Integration

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

We have so far defined the limit of a function and, using the limit of the average rate of change, defined the derivative. In this post we organize integration (the inverse of that process) and examine its various properties.

Antiderivatives

Definition 1 For a function \(f\) defined on an interval \(I\), a differentiable function \(F\) satisfying \(F'(x) = f(x)\) at every point of \(I\) is called an antiderivative of \(f\).

For example, \(F(x) = x^2\) is an antiderivative of \(f(x) = 2x\). However, \(x^2 + 1\) and \(x^2 - 5\) also have derivative \(2x\), so they are antiderivatives as well. That is, adding a constant to an antiderivative still yields an antiderivative. Geometrically, an entire family of curves obtained by shifting the same curve vertically all share the same derivative (the same slope distribution). Antiderivatives are thus not unique, but they can differ only by a constant term.

Proposition 2 If \(F\) is an antiderivative of \(f\) on an interval \(I\), then every antiderivative of \(f\) has the form \(F(x) + C\) for some constant \(C\).

Proof

If \(G\) is also an antiderivative of \(f\), then \((G - F)' = f - f = 0\). By §Mean Value Theorem, ⁋Corollary 5, a function whose derivative is identically \(0\) on an interval is constant, so there exists a constant \(C\) with \(G - F = C\), hence \(G = F + C\).

In view of this proposition, the collection of all antiderivatives of \(f\) is expressed in the single formula

\[\int f(x)\dd{x} = F(x) + C\]

and is called the indefinite integral of \(f\). Here \(C\) is the constant of integration, \(f\) is the integrand, and the symbol \(\dd{x}\) indicates the variable of integration. Because Proposition 2 guarantees that antiderivatives differ only by a constant, the single integration constant \(C\) suffices to represent all of them at once.

The assumption we are implicitly making, that the interval is connected, is not essential in every respect, but the above proposition holds only when the interval is connected. If the domain is disconnected, the constant may differ on each piece. For instance, \(1/x\) is defined separately for \(x > 0\) and \(x < 0\), and although \(F(x) = \ln\lvert x\rvert\) is an antiderivative, adding different constants on the two pieces still yields antiderivatives of \(1/x\) on the whole domain, so the proposition does not literally apply there.

Under the assumption that the interval is connected, the integration constant \(C\) is determined uniquely once an initial condition is given. This is the basic method of solving differential equations by indefinite integration. For example, if \(F'(x) = 3x^2 + 1\) and \(F(1) = 5\), then from the indefinite integral \(F(x) = x^3 + x + C\) we substitute the initial condition to obtain \(1 + 1 + C = 5\), so \(C = 3\) and \(F(x) = x^3 + x + 3\) is uniquely determined.

Properties and Examples of Indefinite Integrals

Meanwhile, in §Differentiation we examined the derivatives of various functions, and because indefinite integration is the reverse of differentiation, we can derive integration formulas from them. Before doing so, let us establish the linearity of indefinite integration.

Proposition 3 (Linearity of indefinite integration) If \(f\) and \(g\) have antiderivatives and \(a, b\) are constants, then

\[\int \bigl(a f(x) + b g(x)\bigr)\dd{x} = a\int f(x)\dd{x} + b\int g(x)\dd{x}.\]

Proof

Let \(F\) and \(G\) be antiderivatives of \(f\) and \(g\) respectively. Then by §Differentiation and Derivatives, ⁋Proposition 4,

\[(aF + bG)' = aF' + bG' = af + bg,\]

so \(aF + bG\) is an antiderivative of \(af + bg\).

Reversing the derivative formulas for various functions from §Differentiation now yields the following basic formulas. That is, differentiating the right-hand side of each formula recovers the integrand.

\[\int x^r\dd{x} = \frac{x^{r+1}}{r+1} + C\ (r \neq -1), \qquad \int \frac{1}{x}\dd{x} = \ln\lvert x\rvert + C,\] \[\int e^x\dd{x} = e^x + C, \qquad \int \cos x\dd{x} = \sin x + C, \qquad \int \sin x\dd{x} = -\cos x + C,\qquad \int \sec^2 x\dd{x} = \tan x + C,\] \[\int \frac{\dd{x}}{1 + x^2} = \arctan x + C, \qquad \int \frac{\dd{x}}{\sqrt{1 - x^2}} = \arcsin x + C.\]

Combining linearity with these formulas allows us to integrate arbitrary sums of basic functions term by term. When the integrand is not in standard form, we first manipulate it using other tools, for instance rewriting via the trigonometric identity

\[\tan^2 x = \sec^2 x - 1\]

or splitting the fraction \((x^2+1)/x\) into \(x + 1/x\) so that each term matches one of the formulas above.

Especially useful are integration by substitution and integration by parts, which are respectively the reverses of §Differentiation, ⁋Theorem 4 (Chain rule) and §Differentiation, ⁋Proposition 3 (Product rule).

Theorem 4 (Integration by substitution) If \(f\) has an antiderivative on an interval \(I\), and \(g\) is differentiable with \(g(x) \in I\) for all \(x\), then

\[\int f(g(x)) g'(x) \dd{x} = \int f(u) \dd{u} \quad (u = g(x)).\]

Proof

Let \(F\) be an antiderivative of \(f\). Then by §Differentiation, ⁋Theorem 4 (Chain rule),

\[\frac{d}{\dd{x}}F(g(x)) = F'(g(x))g'(x) = f(g(x))g'(x),\]

so \(F(g(x))\) is an antiderivative of the left-hand integrand. Therefore \(\int f(g(x))g'(x)\dd{x} = F(g(x)) + C = F(u) + C = \int f(u)\dd{u}\).

In practice, one sets \(u = g(x)\) and \(\dd{u} = g'(x) \dd{x}\), rewrites the expression entirely in terms of \(u\), integrates, and then substitutes back. For example, with \(u = \cos x\),

\[\int \tan x \dd{x} = -\int \frac{\dd{u}}{u} = -\ln\lvert\cos x\rvert + C,\]

and by the same method

\[\int \frac{x}{x^2+1} \dd{x} = \frac{1}{2}\ln(x^2+1) + C\]

is obtained.

Theorem 5 (Integration by parts) If \(u\) and \(v\) are differentiable and their derivatives are continuous, then

\[\int u v' \dd{x} = uv - \int u' v \dd{x}.\]

Proof

By §Differentiation, ⁋Proposition 3 (Product rule), \((uv)' = u'v + uv'\), so \(uv' = (uv)' - u'v\), and integrating both sides gives \(\int (uv)' \dd{x} = uv\), from which the claim follows.

The key is to choose \(u\) as the factor that simplifies upon differentiation, and \(v'\) as the factor that can be integrated. For example, in \(\int x e^x \dd{x}\) we set \(u = x\) to obtain \(xe^x - e^x + C\), while for functions such as logarithms or inverse trigonometric functions whose derivatives actually become simpler, we place them in the \(u\) position with \(v' = 1\) (giving \(\int \ln x \dd{x} = x\ln x - x + C\)). There are also cases where integration by parts does not simplify the integrand but returns to the original integral; then we treat the original integral as an unknown and solve algebraically.

Example 6 \(I = \int e^x\cos x \dd{x}\) returns to itself after two applications of integration by parts. Setting \(u = e^x\) and \(v' = \cos x\),

\[I = e^x\sin x - \int e^x\sin x \dd{x} = e^x\sin x - \Bigl(-e^x\cos x + \int e^x\cos x \dd{x}\Bigr) = e^x(\sin x + \cos x) - I,\]

so \(2I = e^x(\sin x + \cos x)\), hence

\[I = \frac{e^x(\sin x + \cos x)}{2} + C.\]

For rational functions whose denominator factors, partial fraction decomposition reduces each piece to a logarithm or arctangent. For instance, splitting

\[\frac{1}{x^2-1} = \frac{1}{2}\left(\frac{1}{x-1} - \frac{1}{x+1}\right)\]

gives

\[\int \frac{\dd{x}}{x^2 - 1} = \frac{1}{2}\ln\left\lvert\frac{x-1}{x+1}\right\rvert + C,\]

and when the denominator contains an irreducible quadratic, completing the square yields

\[\int \frac{\dd{x}}{x^2 + 2x + 5} = \frac{1}{2}\arctan\frac{x+1}{2} + C.\]

Irrational expressions such as \(\sqrt{a^2 - x^2}\) and \(\sqrt{a^2 + x^2}\) can be cleared by trigonometric substitution. For example, in \(\int \sqrt{1 - x^2} \dd{x}\) setting \(x = \sin\theta\) gives

\[\int \cos^2\theta \dd{\theta} = \frac{1}{2}(\arcsin x + x\sqrt{1-x^2}) + C.\]

Powers of trigonometric functions themselves are reduced in degree by identities or substitution. Odd powers are handled by peeling off one factor and substituting:

\[\int \sin^3 x \dd{x} = -\cos x + \frac{1}{3}\cos^3 x + C,\]

while even powers are reduced via double-angle formulas:

\[\int \sin^2 x \dd{x} = \frac{x}{2} - \frac{\sin 2x}{4} + C.\]

Repeated integration by parts yields recurrence relations that lower the degree one step at a time, allowing systematic treatment of integrals involving mixed powers.

Example 7 (Recurrence relation) For \(I_n = \int x^n e^x \dd{x}\), integration by parts with \(u = x^n\) and \(v' = e^x\) gives

\[I_n = x^n e^x - n I_{n-1}.\]

Starting from \(I_0 = e^x\), one obtains successively

\[I_1 = (x-1)e^x, \quad I_2 = (x^2 - 2x + 2)e^x,\]

lowering the degree at each step. In the same manner, the recurrence

\[\int \sin^n x \dd{x} = -\frac{1}{n} \sin^{n-1}x\cos x + \frac{n-1}{n}\int \sin^{n-2}x \dd{x}\]

is also obtained.

Even with these techniques, some integrals cannot be expressed in terms of elementary functions, such as \(\int e^{-x^2} \dd{x}\) or \(\int (\sin x)/x \dd{x}\). These functions are nevertheless well defined as definite integrals, and defining new functions from them is something we shall see from the Fundamental Theorem of Calculus.

Partitions and Riemann Sums

The integration examined above is defined as the inverse of differentiation, but historically integration arose in a different way, namely from the problem of finding a total accumulated quantity. For example, measuring the area under a curve \(y = f(x)\) above an interval \([a,b]\) falls into this category. The idea of the following definition is to approximate this area by slicing it into thin rectangles and making the subdivision arbitrarily fine.

Definition 8 A partition \(P\) of the closed interval \([a, b]\) is a finite set of points \(a = x_0 < x_1 < \cdots < x_n = b\). The length of each subinterval \([x_{i-1}, x_i]\) is denoted \(\Delta x_i = x_i - x_{i-1}\), and the length of the longest subinterval is called the mesh \(\lVert P\rVert = \max_i \Delta x_i\) of the partition. Choosing a sample point \(c_i \in [x_{i-1}, x_i]\) in each subinterval, the sum

\[S(P, f) = \sum_{i=1}^{n} f(c_i)\Delta x_i\]

is called the Riemann sum of \(f\).

A Riemann sum approximates the area under the curve by rectangles of width \(\Delta x_i\) and height \(f(c_i)\). Taking the sample points \(c_i\) to be the left endpoint, the right endpoint, or points giving the minimum or maximum of the function on each subinterval yields respectively the left, right, lower, and upper Riemann sums. Intuitively, we expect that as the partition is made infinitely fine, this approximation converges to a single value independent of the choice of sample points.

Definition 9 If there exists a real number \(S\) such that for every \(\epsilon > 0\) there is a \(\delta > 0\) with \(\lvert S(P, f) - S\rvert < \epsilon\) for every partition with \(\lVert P\rVert < \delta\) and every choice of sample points, then \(f\) is said to be integrable on \([a,b]\) and \(S\) is called the definite integral, written

\[\int_a^b f(x)\dd{x} = S.\]

The numbers \(a\) and \(b\) are called the lower and upper limits of integration.

This definition can be applied directly to compute definite integrals. Dividing \([0,1]\) into \(n\) equal parts and choosing the right endpoints \(c_i = i/n\), the Riemann sum for

\[\int_0^1 x\dd{x}\]

is

\[\sum_{i=1}^n \frac{i}{n}\cdot\frac1n = \frac{1}{n^2}\cdot\frac{n(n+1)}{2} = \frac{n+1}{2n} \rightarrow \frac12,\]

and similarly

\[\int_0^1 x^2\dd{x} = \lim_{n\rightarrow\infty}\sum_{i=1}^n \frac{i^2}{n^3} = \lim_{n\rightarrow\infty}\frac{n(n+1)(2n+1)}{6n^3} = \frac13.\]

In particular, the first result can also be checked immediately from the area formula for a triangle.

Theorem 10 A function continuous on \([a, b]\) is integrable.

The proof of this theorem also lies beyond our current scope, so we must accept it and move on. The following three facts concerning integrability are at the same level, and we shall use them together in subsequent arguments.

• A function integrable on \([a,b]\) is integrable on every subinterval.
• If \(f\) is integrable, then \(\lvert f\rvert\) is also integrable.
• The product of two integrable functions is integrable.

Properties of the Definite Integral

Since a Riemann sum is defined as a sum and a limit, the definite integral inherits the following properties.

Proposition 11 If \(f\) and \(g\) are integrable on \([a,b]\), then the following hold.

1. For constants \(\alpha, \beta\), \(\int_a^b (\alpha f(x) + \beta g(x))\dd{x} = \alpha\int_a^b f(x)\dd{x} + \beta\int_a^b g(x)\dd{x}\).
2. For \(a < c < b\), \(\int_a^b f(x)\dd{x} = \int_a^c f(x)\dd{x} + \int_c^b f(x)\dd{x}\).
3. If \(f(x) \leq g(x)\) for all \(x\), then \(\int_a^b f(x)\dd{x} \leq \int_a^b g(x)\dd{x}\).

Proof

All three properties hold at the level of Riemann sums and are preserved in the limit. Linearity follows from \(S(P, \alpha f + \beta g) = \alpha S(P, f) + \beta S(P, g)\), and monotonicity follows from \(f(c_i) \leq g(c_i)\) implying \(S(P, f) \leq S(P, g)\). The second result is obtained by considering only partitions that include \(c\) as a subdivision point, so that the Riemann sum splits into the Riemann sums over the two subintervals.

By convention, setting \(\int_a^a f(x)\dd{x} = 0\) and \(\int_b^a f(x)\dd{x} = -\int_a^b f(x)\dd{x}\) makes the second result valid regardless of the ordering of \(a\), \(b\), and \(c\). From the third result, two useful computations follow. First, if \(m \leq f \leq M\), then

\[m(b-a) \leq \int_a^b f(x)\dd{x} \leq M(b-a).\]

Second, applying this to \(-\lvert f\rvert \leq f \leq \lvert f\rvert\) yields the integral version of the triangle inequality:

\[\left\lvert \int_a^b f(x)\dd{x}\right\rvert \leq \int_a^b \lvert f(x)\rvert \dd{x}.\]

Applying the first inequality to continuous functions, one can show that the integral value is attained exactly by the function value at some point.

Proposition 12 (Mean Value Theorem for Integrals) If \(f\) is continuous on \([a,b]\), then there exists \(c \in [a,b]\) such that

\[\int_a^b f(x)\dd{x} = f(c)(b-a).\]

Proof

By §Continuous Functions, ⁋Theorem 4 (Extreme Value Theorem), \(f\) attains a minimum \(m\) and a maximum \(M\) on \([a,b]\). From the above computation, the average value

\[\frac{1}{b-a}\int_a^b f(x)\dd{x}\]

lies in \([m, M]\). If this value equals \(m\) or \(M\), we choose \(c\) to be a point attaining the minimum or maximum (the case \(f\) constant, so \(m = M\), is included here); otherwise we apply §Continuous Functions, ⁋Theorem 5 (Intermediate Value Theorem) to the interval between the points attaining the minimum and maximum to obtain a point \(c\) with \(f(c) = \frac{1}{b-a}\int_a^b f(x)\dd{x}\).

Here

\[\frac{1}{b-a}\int_a^b f(x)\dd{x}\]

is called the average value of \(f\) on \([a,b]\), and Proposition 12 (Mean Value Theorem for Integrals) states that a continuous function actually attains its average value at least at one point.

Area and Applications

The definite integral is most intuitively understood as signed area. For instance, on an interval where \(f < 0\), the terms \(f(c_i)\Delta x_i\) of the Riemann sum are negative, so \(\int_a^b f(x)\dd{x}\) attaches a minus sign to the region enclosed by the \(x\)-axis and \(f\). This perspective becomes especially clear when the sign of the function changes over the interval of integration, distinguishing the case where a single integral cancels positive and negative areas to give \(0\) from the case where the actual area requires absolute values; this is where the integral version of the triangle inequality appears as a strict inequality.

Example 13 The reason \(\int_{-1}^{1} x\dd{x} = 0\) is that the negative area on \([-1,0]\) and the positive area on \([0,1]\) cancel exactly. Each piece is a right triangle with base and height \(1\), so its area is \(\frac{1}{2}\), and splitting using the second result of Proposition 11,

\[\int_{-1}^{1} x\dd{x} = \int_{-1}^{0} x\dd{x} + \int_{0}^{1} x\dd{x} = -\frac{1}{2} + \frac{1}{2} = 0.\]

If the actual area enclosed by the curve and the \(x\)-axis is desired, one must integrate \(\lvert x\rvert\) with the sign removed, giving

\[\int_{-1}^{1} \lvert x\rvert \dd{x} = \int_{-1}^{0} (-x)\dd{x} + \int_{0}^{1} x\dd{x} = \frac{1}{2} + \frac{1}{2} = 1.\]

This is a concrete instance where the integral triangle inequality \(\bigl\lvert\int_a^b f(x) \dd{x}\bigr\rvert \leq \int_a^b \lvert f(x)\rvert \dd{x}\) holds as a strict inequality \(0 < 1\), and the value of each piece is verified immediately from the triangle area. This calculation also agrees with the earlier computation of a triangle’s area via Riemann sums.

Because the Mean Value Theorem for Integrals replaces an integral by a function value at a single point, it is frequently used in dealing with inequalities or inferring average properties. The same theorem holds for weighted integrals provided the weight has constant sign.

Proposition 14 (Weighted Mean Value Theorem for Integrals) If \(f\) is continuous on \([a,b]\) and \(\mu\) is integrable on \([a,b]\) with \(\mu \geq 0\), then there exists \(c \in [a,b]\) such that

\[\int_a^b f(x)\mu(x)\dd{x} = f(c)\int_a^b \mu(x)\dd{x}.\]

Proof

By §Continuous Functions, ⁋Theorem 4 (Extreme Value Theorem), \(f\) attains a minimum \(m\) and a maximum \(M\). Since \(\mu \geq 0\), we have \(m\mu(x) \leq f(x)\mu(x) \leq M\mu(x)\), and integrating using monotonicity and linearity gives

\[m\int_a^b \mu(x)\dd{x} \leq \int_a^b f(x)\mu(x) \dd{x}\leq M\int_a^b \mu(x) \dd{x}.\]

If \(\int_a^b \mu(x)\dd{x} = 0\), then the middle integral is also \(0\) and the equality holds for any \(c\). If \(\int_a^b \mu(x)\dd{x} > 0\), dividing the above inequality by this value yields

\[\frac{\int_a^b f(x)\mu(x) \dd{x}}{\int_a^b \mu(x)\dd{x}} \in [m, M].\]

If this value equals \(m\) or \(M\), we choose \(c\) to be a point attaining the minimum or maximum; otherwise we apply §Continuous Functions, ⁋Theorem 5 (Intermediate Value Theorem) to the interval between the points attaining the minimum and maximum to obtain a point \(c\) taking this value. Multiplying both sides by \(\int_a^b \mu(x)\dd{x}\) then yields the claimed equality.

If we set \(\mu \equiv 1\), the Weighted Mean Value Theorem for Integrals reduces to Proposition 12 (Mean Value Theorem for Integrals); thus Proposition 14 (Weighted Mean Value Theorem for Integrals) is a generalization of the Mean Value Theorem for Integrals, and may be thought of as adding a kind of density.

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References

[Ste] J. Stewart, Calculus, 8th ed., Cengage Learning, 2016.
[Kim] 김홍종, 미적분학 1·2, 제3개정판, 서울대학교출판문화원, 2020.