Definitions [9]
The collection of all anti-derivatives of a function is called its indefinite integral.
Here: f(x) = integrand, dx = variable of integration, C = constant of integration.
A definite integral is connected with finding the area under a curve over a given interval. The chapter introduction presents area as one of the central motivating ideas behind integral calculus.
If the derivative of a function F(x) is f(x), then F(x) is called an antiderivative or integral of f(x). The set of all such antiderivatives is written as:
where C is an arbitrary constant called the constant of integration.
Integration by substitution is a method in which we replace a part of the integral by a new variable to simplify the integration.
General Formula:
If \[x = g(t), \ dx = g'(t) dt\] then \[\int f(x) dx = \int f(g(t))g'(t) dt\]
Integration by partial fractions is a method used to integrate rational functions, that is, functions of the form
If two functions are written in the form uu and dvdv, then integration by parts is based on the product rule of differentiation.
\[\int\left(\mathrm{u.v}\right)\mathrm{dx}=\mathrm{u}\int\mathrm{v}\mathrm{dx}-\int\left(\frac{\mathrm{du}}{\mathrm{dx}}\right).\left(\int\mathrm{v}\mathrm{dx}\right)\mathrm{dx}\]
Integration using trigonometric identities means converting a trigonometric expression into an easier form with the help of standard identities before integrating.
A definite integral represents the value of a function accumulated between two limits.
It can also be interpreted geometrically as the net area between the graph of y = f(x) and the x-axis from x = a to x = b.
If a function f is continuous on an interval, the area function is defined by
This means that A(x) gives the area accumulated from x = a to a variable point x.
Formulae [2]
| No. | Derivatives | Integrals (Anti-derivatives) |
|---|---|---|
| (i) | \[\frac{d}{dx} \left( \frac{x^{n+1}}{n+1} \right) = x^n\]; | \[\int x^n dx = \frac{x^{n+1}}{n+1} + \text{C}, n \neq -1\] |
| \[\frac{d}{dx} (x) = 1\]; | \[\int dx = x + \text{C}\] | |
| (ii) | \[\frac{d}{dx} (\sin x) = \cos x\]; | \[\int \cos x dx = \sin x + \text{C}\] |
| (iii) | \[\frac{d}{dx} (-\cos x) = \sin x\]; | \[\int \sin x dx = -\cos x + \text{C}\] |
| (iv) | \[\frac{d}{dx} (\tan x) = \sec^2 x\]; | \[\int \sec^2 x dx = \tan x + \text{C}\] |
| (v) | \[\frac{d}{dx} (-\cot x) = \text{cosec}^2 x\]; | \[\int \text{cosec}^2 x dx = -\cot x + \text{C}\] |
| (vi) | \[\frac{d}{dx} (\sec x) = \sec x \tan x\]; | \[\int \sec x \tan x dx = \sec x + \text{C}\] |
| (vii) | \[\frac{d}{dx} (-\text{cosec} x) = \text{cosec} x \cot x\]; | \[\int \text{cosec} x \cot x dx = -\text{cosec} x + \text{C}\] |
| (viii) | \[\frac{d}{dx} (\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}}\]; | \[\int \frac{dx}{\sqrt{1-x^2}} = \sin^{-1} x + \text{C}\] |
| (ix) | \[\frac{d}{dx} (-\cos^{-1} x) = \frac{1}{\sqrt{1-x^2}}\]; | \[\int \frac{dx}{\sqrt{1-x^2}} = -\cos^{-1} x + \text{C}\] |
| (x) | \[\frac{d}{dx} (\tan^{-1} x) = \frac{1}{1+x^2}\]; | \[\int \frac{dx}{1+x^2} = \tan^{-1} x + \text{C}\] |
| (xi) | \[\frac{d}{dx} (e^x) = e^x\]; | \[\int e^x dx = e^x + \text{C}\] |
| (xii) | \[\frac{d}{dx}\left(\log|x|\right)=\frac{1}{x};\] | \[\int\frac{1}{x}dx=\log|x|+\mathrm{C}\] |
| (xiii) | \[\frac{d}{dx} \left( \frac{a^x}{\log a} \right) = a^x\]; | \[\int a^x dx = \frac{a^x}{\log a} + \text{C}\] |
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\[\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \log \left| \frac{x - a}{x + a} \right| + C\]
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\[\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a + x}{a - x} \right| + C\]
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\[\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \left(\frac{x}{a}\right) + C\]
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\[\int \frac{dx}{\sqrt{x^2 - a^2}} = \log \left| x + \sqrt{x^2 - a^2} \right| + C\]
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\[\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1} \left(\frac{x}{a}\right) + C\]
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\[\int \frac{dx}{\sqrt{x^2 + a^2}} = \log \left| x + \sqrt{x^2 + a^2} \right| + C\]
Theorems and Laws [3]
Prove that: `int sqrt(a^2 - x^2) * dx = x/2 * sqrt(a^2 - x^2) + a^2/2 * sin^-1(x/a) + c`
Let I = `int sqrt(a^2 - x^2) dx`
= `int sqrt(a^2 - x^2)*1 dx`
= `sqrt(a^2 - x^2)* int 1 dx - int [d/dx (sqrt(a^2 - x^2))* int 1 dx]dx`
= `sqrt(a^2 - x^2)*x - int [1/(2sqrt(a^2 - x^2))*d/dx (a^2 - x^2)*x]dx`
= `sqrt(a^2 - x^2)*x - int 1/(2sqrt(a^2 - x^2))(0 - 2x)*x dx`
= `sqrt(a^2 - x^2)*x - int (-x)/sqrt(a^2 - x^2)*x dx`
= `xsqrt(a^2 - x^2) - int (a^2 - x^2 - a^2)/sqrt(a^2 - x^2)dx`
= `xsqrt(a^2 - x^2) - int sqrt(a^2 - x^2)dx + a^2 int dx/sqrt(a^2 - x^2)`
= `xsqrt(a^2 - x^2) - I + a^2sin^-1(x/a) + c_1`
∴ 2I = `xsqrt(a^2 - x^2) + a^2sin^-1(x/a) + c_1`
∴ I = `x/2 sqrt(a^2 - x^2) + a^2/2 sin^-1(x/a) + c_1/2`
∴ `int sqrt(a^2 - x^2)dx = x/2 sqrt(a^2 - x^2) + a^2/2sin^-1(x/a) + c`, where `c = c_1/2`.
If f is continuous on [a, b] and
If f is continuous on [a, b] and F is any antiderivative of f, then
This is the formula most often used in exams to evaluate definite integrals.
Key Points
- Primitive
Another name for anti-derivative. - Indefinite Integral
The family of all anti-derivatives of a function. - Definite Integral
An integral taken between two fixed limits, commonly used to represent area or total accumulation. - Integral Calculus
The branch of calculus dealing with anti-derivatives, accumulation, and areas under curves.
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Integration is the inverse process of differentiation.
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The result of indefinite integration is called the antiderivative or primitive.
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General form: ∫f(x) dx = F(x) + C.
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The constant CC must always be added in indefinite integrals.
| Property | Formula |
| Reverse of differentiation | \[\frac{d}{dx}\left(\int f(x) dx\right) = f(x)\] |
| Same derivative | \[F'(x) = G'(x) \Rightarrow F(x) = G(x) + C\] |
| Sum rule | \[\int (f + g)dx = \int fdx + \int gdx\] |
| Difference rule | \[\int (f - g)dx = \int fdx - \int gdx\] |
| Constant multiple rule | \[\int k f(x)dx = k \int f(x)dx\] |
| General linearity | \[\int (kf \pm lg)dx = k \int fdx \pm l \int gdx\] |
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Integration by substitution is the reverse process of the chain rule.
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Choose the substitution so that the integral becomes simpler, not more complicated.
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Always rewrite both the function and \(dx\) in terms of the new variable.
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For indefinite integrals, back-substitute and add \(C\).
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For definite integrals, limits should also be changed if the solution is continued in the new variable.
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Trigonometric substitution is mainly used for radicals involving \(a^2-x^2\), \(x^2+a^2\), and \(x^2-a^2\).
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First check whether the rational function is proper or improper.
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Use long division before decomposition if the fraction is improper.
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Factorise the denominator completely before choosing partial fractions.
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For each distinct linear factor, use a constant numerator such as A, B, or C.
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For a repeated linear factor, include every power separately.
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For an irreducible quadratic factor, use a linear numerator of the form Bx + C.
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After decomposition, integrate each term separately using standard formulas.
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Convert the integrand into a known standard form before integrating.
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For \[x^2 - a^2\], factorize and use partial fractions.
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For \[x^2 + a^2\], the answer usually involves \[\tan^{-1}\].
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For \[\sqrt{a^2 - x^2}\], the answer usually involves \[\sin^{-1}\].
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For general quadratics, complete the square first.
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For \[px + q\] in the numerator, relate it to the derivative of the denominator.
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Always write the constant of integration C in the final answer.
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Formula:
\[\int u dv = uv - \int v du\] -
Choose u by LIATE
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For log x and inverse trig, multiply by 1
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Repeated parts may be needed for \[e^x \sin x\], \[e^x \cos x\].
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First inspect the pattern in the integrand.
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Do not integrate complicated trigonometric expressions directly if an identity can simplify them first.
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After simplification, integrate term by term carefully.
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Always add the constant of integration, \(C\).
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Used to find exact accumulated value over a fixed interval.
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Written as \[\int_{a}^{b} f(x) \, dx\].
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Evaluated using \[F(b) - F(a)\].
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Gives a unique numerical value.
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Represents net area geometrically.
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The theorem connects differentiation and integration.
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If \[A(x) = \int_{a}^{x} f(t) \, dt\], then \[A'(x) = f(x)\].
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If F'(x) = f(x), then \[\int_{a}^{b} f(x) \, dx = F(b) - F(a)\].
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The result is used to evaluate definite integrals quickly.
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The function should be continuous on the interval for direct use of the theorem.
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Look for an inner function and its derivative.
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Choose substitution carefully.
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Change the limits immediately.
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Integrate in the new variable.
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Do not add +C in a definite integral.
Concepts [15]
- Introduction of Integrals
- Integration as an Inverse Process of Differentiation
- Geometrical Interpretation of Indefinite Integrals
- Properties of Indefinite Integral
- Comparison Between Differentiation and Integration
- Methods of Integration> Integration by Substitution
- Methods of Integration> Integration Using Partial Fraction
- Integrals of Some Particular Functions
- Methods of Integration> Integration by Parts
- Methods of Integration>Integration Using Trigonometric Identities
- Definite Integrals
- Definite Integral as the Limit of a Sum
- Fundamental Theorem of Integral Calculus
- Evaluation of Definite Integrals by Substitution
- Properties of Definite Integrals
