $\int_1^\infty\frac{dx}{x^2}$. We expect this to converge since $1/x^2$ drops off fast:

$$\int_1^b\frac{dx}{x^2} = \left[-\frac{1}{x}\right]_1^b = 1 - \frac{1}{b}\;\to\; 1 \quad\checkmark$$

Example 2: $\int_0^1\frac{dx}{\sqrt{x}}$. There's a singularity at $x=0$, so we approach it as a limit:

$$\int_\varepsilon^1\frac{dx}{\sqrt{x}} = \left[2\sqrt{x}\right]_\varepsilon^1 = 2 - 2\sqrt{\varepsilon}\;\to\; 2 \quad\checkmark$$

Converges despite the blow-up.

The $p$-test. These two families tell us everything about convergence near $\infty$ and near $0$:

$$\int_1^\infty x^{-p}\,dx \;\text{ converges iff } p > 1$$

$$\int_0^1 x^{-p}\,dx \;\text{ converges iff } p < 1$$