Applications of Integration

How do we measure the area of a region whose top boundary is a curve rather than a straight line? The idea, going back to Archimedes, is to approximate the region by many thin rectangles, add up their areas, and then let the rectangles become infinitely thin.

Take a continuous function $f$ on $[a, b]$. Split the interval into $n$ equal pieces, each of width $\Delta x = \dfrac{b-a}{n}$, with sample points $x_1, x_2, \dots, x_n$. The total area of the rectangles is the Riemann sum $\sum_{i=1}^{n} f(x_i)\,\Delta x$.

Here $a$ and $b$ are the lower and upper limits, $f(x)$ is the integrand, and $dx$ is the infinitesimal width the $\Delta x$ becomes in the limit. When $f(x) \ge 0$, this number is exactly the area under the curve between $x = a$ and $x = b$.

Before we shortcut everything with antiderivatives, it is worth computing an integral straight from the definition once, so the limit-of-a-sum idea feels concrete. The working formula, with $a$ and $b$ the limits and $h = \dfrac{b-a}{n}$, is:

The two sums you need are $\displaystyle\sum_{r=1}^{n} r = \dfrac{n(n+1)}{2}$ and $\displaystyle\sum_{r=1}^{n} r^3 = \left[\dfrac{n(n+1)}{2}\right]^2$. After substituting, divide through by the powers of $n$ and let $n \to \infty$.

Computing that limit directly is painful. The Fundamental Theorems rescue us by linking integration to differentiation — the two turn out to be inverse operations.

The First Fundamental Theorem says that integration builds an antiderivative. If we define $F(x) = \int_a^x f(t)\,dt$, then $F$ is differentiable and $F'(x) = f(x)$.

The Second Fundamental Theorem turns this into an evaluation recipe: to compute a definite integral, find any antiderivative $F$ (so $F' = f$) and subtract its values at the two limits.

A handful of properties, all consequences of the definition, let you simplify integrals before evaluating them — and sometimes evaluate them with almost no work.

For integrals symmetric about the origin, the parity of $f$ gives an instant answer:

When integrating a product $u \cdot v$ where $u$ is a polynomial (so its derivatives eventually vanish), Bernoulli's formula packages repeated integration by parts into a single alternating-sign expression.

Let $u', u'', u''', \dots$ be the successive derivatives of $u$, and let $v_1, v_2, v_3, \dots$ be the successive integrals of $v$ (so $v_1 = \int v\,dx$, $v_2 = \int v_1\,dx$, and so on). Then:

The geometric meaning of the definite integral is area. For a curve $y = f(x)$ lying above the $x$-axis between $x = a$ and $x = b$, the area of the enclosed region is simply the integral.

When a region is described more naturally in terms of $y$ — bounded on the left by the $y$-axis and by horizontal lines $y = c$ and $y = d$ — integrate $x$ with respect to $y$:

To find the area of a region trapped between two curves $y = f(x)$ (upper) and $y = g(x)$ (lower), integrate the gap between them across the interval where they overlap.

The limits $a$ and $b$ are usually the $x$-coordinates where the curves meet, found by solving $f(x) = g(x)$. Always subtract lower from upper so the height is non-negative.

Revolve the region under $y = f(x)$ about the $x$-axis and it sweeps out a solid. Slice the solid perpendicular to the axis: each thin slice is a disk of radius $y$ and thickness $dx$, with volume $\pi y^2\,dx$. Summing these disks gives the volume.

When the region is revolved about the $y$-axis instead, swap the roles of the variables:

An improper integral has an infinite limit or an integrand that blows up inside the range. We handle it as a limit — replace the troublesome bound by a variable and let it run to its target.

If the limit exists and is finite, the improper integral converges; otherwise it diverges.

So $\Gamma(1) = 0! = 1$, $\Gamma(5) = 4! = 24$. A useful non-integer value is $\Gamma\!\left(\tfrac{1}{2}\right) = \sqrt{\pi}$. A closely related standard result is $\int_0^{\infty} e^{-ax} x^n\,dx = \dfrac{n!}{a^{\,n+1}}$.

A reduction formula expresses an integral $I_n$ that depends on an integer $n$ in terms of a lower one. Applying it repeatedly drops the exponent until a known base case remains. The headline example is the power of sine over a quarter period:

The base cases are $I_0 = \int_0^{\pi/2} 1\,dx = \tfrac{\pi}{2}$ and $I_1 = \int_0^{\pi/2}\sin x\,dx = 1$. Cascading the formula gives Wallis' formula — the same result holds for $\cos^n x$: