---
title: Approximate and Improper Integrals
module: Techniques of Integration
moduleNumber: 7
lessonNumber: 4
order: 704
summary: >
  Two definite integrals the Fundamental Theorem cannot reach. With no
  antiderivative available, the Midpoint, Trapezoidal, and Simpson rules
  approximate the integral from sample values, each carrying a provable error
  bound. With an infinite interval or an integrand that blows up, the improper
  integral is defined as a limit that either converges or diverges; the
  Comparison Test settles which without evaluating it.
topics: [Techniques of Integration]
sources:
  - book: Stewart
    ref: "Ch. 7 — Techniques of Integration; §7.7 Approximate Integration, §7.8 Improper Integrals"
---

The [Fundamental Theorem](/calculus/integrals/the-fundamental-theorem-of-calculus)
evaluates $\int_a^b f(x)\,\d x$ from an antiderivative of $f$. Two situations put
that out of reach. Either $f$ has no elementary antiderivative — as with
$e^{x^2}$ or $(\sin x)/x$ — or $f$ is known only as measured data with no formula
at all. In both cases the integral is approximated numerically. A separate
failure occurs when the interval runs to infinity or $f$ has an infinite
discontinuity; there the integral is redefined as a limit, and the question
becomes whether that limit exists.

## Numerical approximation

The definite integral is a limit of Riemann sums, so any Riemann sum
approximates it. Divide $[a,b]$ into $n$ subintervals of width
$\Delta x = (b-a)/n$ with endpoints $x_0, x_1, \dots, x_n$. Sampling the left or
right endpoint of each subinterval gives the crude approximations $L_n$ and
$R_n$. Two better sampling choices follow.

> **Definition (Midpoint Rule).** With $\overline{x}_i$ the midpoint of the
> $i$-th subinterval,
> $$
> \int_a^b f(x)\,\d x \approx M_n = \Delta x\,\bigl[f(\overline{x}_1) + f(\overline{x}_2) + \cdots + f(\overline{x}_n)\bigr].
> $$

> **Definition (Trapezoidal Rule).** Averaging the left and right endpoint sums
> gives
> $$
> \int_a^b f(x)\,\d x \approx T_n = \frac{\Delta x}{2}\,\bigl[f(x_0) + 2f(x_1) + 2f(x_2) + \cdots + 2f(x_{n-1}) + f(x_n)\bigr].
> $$

The Midpoint Rule sums rectangles whose height is read at the middle of each
strip; the Trapezoidal Rule replaces each strip's top by the straight chord
joining its endpoints, so the area element is a trapezoid. Both are exact for
straight-line $f$ and err in proportion to how much $f$ curves.

$$
% caption: The Midpoint Rule: each rectangle's height is the value of $f$ at the center of its subinterval, marked by a dot on the curve; the over- and under-shoots on the two halves of a strip partly cancel.
\begin{tikzpicture}[scale=1.0,>=stealth,font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\draw[acc,fill=acc!12] (0,0) rectangle (1,1.55);
\draw[acc,fill=acc!12] (1,0) rectangle (2,2.20);
\draw[acc,fill=acc!12] (2,0) rectangle (3,2.02);
\draw[acc,fill=acc!12] (3,0) rectangle (4,1.66);
\draw[black,very thick] (0,1.15) .. controls (0.5,1.55) and (1.1,2.2) .. (1.5,2.20) .. controls (2.0,2.2) and (2.3,2.02) .. (2.5,2.02) .. controls (3.0,2.0) and (3.3,1.66) .. (3.5,1.66) .. controls (3.8,1.66) and (3.9,1.6) .. (4,1.55);
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\fill[acc] (2.5,2.02) circle (1.7pt);
\fill[acc] (3.5,1.66) circle (1.7pt);
\draw[black] (0,0) -- (4,0);
\node[black,font=\footnotesize] at (2,-0.5){Midpoint: rectangles at strip centers};
\end{tikzpicture}
$$

### Simpson's Rule

Fitting a straight chord ignores curvature. Fitting a **parabola** through three
consecutive points captures it. Group the subintervals in pairs (so $n$ must be
even) and pass a parabola through each triple of points; the area under one
parabola over $[x_{i-1}, x_{i+1}]$ works out to
$\tfrac{\Delta x}{3}\bigl(y_{i-1} + 4y_i + y_{i+1}\bigr)$. Summing gives the rule.

> **Definition (Simpson's Rule).** For even $n$,
> $$
> \int_a^b f(x)\,\d x \approx S_n = \frac{\Delta x}{3}\,\bigl[f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + \cdots + 4f(x_{n-1}) + f(x_n)\bigr],
> $$
> with coefficient pattern $1, 4, 2, 4, 2, \dots, 4, 1$.

$$
% caption: The Trapezoidal Rule tops each strip with a straight chord; Simpson's Rule tops each pair of strips with a parabola through three points, following the curve far more closely.
\begin{tikzpicture}[scale=1.0,>=stealth,font=\small]
\definecolor{acc}{HTML}{4A6FA5}
% ---- left panel: trapezoid ----
\begin{scope}
  \fill[acc!10] (0,0) -- (0,1.0) -- (2,2.2) -- (4,1.6) -- (4,0) -- cycle;
  \draw[black,very thick] (0,1.0) .. controls (1.0,2.0) and (1.4,2.3) .. (2,2.2) .. controls (2.9,2.05) and (3.4,1.75) .. (4,1.6);
  \draw[acc,thick] (0,1.0) -- (2,2.2) -- (4,1.6);
  \draw[black] (0,0) -- (4,0);
  \draw[black,dashed] (0,0)--(0,1.0); \draw[black,dashed] (2,0)--(2,2.2); \draw[black,dashed] (4,0)--(4,1.6);
  \node[black,font=\footnotesize] at (2,-0.5){Trapezoidal: chords};
\end{scope}
% ---- right panel: parabola ----
\begin{scope}[xshift=6cm]
  \fill[acc!10] (0,0) -- (0,1.0) .. controls (1.0,2.05) and (1.4,2.28) .. (2,2.2) .. controls (2.9,2.08) and (3.4,1.72) .. (4,1.6) -- (4,0) -- cycle;
  \draw[black,very thick] (0,1.0) .. controls (1.0,2.0) and (1.4,2.3) .. (2,2.2) .. controls (2.9,2.05) and (3.4,1.75) .. (4,1.6);
  \draw[acc,thick] (0,1.0) .. controls (1.0,2.05) and (1.4,2.28) .. (2,2.2) .. controls (2.9,2.08) and (3.4,1.72) .. (4,1.6);
  \draw[black] (0,0) -- (4,0);
  \draw[black,dashed] (0,0)--(0,1.0); \draw[black,dashed] (2,0)--(2,2.2); \draw[black,dashed] (4,0)--(4,1.6);
  \node[black,font=\footnotesize] at (2,-0.5){Simpson: parabola};
\end{scope}
\end{tikzpicture}
$$

> **Worked example.** Approximate $\displaystyle\int_1^2 \frac{1}{x}\,\d x$ with
> $n = 5$ by the Trapezoidal and Midpoint Rules.
>
> Here $\Delta x = 0.2$, with endpoints $1, 1.2, 1.4, 1.6, 1.8, 2$ and midpoints
> $1.1, 1.3, 1.5, 1.7, 1.9$. The Trapezoidal Rule weights the endpoints
> $1, 2, 2, 2, 2, 1$:
>
> $$
> T_5 = \frac{0.2}{2}\left[\frac{1}{1} + \frac{2}{1.2} + \frac{2}{1.4} + \frac{2}{1.6} + \frac{2}{1.8} + \frac{1}{2}\right] \approx 0.695635.
> $$
>
> The Midpoint Rule sums the midpoint heights:
>
> $$
> M_5 = 0.2\left[\frac{1}{1.1} + \frac{1}{1.3} + \frac{1}{1.5} + \frac{1}{1.7} + \frac{1}{1.9}\right] \approx 0.691908.
> $$
>
> The true value is $\ln 2 \approx 0.693147$, so $T_5$ overshoots by about
> $0.0025$ and $M_5$ undershoots by about $0.0012$: the midpoint error is roughly
> half the trapezoidal error and opposite in sign, as the bounds predict.

### Error bounds

Each rule comes with a provable bound on the error, the amount that must be added
to the approximation to make it exact. The bounds depend on how large a
derivative of $f$ can be on $[a,b]$.[^errors]

> **Theorem (Error bounds).** Suppose $|f''(x)| \le K$ on $[a,b]$. Then the
> Trapezoidal and Midpoint errors satisfy
> $$
> |E_T| \le \frac{K(b-a)^3}{12n^2}, \qquad |E_M| \le \frac{K(b-a)^3}{24n^2}.
> $$
> If instead $|f^{(4)}(x)| \le K$, Simpson's error satisfies
> $$
> |E_S| \le \frac{K(b-a)^5}{180\,n^4}.
> $$

The Midpoint error bound is half the Trapezoidal, and
in practice their errors have opposite signs — Simpson's Rule combines the two as
the weighted average $S_{2n} = \tfrac{1}{3}T_n + \tfrac{2}{3}M_n$. Doubling $n$ cuts
the trapezoidal and midpoint errors by about $4$ (the $n^2$), but cuts Simpson's
error by about $16$ (the $n^4$). Because $f^{(4)}$ vanishes for any cubic,
Simpson's Rule is exact on polynomials of degree $3$ or lower.

The rates show clearly when the errors for $\int_1^2 \tfrac{1}{x}\,\d x$ (true value
$\ln 2 \approx 0.693147$) are tabulated against $n$.

| $n$ | $T_n$ error | $M_n$ error | $S_n$ error |
| --- | --- | --- | --- |
| $4$ | $-1.9\times10^{-3}$ | $9.6\times10^{-4}$ | $-7.3\times10^{-6}$ |
| $8$ | $-4.9\times10^{-4}$ | $2.4\times10^{-4}$ | $-4.7\times10^{-7}$ |
| $16$ | $-1.2\times10^{-4}$ | $6.1\times10^{-5}$ | $-3.0\times10^{-8}$ |

$$
% caption: Error against the number of subintervals on log axes: the trapezoidal and midpoint errors fall as $1/n^2$ (slope $-2$), Simpson's as $1/n^4$ (slope $-4$).
\begin{tikzpicture}[scale=1.0,>=stealth,font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->,black] (0,0) -- (5.4,0) node[right,black,font=\footnotesize]{$\log n$};
\draw[->,black] (0,0) -- (0,4.6) node[above,black,font=\footnotesize]{$\log |E|$};
% trapezoid line slope -2 (steeper down): from (1,4) to (5,2)
\draw[black,thick] (1,4.0) -- (5,2.0);
\fill[black] (1,4.0) circle (1.6pt); \fill[black] (3,3.0) circle (1.6pt); \fill[black] (5,2.0) circle (1.6pt);
\node[black,font=\footnotesize,anchor=west] at (5.05,2.05){trapezoid};
% midpoint line, parallel, lower
\draw[black,thick,dashed] (1,3.4) -- (5,1.4);
\node[black,font=\footnotesize,anchor=west] at (3.6,1.5){midpoint};
% simpson line slope -4 (steeper): from (1,3.5) to (5,-0.5) but keep positive: (1,3.6) to (4,0.6)
\draw[acc,thick] (1,3.6) -- (4.2,0.4);
\fill[acc] (1,3.6) circle (1.6pt); \fill[acc] (2.6,2.0) circle (1.6pt); \fill[acc] (4.2,0.4) circle (1.6pt);
\node[acc,font=\footnotesize,anchor=west] at (2.9,2.15){Simpson (slope $-4$)};
\end{tikzpicture}
$$

To guarantee accuracy within a tolerance, solve the bound for $n$. For
$\int_1^2 \tfrac{1}{x}\,\d x$, here $f''(x) = 2/x^3$ so $K = 2$ on $[1,2]$; the
Trapezoidal bound $2/(12n^2) < 0.0001$ forces $n \ge 41$, while Simpson's
$2/(180n^4) < 0.0001$ needs only $n \ge 8$.

The bound also caps the error of a computed approximation before the true value
is known.

> **Worked example.** Bound the Midpoint error for
> $\displaystyle\int_0^1 e^{x^2}\,\d x$ with $n = 10$.
>
> The rule gives $M_{10} \approx 1.460393$. For the bound, differentiate twice:
> $f'(x) = 2x e^{x^2}$ and $f''(x) = (2 + 4x^2)e^{x^2}$. On $[0,1]$ both $x^2 \le 1$
> and $e^{x^2} \le e$, so
>
> $$
> |f''(x)| = (2 + 4x^2)e^{x^2} \le 6e,
> $$
>
> giving $K = 6e$. With $a = 0$, $b = 1$, $n = 10$,
>
> $$
> |E_M| \le \frac{6e\,(1)^3}{24 \cdot 10^2} = \frac{e}{400} \approx 0.007.
> $$
>
> The approximation is therefore correct to about two decimal places; the actual
> error is near $0.0023$, well inside the worst-case bound.

## Improper integrals over infinite intervals

The region under $y = 1/x^2$ to the right of $x = 1$ extends forever, yet its area
is finite. The area out to $x = t$ is

$$
\int_1^t \frac{1}{x^2}\,\d x = 1 - \frac{1}{t},
$$

which approaches $1$ as $t \to \infty$. Defining the integral over $[1,\infty)$ as
this limit gives it the value $1$.

> **Definition (Type 1 improper integral).** If $\int_a^t f(x)\,\d x$ exists for all
> $t \ge a$, then
> $$
> \int_a^{\infty} f(x)\,\d x = \lim_{t \to \infty} \int_a^t f(x)\,\d x,
> $$
> and the integral is **convergent** if this limit exists as a finite number,
> **divergent** otherwise. The lower-infinite and doubly-infinite cases are
> defined the same way, the latter split as
> $\int_{-\infty}^{\infty} = \int_{-\infty}^{a} + \int_a^{\infty}$.

The nearly identical curve $y = 1/x$ behaves oppositely:

$$
\int_1^t \frac{1}{x}\,\d x = \ln t \to \infty,
$$

so $\int_1^{\infty} \tfrac{1}{x}\,\d x$ diverges. The two curves look alike for
large $x$ and both tend to $0$, but $1/x^2$ decays fast enough for finite area
while $1/x$ does not.

$$
% caption: The regions under $1/x^2$ and $1/x$ to the right of $x=1$: the faster decay of $1/x^2$ gives finite area, while $1/x$ encloses infinite area.
\begin{tikzpicture}[scale=1.0,>=stealth,font=\small]
\definecolor{acc}{HTML}{4A6FA5}
% left: 1/x^2 finite
\begin{scope}
  \draw[->,black] (0,0) -- (4.4,0) node[right,black,font=\footnotesize]{$x$};
  \draw[->,black] (0,0) -- (0,3.2) node[above,black,font=\footnotesize]{$y$};
  \fill[acc!12] (1,0) -- (1,2.4) .. controls (1.5,1.07) and (2.2,0.5) .. (4.1,0.14) -- (4.1,0) -- cycle;
  \draw[black,very thick] (0.63,3.0) .. controls (0.85,2.4) and (1.0,2.4) .. (1,2.4) .. controls (1.5,1.07) and (2.2,0.5) .. (4.1,0.14);
  \draw[black,dashed] (1,0) -- (1,2.4);
  \node[below,black,font=\footnotesize] at (1,0){$1$};
  \node[black,font=\footnotesize] at (2.4,1.4){finite area};
  \node[black,font=\footnotesize,anchor=west] at (2.4,2.4){$y=1/x^2$};
\end{scope}
% right: 1/x infinite
\begin{scope}[xshift=6.2cm]
  \draw[->,black] (0,0) -- (4.4,0) node[right,black,font=\footnotesize]{$x$};
  \draw[->,black] (0,0) -- (0,3.2) node[above,black,font=\footnotesize]{$y$};
  \fill[acc!12] (1,0) -- (1,2.4) .. controls (1.8,1.33) and (2.8,0.86) .. (4.1,0.585) -- (4.1,0) -- cycle;
  \draw[black,very thick] (0.8,3.0) .. controls (0.9,2.66) and (1,2.4) .. (1,2.4) .. controls (1.8,1.33) and (2.8,0.86) .. (4.1,0.585);
  \draw[black,dashed] (1,0) -- (1,2.4);
  \node[below,black,font=\footnotesize] at (1,0){$1$};
  \node[black,font=\footnotesize] at (2.5,1.55){infinite area};
  \node[black,font=\footnotesize,anchor=west] at (2.6,2.4){$y=1/x$};
\end{scope}
\end{tikzpicture}
$$

The pattern generalizes to a benchmark family, the **$p$-integrals**, obtained by
the same limit computation for a general exponent.

> **Theorem ($p$-integral).** The integral $\displaystyle\int_1^{\infty} \frac{1}{x^p}\,\d x$
> converges if $p > 1$ and diverges if $p \le 1$.

A doubly-infinite integral splits at any finite point and converges only if both
halves do.

> **Worked example.** Evaluate $\displaystyle\int_{-\infty}^{\infty} \frac{\d x}{1 + x^2}$.
>
> Split at $0$ and take each half as a limit. With antiderivative $\tan^{-1}x$,
>
> $$
> \int_0^{\infty} \frac{\d x}{1 + x^2} = \lim_{t \to \infty}\bigl[\tan^{-1}x\bigr]_0^t = \lim_{t \to \infty}\tan^{-1}t = \frac{\pi}{2},
> $$
>
> and by symmetry $\int_{-\infty}^0 \tfrac{\d x}{1 + x^2} = \tfrac{\pi}{2}$ as well.
> Both halves are finite, so
>
> $$
> \int_{-\infty}^{\infty} \frac{\d x}{1 + x^2} = \frac{\pi}{2} + \frac{\pi}{2} = \pi.
> $$

## Improper integrals with a discontinuous integrand

The second kind of impropriety comes from the integrand rather than the interval.
If $f$ has a vertical asymptote at an endpoint, integrate up to a movable bound and
take the limit toward the bad point.

> **Definition (Type 2 improper integral).** If $f$ is continuous on $[a,b)$ and
> discontinuous at $b$, then
> $$
> \int_a^b f(x)\,\d x = \lim_{t \to b^-} \int_a^t f(x)\,\d x,
> $$
> convergent if the limit is finite. A discontinuity at $a$ is handled
> symmetrically; an interior discontinuity at $c$ splits the integral into
> $\int_a^c + \int_c^b$, and the whole converges only if **both** pieces do.

For $\displaystyle\int_2^5 \frac{\d x}{\sqrt{x-2}}$, the integrand blows up at the
left endpoint $x = 2$, so

$$
\int_2^5 \frac{\d x}{\sqrt{x-2}} = \lim_{t \to 2^+} \Bigl[2\sqrt{x-2}\Bigr]_t^5 = \lim_{t\to 2^+}\bigl(2\sqrt{3} - 2\sqrt{t-2}\bigr) = 2\sqrt{3},
$$

a finite value: the region is unbounded in height but has finite area. A
logarithmic example converges too:
$\int_0^1 \ln x\,\d x = \lim_{t\to0^+}\bigl[x\ln x - x\bigr]_t^1 = -1$.

$$
% caption: The region under $1/\sqrt{x-2}$ from $x=2$ to $5$: the integrand shoots to infinity at the left endpoint, but the tail is thin enough that the area is finite, equal to $2\sqrt{3}$.
\begin{tikzpicture}[scale=1.0,>=stealth,font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->,black] (-0.3,0) -- (5.4,0) node[right,black]{$x$};
\draw[->,black] (0,-0.3) -- (0,3.5) node[above,black]{$y$};
\draw[black,dashed] (1,0) -- (1,3.35);
\fill[acc!12] (1,0) -- (1,3.1) .. controls (1.35,1.55) and (2.5,0.9) .. (4.6,0.62) -- (4.6,0) -- cycle;
\draw[black,very thick] (1.06,3.15) .. controls (1.35,1.55) and (2.5,0.9) .. (4.6,0.62);
\node[below,black,font=\footnotesize] at (1,0){$2$};
\node[below,black,font=\footnotesize] at (4.6,0){$5$};
\node[black,font=\footnotesize] at (2.7,0.55){bounded area};
\end{tikzpicture}
$$

The interior case demands care. The integrand of
$\int_0^3 \dfrac{\d x}{x-1}$ has an asymptote at $x = 1$ **inside** the interval.
Splitting at $c = 1$, the piece $\int_0^1 \tfrac{\d x}{x-1}$ already diverges, so the
whole integral diverges. Treating it as an ordinary integral and writing
$\ln|x-1|\big|_0^3 = \ln 2$ is wrong: it ignores the discontinuity. Always scan
the interval for points where the integrand is undefined before evaluating.

## The Comparison Test

When an improper integral has no elementary antiderivative, its value cannot be
computed directly, but convergence can still be settled by comparison with a
simpler integral of known behavior.

> **Theorem (Comparison Test).** Suppose $f$ and $g$ are continuous with
> $f(x) \ge g(x) \ge 0$ for $x \ge a$.
>
> - If $\displaystyle\int_a^{\infty} f(x)\,\d x$ converges, so does
>   $\displaystyle\int_a^{\infty} g(x)\,\d x$.
> - If $\displaystyle\int_a^{\infty} g(x)\,\d x$ diverges, so does
>   $\displaystyle\int_a^{\infty} f(x)\,\d x$.

A finite area under the upper curve caps the area under the lower one; an infinite
area under the lower curve forces the upper one to be infinite too.

$$
% caption: The Comparison Test: with $0\le g\le f$, finite area under $f$ bounds the area under $g$, and infinite area under $g$ forces the same for $f$.
\begin{tikzpicture}[scale=1.0,>=stealth,font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->,black] (0,0) -- (6.4,0) node[right,black,font=\footnotesize]{$x$};
\draw[->,black] (0,0) -- (0,3.4) node[above,black,font=\footnotesize]{$y$};
\fill[acc!10] (1,0) -- (1,1.4) .. controls (2.6,0.7) and (4.2,0.42) .. (6.1,0.28) -- (6.1,0) -- cycle;
\draw[acc,very thick] (1,2.6) .. controls (2.6,1.6) and (4.2,1.05) .. (6.1,0.85) node[right,black,font=\footnotesize]{$f$};
\draw[black,thick] (1,1.4) .. controls (2.6,0.7) and (4.2,0.42) .. (6.1,0.28) node[right,black,font=\footnotesize]{$g$};
\draw[black,dashed] (1,0) -- (1,2.6);
\node[below,black,font=\footnotesize] at (1,0){$a$};
\end{tikzpicture}
$$

The classic application is the Gaussian integral. The antiderivative of
$e^{-x^2}$ is not elementary, yet its convergence is easy. For $x \ge 1$,
$x^2 \ge x$, so $e^{-x^2} \le e^{-x}$, and

$$
\int_1^{\infty} e^{-x}\,\d x = \lim_{t\to\infty}\bigl(e^{-1} - e^{-t}\bigr) = e^{-1}
$$

converges. By comparison $\int_1^{\infty} e^{-x^2}\,\d x$ converges, and adding the
ordinary integral over $[0,1]$ shows $\int_0^{\infty} e^{-x^2}\,\d x$ converges. Its
exact value $\tfrac{1}{2}\sqrt{\pi}$ comes later, from multivariable methods, and
underlies the normal distribution used in
[probability](/calculus/applications-of-integration/physics-economics-and-probability).
The opposite conclusion follows from a divergent lower bound: since
$\dfrac{1 + e^{-x}}{x} \ge \dfrac{1}{x}$ and $\int_1^{\infty} \tfrac{1}{x}\,\d x$
diverges, so does $\int_1^{\infty} \dfrac{1 + e^{-x}}{x}\,\d x$.

| Impropriety | Cause | Definition | Decide by |
| --- | --- | --- | --- |
| Type 1 | infinite interval | limit as bound $\to \pm\infty$ | direct limit, $p$-integral, comparison |
| Type 2 | infinite discontinuity | limit toward the bad point | direct limit, comparison |
| interior discontinuity | asymptote inside $[a,b]$ | split at $c$; both pieces must converge | evaluate each piece |

Between the exact methods and the approximation and limit definitions,
every definite integral that arises in practice can be either evaluated
exactly, approximated to a stated tolerance, or bounded for
convergence.[^improper]

[^errors]: **Stewart**, _Calculus_, §7.7 — Approximate Integration: the Midpoint, Trapezoidal, and Simpson rules with their error bounds and the observation that Simpson is exact on cubics.
[^improper]: **Stewart**, _Calculus_, §7.8 — Improper Integrals: the Type 1 and Type 2 definitions, the $p$-integral, and the Comparison Test.