Definite Integrals

Don’t rush things

1.Adrien-Marie Legendre is one of the best-known and most influential mathematicians of the eighteenth century. He was born in Paris in 1752, and around the age of thirty he began to take an interest in the theory of elliptic integrals, mathematical objects used to compute the length of an arc of an ellipse.

2.The problem is not trivial, and it was even less so at the time, when no established theory for solving it yet existed. For an ellipse with semi-major axis $a$ and eccentricity $e\in[0,1),$ the perimeter is given by the following complete elliptic integral of the second kind:

$$ L=4aE(e)=4a\int_{0}^{\pi/2}\sqrt{1-e^2\sin^2\theta} \ d\theta $$

3.We will not deal with such complicated objects in our exposition, but will focus on the kinds of definite integrals normally covered in secondary school and in calculus courses during the first years of university. Be aware, though, that monsters of this sort exist, and that they call for methods of solution that are anything but trivial.

4.In the formula, $\pi/2$ and $0$ are the limits of integration, which distinguish definite integrals from indefinite ones, at least as far as notation goes.

5.Returning to our story, the sources give 1786 as the year of Legendre's first publications on the theory of elliptic integrals. His research between 1811 and 1819 led to the publication of further volumes devoted to the development of the theory and to its applications. Dissatisfied with his own work, a few years later he ended up rewriting the entire corpus from scratch, and he kept it up to date until the last years of his life.

6.One morning in early August, Legendre received a letter from Carl Gustav Jacobi, then twenty-two years old, which offered a new perspective on the subject and put forward a new theory that made the one Legendre had worked on for nearly forty years obsolete. The young Norwegian Niels Henrik Abel arrived at the same results.

7.The sources report that Legendre openly supported the work of the two young men, even though their work had rendered four decades of his research almost entirely obsolete. Abel died at 26. Jacobi at 46. Legendre, on the other hand, lived to be eighty.

8.To come back to our subject, we will not need forty years to learn how to evaluate definite integrals. It will take us far less. If you have already understood what an indefinite integral is and have memorised its main properties together with the basic antiderivatives, you are well on your way. Otherwise, I advise you to stop here and take the necessary step back.

9.In short, a definite integral represents the accumulation of a quantity given by a function $y=f(x)$ over an interval $[a, b]$. In less precise but still valid terms, it measures the signed area between a given curve and the $x$-axis, where the portions lying above the $x$-axis are added and those below are subtracted.

10.This entry deals with the main definitions and properties of definite integrals, while the methods and integration strategies, which form the problem-solving core of integral calculus, are covered in their own sections.

Area under a function: from curve to integral

11.Consider a function $f(x)$ defined on a closed interval $[a, b]$, with $a < b$. If $f(x)$ is continuous and nonnegative on $[a, b]$, its graph, the $x$-axis, and the vertical lines $x = a$ and $x = b$ bound a curvilinear trapezoid. The area of this region is given by the definite integral:

$$\int_{a}^{b} f(x) \ dx$$

12.The standard formulas of elementary geometry do not apply directly to a general curvilinear trapezoid, because one of its boundaries is a curve rather than a straight segment. The area of the curvilinear trapezoid can be approximated by dividing the interval $[a, b]$ into $n$ subintervals of equal width:

$$\Delta x = \frac{b - a}{n}$$

13.Over each subinterval, the region is approximated by a rectangle, and the sum of the areas of these rectangles provides an estimate of the total area.

14.Write the partition points as $x_i = a + i\Delta x,$ for $i=0,\ldots,n.$ Then $x_0=a$ and $x_n=b,$ and the $i$-th subinterval is $[x_{i-1},x_i].$ Denoting by $m_i$ and $M_i$ the infimum and supremum of $f(x)$ on this subinterval, the lower and upper sums are defined as:

$$s_n^{-} = \sum_{i=1}^{n} m_i \Delta x \qquad s_n^{+} = \sum_{i=1}^{n} M_i \Delta x$$

15.The lower sum $s_n^{-}$ approximates the area from below, while the upper sum $s_n^{+}$ approximates it from above.

16.The data for the lower and upper rectangles follow the same pattern on every subinterval:

17.The equal-width construction is a special case of a partition whose subintervals need not have equal widths. For an arbitrary partition $P$ given by $a = x_0 < x_1 < \cdots < x_n = b,$ the lower and upper sums are:

$$ \begin{align} L(f, P) &= \sum_{i=1}^{n} m_i(x_i - x_{i-1}) \\[6pt] U(f, P) &= \sum_{i=1}^{n} M_i(x_i - x_{i-1}) \end{align} $$

18.If a partition $P'$ refines $P$ by adding division points, the lower sum cannot decrease and the upper sum cannot increase. Consequently:

$$L(f, P) \leq L(f, P') \leq U(f, P') \leq U(f, P)$$

19.For a bounded function $f(x)$ on $[a, b],$ the lower and upper integrals collect the best estimates obtained from all possible partitions:

$$ \begin{align} L(f, [a, b]) &= \sup_P L(f, P) \\[6pt] U(f, [a, b]) &= \inf_P U(f, P) \end{align} $$

20.Every lower sum is no greater than every upper sum, so $L(f, [a, b]) \leq U(f, [a, b]).$ The function $f(x)$ is Riemann integrable precisely when these two values coincide. Their common value is the definite integral:

$$L(f, [a, b]) = U(f, [a, b]) = \int_{a}^{b} f(x) \ dx$$

21.An equivalent definition of the definite integral uses tagged Riemann sums. For a partition $P$ given by $a = x_0 < x_1 < \cdots < x_n = b,$ choose a point $\xi_i \in [x_{i-1},x_i]$ in each subinterval and set $\Delta x_i = x_i - x_{i-1}.$ The point $\xi_i$ is called the tag of the $i$-th subinterval, and the mesh of $P$ is defined by:

$$ \|P\| = \max_{1 \leq i \leq n} \Delta x_i $$

22.A bounded function $f$ is Riemann integrable with integral $I$ if, for every $\varepsilon>0,$ there exists $\delta>0$ such that every partition $P$ and every choice of tags satisfy:

$$ \|P\|<\delta \implies \left|\sum_{i=1}^{n}f(\xi_i)\Delta x_i-I\right|<\varepsilon $$

23.This condition requires all tagged Riemann sums with sufficiently small mesh to approach the same value. It is commonly abbreviated by the notation:

$$ \int_a^b f(x) \ dx = \lim_{\|P\| \to 0} \sum_{i=1}^{n} f(\xi_i)\Delta x_i $$

24.The displayed limit therefore ranges over all tagged partitions rather than over a single prescribed sequence. This entry will use the Darboux formulation through $L(f,P)$ and $U(f,P).$ Every tagged Riemann sum lies between the corresponding lower and upper sums. Together with estimates for sufficiently fine partitions, this bound proves that a bounded function on $[a,b]$ is integrable under one definition if and only if it is integrable under the other, and both definitions assign the same value.

25.

John K. Hunter presents the formulation through tagged partitions and proves its equivalence with the Darboux definition in Introduction to Analysis, listed in the bibliography.

26.Every continuous real-valued function on $[a, b]$ is Riemann integrable. Continuity on this interval implies uniform continuity, which makes the oscillation $M_i - m_i$ uniformly small when the subintervals are sufficiently short. The values $a$ and $b$ are the lower and upper limits of integration, and $f(x)$ is the integrand. The notation $f(x) \ dx$ is suggested by the area $f(x)\Delta x$ of each approximating rectangle. The symbol $dx$ identifies $x$ as the integration variable and records the limiting role of the subinterval widths.

Computing definite integrals

27.If $f(x)$ is continuous on $[a, b]$ and $F(x)$ is any antiderivative of $f(x)$, then the definite integral is given by the difference of the antiderivative at the endpoints:

$$\int_{a}^{b} f(x) \ dx = F(b) - F(a)$$

28.The two quantities appearing in this expression have the following meaning:

30.This formula is the conclusion of the second Fundamental Theorem of Calculus. The first part of the theorem establishes that the function obtained by integrating from a fixed point to a variable endpoint is differentiable, with derivative equal to the integrand. Define the accumulation function by:

$$F(x) = \int_{a}^{x} f(t) \ dt$$

31.For every $x \in (a, b)$, this function satisfies $F'(x) = f(x)$, which makes differentiation and integration mutually inverse operations in a precise sense. Both results are treated in detail on the dedicated page on the Fundamental Theorem of Calculus.

Properties

32.When the two endpoints of integration coincide, the integral vanishes:

$$\int_{a}^{a} f(x) \ dx = 0$$

33.The identity follows directly from the definition: an interval of zero width contributes no area. Reversing the limits of integration changes the sign of the integral:

$$\int_{a}^{b} f(x) \ dx = -\int_{b}^{a} f(x) \ dx$$

34.This reflects the oriented nature of the definite integral: traversing the interval in the opposite direction reverses the sign of the accumulated area. If $f(x) = k$ is constant on $[a, b]$, its integral is the constant value multiplied by the length of the interval:

$$\int_{a}^{b} k \ dx = k(b - a)$$

35.A constant factor can be moved outside the integral sign:

$$\int_{a}^{b} kf(x) \ dx = k \int_{a}^{b} f(x) \ dx$$

36.The integral is additive over sums of functions:

$$\int_{a}^{b} (f(x) + g(x)) \ dx = \int_{a}^{b} f(x) \ dx + \int_{a}^{b} g(x) \ dx$$

37.Together, the previous two properties make the definite integral a linear operator. The integral is also additive over adjacent intervals: for any three points $a$, $b$, $c$ in the domain of $f$, the following identity holds:

$$\int_{a}^{c} f(x) \ dx = \int_{a}^{b} f(x) \ dx + \int_{b}^{c} f(x) \ dx$$

38.This additivity allows us to integrate a piecewise function by splitting the interval at its junction points.

39.If $f(x) \leq g(x)$ for every $x \in [a, b]$, the same inequality propagates to the integrals:

$$\int_{a}^{b} f(x) \ dx \leq \int_{a}^{b} g(x) \ dx$$

40.

This is the comparison property of integrals. The vertical difference $g(x) - f(x)$ is nonnegative throughout the interval, so its integral is also nonnegative.

41.For a bounded function $f(x)$ that is Riemann integrable on $[a, b],$ applying the comparison property to the constant functions equal to its infimum and supremum gives the bounds:

$$ (b - a)\inf_{x \in [a, b]} f(x) \leq \int_{a}^{b} f(x) \ dx \leq (b - a)\sup_{x \in [a, b]} f(x) $$

42.If $f(x)$ is Riemann integrable, then $|f(x)|$ is also Riemann integrable. The inequalities $-|f(x)| \leq f(x) \leq |f(x)|$ and the comparison property imply:

$$\left|\int_{a}^{b} f(x) \ dx\right| \leq \int_{a}^{b} |f(x)| \ dx$$

Mean value theorem for integrals

43.The mean value theorem for integrals states that if $f(x)$ is continuous on $[a, b]$, then there exists at least one point $c \in (a, b)$ such that:

$$\int_{a}^{b} f(x) \ dx = f(c)(b - a)$$

44.The value $f(c)$ is the average value of the function over the interval. Geometrically, the theorem asserts the existence of a rectangle with base $b - a$ and height $f(c)$ whose oriented area equals the definite integral. The theorem guarantees the existence of such a point without providing a method to locate it. Solving for $f(c)$, the average value of $f$ over $[a, b]$ can be written as:

$$f(c) = \frac{1}{b - a} \int_{a}^{b} f(x) \ dx$$

45.

The mean value theorem for integrals is the integral counterpart of Lagrange's mean value theorem. The differential statement guarantees a point where the instantaneous rate of change equals the average rate of change, while the integral statement guarantees a point where the function value equals the average value over the interval.

Example 1

46.Compute the following definite integral:

$$\int_{0}^{3} (3x - x^2) \ dx$$

47.Applying linearity and moving the constant factor outside the first integral gives:

$$3\int_{0}^{3} x \ dx - \int_{0}^{3} x^2 \ dx$$

48.The antiderivative of each term follows from the power rule discussed in the page on indefinite integrals:

$$F(x) = \frac{3x^2}{2} - \frac{x^3}{3}$$

49.Evaluating $F(3) - F(0)$ yields:

$$ \begin{align} F(3) - F(0) &= \left(\frac{3 \cdot 9}{2} - \frac{27}{3}\right) - \left(\frac{3 \cdot 0}{2} - \frac{0}{3}\right) \\[6pt] &= \frac{27}{2} - 9 \\[6pt] &= \frac{27 - 18}{2} \\[6pt] &= \frac{9}{2} \end{align} $$

50.The area of the region bounded by the graph of $f(x) = 3x - x^2$ and the $x$-axis over $[0, 3]$ is therefore:

$$\int_{0}^{3} (3x - x^2) \ dx = \frac{9}{2}$$

51.

When an antiderivative is not immediately recognizable, techniques such as integration by substitution and integration by parts may produce it. If no elementary antiderivative exists, the definite integral may require numerical methods or special functions.

Example 2

52.Compute the following definite integral:

$$\int_{0}^{\pi} (x + \sin x) \ dx$$

53.

The integrand combines a polynomial term with a trigonometric function. The relevant antiderivatives are collected on the page on integrals of trigonometric functions.

54.Applying linearity, the integral splits as:

$$\int_{0}^{\pi} x \ dx + \int_{0}^{\pi} \sin x \ dx$$

55.The antiderivative of each term gives:

$$F(x) = \frac{x^2}{2} - \cos x$$

56.Evaluating $F(\pi) - F(0)$ yields:

$$ \begin{align} F(\pi) - F(0) &= \left(\frac{\pi^2}{2} - \cos\pi\right) - \left(\frac{0}{2} - \cos 0\right) \\[6pt] &= \left(\frac{\pi^2}{2} + 1\right) - (0 - 1) \\[6pt] &= \frac{\pi^2}{2} + 2 \end{align} $$

57.The area of the region bounded by the graph of $f(x) = x + \sin x$ and the $x$-axis over $[0, \pi]$ is:

$$\int_{0}^{\pi} (x + \sin x) \ dx = \frac{\pi^2}{2} + 2$$

Example 3

58.Compute the area under the exponential curve $f(x)=e^{2x}$ over $[0,2]$ directly from right-endpoint Riemann sums. Divide $[0,2]$ into $n$ subintervals of equal width. The width and the right endpoint of the $k$-th subinterval are:

$$\Delta x = \frac{2}{n} \qquad x_k = \frac{2k}{n}$$

59.The function $f(x)=e^{2x}$ is increasing, so the right-endpoint rectangles give upper sums. The height of the $k$-th rectangle is:

$$f(x_k) = e^{2x_k} = e^{4k/n}$$

60.Hence the sum of the rectangle areas is:

$$R_n = \sum_{k=1}^{n} f(x_k)\Delta x = \frac{2}{n}\sum_{k=1}^{n} e^{4k/n}$$

61.Set $q_n=e^{4/n}.$ The terms $e^{4k/n}=q_n^k$ form a finite geometric progression, with $q_n^n=e^4.$ The finite-sum formula gives:

$$ \begin{align} R_n &= \frac{2}{n}\sum_{k=1}^{n}q_n^k \\[6pt] &= \frac{2}{n}\frac{q_n(q_n^n-1)}{q_n-1} \\[6pt] &= \frac{2q_n(e^4-1)}{n(q_n-1)} \end{align} $$

62.As $n$ tends to infinity, $q_n$ tends to $1.$ Applying the remarkable limit for the exponential function gives:

$$ \lim_{n \to \infty}n(q_n-1) = \lim_{n \to \infty}4\left(\frac{e^{4/n}-1}{4/n}\right) = 4 $$

63.Taking the limit of the upper sums gives the definite integral:

$$ \int_{0}^{2}e^{2x} \ dx = \lim_{n \to \infty}R_n = \frac{e^4-1}{2} $$

64.Since $e^{2x}$ is positive on $[0,2],$ this integral is the geometric area under the curve, approximately $26.799.$

Handling definite integrals with positive and negative areas

65.The interpretation of the definite integral as an area holds when $f(x) \geq 0$ throughout $[a, b]$. When $f(x)$ changes sign within the interval, the integral assigns a negative value to the portions of the region lying below the $x$-axis, and the result is an oriented area rather than a purely geometric one.

66.To recover the geometric area, the interval $[a, b]$ is divided into subintervals over which $f(x)$ maintains constant sign. If $f(x) \geq 0$ on $[a, c]$ and $f(x) \leq 0$ on $[c, b]$, additivity gives the oriented integral:

$$\int_{a}^{b} f(x) \ dx = \int_{a}^{c} f(x) \ dx + \int_{c}^{b} f(x) \ dx$$

67.The geometric area is obtained by changing the sign of the negative contribution:

$$S = \int_{a}^{c} f(x) \ dx - \int_{c}^{b} f(x) \ dx = \int_{a}^{b} |f(x)| \ dx$$

68.For an even function, symmetry about the $y$-axis implies that the contributions from $[-a, 0]$ and $[0, a]$ are equal. Hence:

$$\int_{-a}^{a} f(x) \ dx = 2\int_{0}^{a} f(x) \ dx$$

69.For an odd function, symmetry about the origin implies that the contributions from $[-a, 0]$ and $[0, a]$ are equal in magnitude but opposite in sign. Hence:

$$\int_{-a}^{a} f(x) \ dx = 0$$

70.In both cases, the geometric area enclosed between the graph of $f(x)$ and the $x$-axis over $[-a, a]$ is obtained by integrating the absolute value of the function. Since $|f(x)|$ is even whenever $f(x)$ is even or odd, the area is:

$$S = 2\int_{0}^{a} |f(x)| \ dx$$

71.If $f(x)$ is even and nonnegative on $[0, a]$, this formula reduces to:

$$S = 2\int_{0}^{a} f(x) \ dx$$

72.Further examples of geometric area calculations are given on the page on finding areas by integration.

Improper integrals

73.The Riemann integral considered above requires a bounded interval and a bounded integrand. If the interval is unbounded or the integrand becomes unbounded near an endpoint or an interior point, an improper integral may be defined by replacing the problematic bound with a parameter and taking a limit. For example, if $f(x)$ is Riemann integrable on every interval $[a,t]$ with $t>a,$ then:

$$\int_a^{+\infty}f(x) \ dx:=\lim_{t\to+\infty}\int_a^t f(x) \ dx$$

74.The integral converges only when its defining limit exists and is finite. The dedicated page gives the definitions for the other types of improper integral, the convergence criteria, and worked examples.