Integration

The definite integral measures accumulated quantity: area under a curve, total probability, an average value. The fundamental theorem of calculus turns its computation into finding an antiderivative. The working tasks are evaluating a definite or improper integral (a symmetry property often makes this quick), differentiating an integral with variable limits, finding an area between curves, and making a trapezoidal or Simpson estimate.

$F$ is an antiderivative of $f$ if $F' = f$. Any two differ by a constant, written $\int f(x)\,dx = F(x) + C$. Standard forms:

Split $[a, b]$ into $n$ pieces of width $h = (b - a)/n$ and pick a point $x_k^*$ in each. The Riemann sum $\sum f(x_k^*)h$ adds rectangle areas. If these sums tend to the same limit for every choice as $h \to 0$, that limit is $\int_a^b f(x)\,dx$. Every continuous function on $[a, b]$ is integrable, and so is every bounded function with finitely many jumps.

Area counts with sign: parts of the graph below the axis contribute negatively. For the geometric area use $\int_a^b |f(x)|\,dx$.

Part 1. If $f$ is continuous on $[a, b]$, then $G(x) = \int_a^x f(t)\,dt$ is differentiable and $G'(x) = f(x)$.

Part 2. If $F$ is any antiderivative of continuous $f$, then $\int_a^b f(x)\,dx = F(b) - F(a)$.

Variable limits (Leibniz rule). By Part 1 and the chain rule,

An integral with an infinite limit or an unbounded integrand is defined as a limit, for example $\int_1^\infty f = \lim_{R \to \infty}\int_1^R f$. It converges if the limit is finite.

Convergence trap. $\int_{-1}^{1} dx/x^2$ is improper (unbounded at 0) and diverges. Applying the antiderivative blindly gives $-2$, which is wrong.

Gamma integral. Let $I_n = \int_0^\infty x^n e^{-x}\,dx$, so $I_0 = 1$. For $n \ge 1$, integrate by parts with $u = x^n$ and $dv = e^{-x}\,dx$:

because $x^n e^{-x} \to 0$ as $x \to \infty$ and the bracket is 0 at $x = 0$. Hence $I_n = n!$. Substituting $x = at$ gives $\int_0^\infty t^n e^{-at}\,dt = n!/a^{n+1}$ for $a > 0$, the source of the exponential moments $E[X^n] = n!/\lambda^n$ in Continuous Distributions. The gamma function $\Gamma(s) = \int_0^\infty x^{s-1} e^{-x}\,dx$ extends the factorial: $\Gamma(n + 1) = n!$.

Gaussian integral. Let $G = \int_{-\infty}^{\infty} e^{-x^2}\,dx$. It converges, since $x^2 \ge |x| - \frac14$ gives $e^{-x^2} \le e^{1/4} e^{-|x|}$. Square it, write the square as a double integral over the plane, and change to polar coordinates ($x = r\cos\theta$, $y = r\sin\theta$, area element $r\,dr\,d\theta$):

because $\int_0^\infty r e^{-r^2}\,dr = \big[-\tfrac12 e^{-r^2}\big]_0^\infty = \tfrac12$. Since $G > 0$, $G = \sqrt{\pi}$. To make the limit step honest, note that the square $[-R, R]^2$ lies between the discs of radius $R$ and $R\sqrt{2}$, so

and both bounds tend to $\pi$. Consequences: $\int_0^\infty e^{-x^2}\,dx = \sqrt{\pi}/2$ (even integrand); $\int_{-\infty}^{\infty} e^{-t^2/2}\,dt = \sqrt{2\pi}$ (substitute $x = t/\sqrt{2}$), the normalising constant of the normal density; and $\Gamma(1/2) = \int_0^\infty x^{-1/2} e^{-x}\,dx = \sqrt{\pi}$ (substitute $x = t^2$).

With $n$ strips of width $h$ and values $y_0, \dots, y_n$:

For $x^2$ on $[0, 2]$ with $n$ right-endpoint rectangles, $h = 2/n$ and the sum is $h^3\sum_{k=1}^{n} k^2 = \frac{8(n + 1)(2n + 1)}{6n^2}$. For $n = 4$ this is 3.75; as $n \to \infty$ it tends to $16/6 = 8/3$, matching $[x^3/3]_0^2$.

$\frac{d}{dx}\int_0^{x^2}\sin t\,dt = 2x\sin(x^2)$. A limit that needs it: $\lim_{x \to 0}\frac{1}{x^3}\int_0^x \sin(t^2)\,dt$ is $0/0$, and l'Hôpital gives $\lim \frac{\sin(x^2)}{3x^2} = \frac{1}{3}$.

$\int_0^1 2xe^{x^2}\,dx$: with $u = x^2$, the limits become 0 and 1, so the value is $\int_0^1 e^u\,du = e - 1 \approx 1.718$.

$\int_0^1 xe^x\,dx = [xe^x]_0^1 - \int_0^1 e^x\,dx = e - (e - 1) = 1$. Similarly $\int_1^e \ln x\,dx = [x\ln x - x]_1^e = 0 - (-1) = 1$.

$I = \int_0^{\pi/2}\frac{\sin x}{\sin x + \cos x}\,dx$. Replacing $x$ by $\pi/2 - x$ swaps sine and cosine, so $I$ also equals the integral of $\frac{\cos x}{\sin x + \cos x}$. Adding, $2I = \int_0^{\pi/2} 1\,dx = \pi/2$, so $I = \pi/4$.

$\int_{-1}^{1} x^3\cos x\,dx = 0$ (odd integrand). $\int_{-2}^{2} |x|\,dx = 2\int_0^2 x\,dx = 4$. $\int_{-1}^{1}(x^3 + x^2)\,dx = 0 + 2/3 = 2/3$.

$\int_1^\infty x^{-2}\,dx = 1$; $\int_1^\infty x^{-1}\,dx$ diverges ($\ln R \to \infty$); $\int_0^1 x^{-1/2}\,dx = 2$. $\int_0^\infty xe^{-x}\,dx = 1! = 1$: the mean of an exponential random variable with rate 1.

$y = x$ and $y = x^2$ meet at $x = 0$ and $x = 1$, with $x \ge x^2$ between. Area $= \int_0^1 (x - x^2)\,dx = 1/2 - 1/3 = 1/6$.

$\int_0^1\frac{dx}{(x + 1)(x + 2)} = \left[\ln\frac{x + 1}{x + 2}\right]_0^1 = \ln\frac{2}{3} - \ln\frac{1}{2} = \ln\frac{4}{3} \approx 0.2877$.

$\int_0^1\frac{dx}{1 + x} = \ln 2 \approx 0.693147$. With $h = 0.5$, the values are $1, 2/3, 1/2$.

The average of $\sin x$ on $[0, \pi]$ is $\frac{1}{\pi}\int_0^\pi \sin x\,dx = \frac{2}{\pi} \approx 0.637$.