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Integrals
Introduction
Integration as an Inverse Process of Differentiation
Practice: Antiderivatives by Inspection
Methods of Integration
Practice: Integration by Substitution
Practice: Trigonometric Integrals
Integrals of Some Particular Functions
Practice: Standard Forms and Completing the Square
Integration by Partial Fractions
Practice: Partial Fractions
Integration by Parts
Practice: Integration by Parts
Practice: Integrals of Square-Root Quadratics
Definite Integral
Fundamental Theorem of Calculus
Practice: Evaluating Definite Integrals
Evaluation of Definite Integrals by Substitution
Practice: Definite Integrals via Substitution
Some Properties of Definite Integrals
Practice: Properties of Definite Integrals
Miscellaneous Examples
Chapter Review: Mixed Practice
Summary
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Glossary

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IntegralsIntegration as an Inverse Process of Differentiation

Reading time: ~35 min

Here is a guessing game. I took a function, differentiated it, and got x^2. What was my function? \frac{x^3}{3} works, because \frac{d}{dx}\left(\frac{x^3}{3}\right) = x^2. But so does \frac{x^3}{3} + 5, and so does \frac{x^3}{3} - 7. The graph shows a whole family of answers, y = \frac{x^3}{3} + C; the bold green curve is the one you control.

Slide C until your curve passes through the ringed points P and Q. Keep an eye on the orange dashes: they show the slope of your curve at x = −2, −1, 1 and 2.

C = ${c}

The family y = \frac{x^3}{3} + C. Every member has the same derivative, x^2.

The orange dashes never turned: sliding a curve up or down does not change how steep it is anywhere. In symbols, the derivative of a constant is , so \frac{d}{dx}\left(\frac{x^3}{3} + C\right) = x^2 for every real number C.

The same thing happens with every derivative you know. \frac{d}{dx}(\sin x) = \cos x, so \sin x, \sin x + 1 and \sin x - 100 are all anti derivatives of \cos x. And \frac{d}{dx}(e^x) = e^x, so e^x + C is an anti derivative of e^x for any C. An anti derivative is never : there are infinitely many, one for each value of C.

Because C can be any real number, it is called an arbitrary constant or the constant of integration. Each value of C picks out one member of the family, so C works like a dial, a parameter.

In general, if \frac{d}{dx}F(x) = f(x) for every x in an interval I, then

\frac{d}{dx}\left[F(x) + C\right] = f(x), \quad x \in I,

so \{F + C : C \in \mathbf{R}\} is a family of anti derivatives of f. Could there be some other anti derivative hiding outside this family? No:

Remark: same derivative, differ by a constant

Suppose g and h have the same derivative on an interval I. Let f(x) = g(x) - h(x). Then

f'(x) = g'(x) - h'(x) = 0 for every x \in I.

A function whose rate of change is zero everywhere on an interval never rises and never falls, so it is constant. Hence g(x) = h(x) + C for some constant C.

So the family \{F + C : C \in \mathbf{R}\} contains every anti derivative of f, and we give it one symbol:

\int f(x)\, dx = F(x) + C,

read as the indefinite integral of f with respect to x. Equivalently, if \frac{dy}{dx} = f(x), we write y = \int f(x)\, dx.

Tip: never drop the + C

\sin x is an anti derivative of \cos x, but \int \cos x\, dx is the whole family \sin x + C. Leaving out the C in an indefinite integral costs marks in the board exam.

Table 7.1 — the words that come with the symbol.

Symbol, term or phraseMeaning
\int f(x)\, dxthe integral of f with respect to x
f(x) in \int f(x)\, dxthe integrand
x in \int f(x)\, dxthe variable of integration
integratefind the integral
an integral of fa function F with F'(x) = f(x)
integrationthe process of finding the integral
constant of integrationany real number C, treated as a constant function

The sign \int is a stretched letter S, for summa (sum): Leibniz chose it because, as the metro showed, an integral adds up many small pieces. The dx at the end is a label naming the variable, just as in \frac{dy}{dx}.

Every derivative formula you learnt in the previous chapter can now be read backwards. The left column below is a derivative you know; the right column is the integral it gives you for free.

DerivativeIntegral (anti derivative)
(i) \frac{d}{dx}\left(\frac{x^{n+1}}{n+1}\right) = x^n\int x^n\, dx = \frac{x^{n+1}}{n+1} + C,\ n \neq -1
in particular \frac{d}{dx}(x) = 1\int dx = x + C
(ii) \frac{d}{dx}(\sin x) = \cos x\int \cos x\, dx = \sin x + C
(iii) \frac{d}{dx}(-\cos x) = \sin x\int \sin x\, dx = -\cos x + C
(iv) \frac{d}{dx}(\tan x) = \sec^2 x\int \sec^2 x\, dx = \tan x + C
(v) \frac{d}{dx}(-\cot x) = \operatorname{cosec}^2 x\int \operatorname{cosec}^2 x\, dx = -\cot x + C
(vi) \frac{d}{dx}(\sec x) = \sec x \tan x\int \sec x \tan x\, dx = \sec x + C
(vii) \frac{d}{dx}(-\operatorname{cosec} x) = \operatorname{cosec} x \cot x\int \operatorname{cosec} x \cot x\, dx = -\operatorname{cosec} x + C
(viii) \frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}}\int \frac{dx}{\sqrt{1-x^2}} = \sin^{-1} x + C
(ix) \frac{d}{dx}(-\cos^{-1} x) = \frac{1}{\sqrt{1-x^2}}\int \frac{dx}{\sqrt{1-x^2}} = -\cos^{-1} x + C
(x) \frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}\int \frac{dx}{1+x^2} = \tan^{-1} x + C
(xi) \frac{d}{dx}(e^x) = e^x\int e^x\, dx = e^x + C
(xii) \frac{d}{dx}(\log \lvert x \rvert) = \frac{1}{x}\int \frac{1}{x}\, dx = \log \lvert x \rvert + C
(xiii) \frac{d}{dx}\left(\frac{a^x}{\log a}\right) = a^x\int a^x\, dx = \frac{a^x}{\log a} + C

So the power rule says: raise the power by one and divide by the new power. For \int x^5\, dx that gives .

The power rule fails for exactly one power. Putting n = -1 would mean dividing by n + 1 = , which is why n \neq -1 is written beside it. That case is not lost: it is row (xii), \int \frac{1}{x}\, dx = \log \lvert x \rvert + C.

Note

Every row of the table can be checked by differentiating the right-hand side, and the same goes for every integral in this chapter: differentiate your answer, and if the integrand comes back, you are right. We also usually leave out the interval on which each function is defined, but in a particular problem you must keep it in mind, as the modulus in \log \lvert x \rvert reminds us.

Some Properties of Indefinite Integral

Integration and differentiation undo each other. That single fact gives every property of the indefinite integral.

(I) \frac{d}{dx}\int f(x)\, dx = f(x) and \int f'(x)\, dx = f(x) + C.

Proof. Let F be any anti derivative of f, so \int f(x)\, dx = F(x) + C. Then

\frac{d}{dx}\int f(x)\, dx

= \frac{d}{dx}\left[F(x) + C\right]

= F'(x)

= f(x).

For the second statement: f is an anti derivative of f', so \int f'(x)\, dx = f(x) + , where C is the constant of integration.

(II) Two indefinite integrals with the same derivative are the same family of curves, so they are equivalent.

Proof. Suppose \frac{d}{dx}\int f(x)\, dx = \frac{d}{dx}\int g(x)\, dx. Then

\frac{d}{dx}\left[\int f(x)\, dx - \int g(x)\, dx\right] = 0,

and by the Remark above, a function with zero derivative is a constant:

\int f(x)\, dx - \int g(x)\, dx = C.

So the families \left\{\int f(x)\, dx + C_1\right\} and \left\{\int g(x)\, dx + C_2\right\} contain exactly the same curves. We record this by writing \int f(x)\, dx = \int g(x)\, dx, without mentioning the constants.

(III) \int \left[f(x) + g(x)\right] dx = \int f(x)\, dx + \int g(x)\, dx

Proof. By Property (I), the left side has derivative f(x) + g(x). The right side has derivative

\frac{d}{dx}\int f(x)\, dx + \frac{d}{dx}\int g(x)\, dx

= f(x) + g(x)

as well. Same derivative, so by Property (II) the two sides are equal.

(IV) For any real number k, \int k\, f(x)\, dx = k \int f(x)\, dx.

Proof. Both sides have derivative , by Property (I), so Property (II) makes them equal.

(V) Putting (III) and (IV) together, for functions f_1, f_2, \ldots, f_n and real numbers k_1, k_2, \ldots, k_n:

\int \left[k_1 f_1(x) + k_2 f_2(x) + \cdots + k_n f_n(x)\right] dx

= k_1 \int f_1(x)\, dx + k_2 \int f_2(x)\, dx + \cdots + k_n \int f_n(x)\, dx.

So we may integrate a sum term by term and pull constants out in front. That is all: there is no product rule or quotient rule of this kind. \int f(x)\, g(x)\, dx is not \int f(x)\, dx \cdot \int g(x)\, dx. Products need a separate technique, integration by parts, later in this chapter.

Finding an anti derivative by recognising a derivative you already know is called the method of inspection. The next examples use it.

Example 1 : Write an anti derivative for each of the following functions using the method of inspection:

(i) \cos 2x    (ii) 3x^2 + 4x^3    (iii) \frac{1}{x},\ x \neq 0

Solution :

(i) We want a function whose derivative is \cos 2x. The chain rule gives \frac{d}{dx}(\sin 2x) = 2\cos 2x, which is twice too much. So divide by 2:

\frac{d}{dx}\left(\frac{1}{2}\sin 2x\right) = \cos 2x, and an anti derivative of \cos 2x is .

(ii) \frac{d}{dx}\left(x^3\right) = 3x^2 and \frac{d}{dx}\left(x^4\right) = 4x^3, so an anti derivative of 3x^2 + 4x^3 is .

(iii) For x > 0, \frac{d}{dx}(\log x) = \frac{1}{x}. For x < 0, \frac{d}{dx}\left[\log(-x)\right] = \frac{1}{-x} \cdot (-1) = \frac{1}{x}. The two cases combine into one formula:

\frac{d}{dx}\left(\log \lvert x \rvert\right) = \frac{1}{x},\quad x \neq 0.

So \log \lvert x \rvert is one of the anti derivatives of \frac{1}{x}, and \int \frac{1}{x}\, dx = \log \lvert x \rvert + C.

Example 2 : Find the following integrals:

(i) \int \frac{x^3 - 1}{x^2}\, dx    (ii) \int \left(x^{\frac{2}{3}} + 1\right) dx    (iii) \int \left(x^{\frac{3}{2}} + 2e^x - \frac{1}{x}\right) dx

Solution :

(i) Split the fraction first: \frac{x^3 - 1}{x^2} = x - x^{-2}. By Property (V),

\int \frac{x^3 - 1}{x^2}\, dx

= \int x\, dx - \int x^{-2}\, dx

= \left(\frac{x^2}{2} + C_1\right) - \left(\frac{x^{-1}}{-1} + C_2\right)

= \frac{x^2}{2} + \frac{1}{x} + C_1 - C_2.

C_1 - C_2 is just another constant, so we write the answer as . From now on we write a single constant in the final answer.

(ii) \int \left(x^{\frac{2}{3}} + 1\right) dx

= \frac{x^{\frac{5}{3}}}{\frac{5}{3}} + x + C = x^{\frac{5}{3}} + x + C.

(iii) Term by term:

\int x^{\frac{3}{2}}\, dx + 2\int e^x\, dx - \int \frac{1}{x}\, dx

= \frac{x^{\frac{5}{2}}}{\frac{5}{2}} + 2e^x - \log \lvert x \rvert + C

= \frac{2}{5}x^{\frac{5}{2}} + 2e^x - \log \lvert x \rvert + C.

Example 3 : Find the following integrals:

(i) \int (\sin x + \cos x)\, dx    (ii) \int \operatorname{cosec} x\,(\operatorname{cosec} x + \cot x)\, dx    (iii) \int \frac{1 - \sin x}{\cos^2 x}\, dx

Solution :

(i) \int \sin x\, dx + \int \cos x\, dx = .

(ii) Multiply out: \int \operatorname{cosec}^2 x\, dx + \int \operatorname{cosec} x \cot x\, dx = .

(iii) Split the fraction and spot two standard forms:

\int \frac{1}{\cos^2 x}\, dx - \int \frac{\sin x}{\cos^2 x}\, dx

= \int \sec^2 x\, dx - \int \sec x \tan x\, dx

= \tan x - \sec x + C.

Example 4 : Find the anti derivative F of f defined by f(x) = 4x^3 - 6, where F(0) = 3.

This is the family-of-curves picture again: the family is known, and one point on the curve picks out the member we want.

Solution : \frac{d}{dx}\left(x^4 - 6x\right) = 4x^3 - 6, so every anti derivative has the form

F(x) = x^4 - 6x + C.

We are told F(0) = 3. Putting x = 0: 3 = 0 - 0 + C, so C = .

Hence the required anti derivative is the single function F(x) = x^4 - 6x + 3.

Remarks

(i) If F is an anti derivative of f, so is F + C for every constant C. In applications an extra condition, like F(0) = 3 here, usually fixes C and gives one particular anti derivative.

(ii) Some anti derivatives cannot be written with the elementary functions (polynomials, logarithms, exponentials, trigonometric functions and their inverses). For example, no elementary function has derivative e^{-x^2}, so \int e^{-x^2}\, dx cannot be found by inspection. Its curve is the bell curve of statistics, and areas under it are read from tables instead.

(iii) The letter used for the variable does not matter: \int y^4\, dy = \frac{y^5}{5} + C.

Speed Round: Undo the Derivative

Each card shows an integrand. Pick its anti derivative, then check yourself by differentiating in your head.

Get five right to move on.

Sina