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Mathematics · Ch 16 — Integration

Integration as an Inverse Process of Differentiation

16.2

Integration as an Inverse Process of Differentiation

7.2 Integration as an Inverse Process of Differentiation

The Fundamental Idea

Differentiation gives us the rate at which a function changes. Integration reverses this: we start with the derivative and ask, "What original function could have produced this?" This reverse process is called anti-differentiation or integration.

Consider three familiar derivatives:

ddx(sin⁡x)=cos⁡x\frac{d}{dx}(\sin x) = \cos x

ddx(x33)=x2\frac{d}{dx}\left(\frac{x^3}{3}\right) = x^2

ddx(ex)=ex\frac{d}{dx}(e^x) = e^x

In the first case, cos⁡x\cos x is the derivative of sin⁡x\sin x. We therefore say that sin⁡x\sin x is an anti-derivative (or an integral) of cos⁡x\cos x. Similarly, x33\frac{x^3}{3} is an anti-derivative of x2x^2, and exe^x is an anti-derivative of itself.

The Constant of Integration

Here is a crucial observation: the derivative of any constant function is zero. Therefore, if we add any constant CC to sin⁡x\sin x, the derivative remains cos⁡x\cos x:

ddx(sin⁡x+C)=cos⁡x\frac{d}{dx}(\sin x + C) = \cos x

Similarly:

ddx(x33+C)=x2\frac{d}{dx}\left(\frac{x^3}{3} + C\right) = x^2

ddx(ex+C)=ex\frac{d}{dx}(e^x + C) = e^x

This means anti-derivatives are not unique. For any given function, there exist infinitely many anti-derivatives, all differing by a constant. The constant CC is called the constant of integration (or an arbitrary constant), and it can be any real number.

Important

If F′(x)=f(x)F'(x) = f(x) for all xx in an interval II, then for any real number CC:

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

The collection {F+C:C∈R}\{F + C : C \in \mathbb{R}\} forms the family of all anti-derivatives of ff.

Why Functions with the Same Derivative Differ by a Constant

›Proof

Let gg and hh be two functions that have the same derivative on an interval II. Define f(x)=g(x)−h(x)f(x) = g(x) - h(x) for all x∈Ix \in I.

Differentiating:

f′(x)=g′(x)−h′(x)f'(x) = g'(x) - h'(x)

Since g′(x)=h′(x)g'(x) = h'(x) by hypothesis:

f′(x)=0for all x∈If'(x) = 0 \quad \text{for all } x \in I

This means the rate of change of ff with respect to xx is zero everywhere on II. Therefore ff must be constant on II. Hence g(x)−h(x)=Cg(x) - h(x) = C for some constant CC, or g(x)=h(x)+Cg(x) = h(x) + C.

This result justifies the statement that {F+C:C∈R}\{F + C : C \in \mathbb{R}\} gives all possible anti-derivatives of ff.

Notation for Indefinite Integrals

We introduce a special symbol to represent the entire family of anti-derivatives:

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

This is read as "the indefinite integral of ff with respect to xx". The symbol ∫\int is the integral sign, f(x)f(x) is the integrand, xx is the variable of integration, and CC is the constant of integration.

Note

If we are given dydx=f(x)\frac{dy}{dx} = f(x), we write y=∫f(x) dxy = \int f(x)\,dx.

Standard Integrals from Known Derivatives

Since integration reverses differentiation, every derivative formula gives us an integral formula. The following table lists the standard results we will use to find integrals of other functions.

Standard Integrals (Anti-derivatives)

Derivative FormulaCorresponding Integral Formula
ddx(xn+1n+1)=xn\frac{d}{dx}\left(\frac{x^{n+1}}{n+1}\right) = x^n∫xn dx=xn+1n+1+C\int x^n\,dx = \frac{x^{n+1}}{n+1} + C, n≠−1n \neq -1
ddx(x)=1\frac{d}{dx}(x) = 1∫dx=x+C\int dx = x + C
ddx(sin⁡x)=cos⁡x\frac{d}{dx}(\sin x) = \cos x∫cos⁡x dx=sin⁡x+C\int \cos x\,dx = \sin x + C
ddx(−cos⁡x)=sin⁡x\frac{d}{dx}(-\cos x) = \sin x∫sin⁡x dx=−cos⁡x+C\int \sin x\,dx = -\cos x + C
ddx(tan⁡x)=sec⁡2x\frac{d}{dx}(\tan x) = \sec^2 x∫sec⁡2x dx=tan⁡x+C\int \sec^2 x\,dx = \tan x + C
ddx(−cot⁡x)=csc⁡2x\frac{d}{dx}(-\cot x) = \csc^2 x∫csc⁡2x dx=−cot⁡x+C\int \csc^2 x\,dx = -\cot x + C
ddx(sec⁡x)=sec⁡xtan⁡x\frac{d}{dx}(\sec x) = \sec x \tan x∫sec⁡xtan⁡x dx=sec⁡x+C\int \sec x \tan x\,dx = \sec x + C
ddx(−csc⁡x)=csc⁡xcot⁡x\frac{d}{dx}(-\csc x) = \csc x \cot x∫csc⁡xcot⁡x dx=−csc⁡x+C\int \csc x \cot x\,dx = -\csc x + C
ddx(sin⁡−1x)=11−x2\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}}∫dx1−x2=sin⁡−1x+C\int \frac{dx}{\sqrt{1-x^2}} = \sin^{-1} x + C
ddx(−cos⁡−1x)=11−x2\frac{d}{dx}(-\cos^{-1} x) = \frac{1}{\sqrt{1-x^2}}∫dx1−x2=−cos⁡−1x+C\int \frac{dx}{\sqrt{1-x^2}} = -\cos^{-1} x + C
ddx(tan⁡−1x)=11+x2\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}∫dx1+x2=tan⁡−1x+C\int \frac{dx}{1+x^2} = \tan^{-1} x + C
ddx(ex)=ex\frac{d}{dx}(e^x) = e^x∫ex dx=ex+C\int e^x\,dx = e^x + C
Table 7.1Some Standard Symbols, Terms and Phrases
Symbols / Terms / PhrasesMeaning
∫f(x) dx\displaystyle\int f(x)\,dxIntegral of ff with respect to xx
f(x)f(x) in ∫f(x) dx\displaystyle\int f(x)\,dxIntegrand
xx in ∫f(x) dx\displaystyle\int f(x)\,dxVariable of integration
IntegrateFind the integral
An integral of ffA function FF such that F′(x)=f(x)F'(x)=f(x)
IntegrationThe process of finding the integral