Notes On Integration - CBSE Class 12 Maths
The slope of a tangent to the curve y = f(x) at the point (xo,yo) is given by dy/dx|(xo,yo) Differential calculus To find derivatives, defining tangent lines to the graphs of the functions, calculating the slopes of lines and rate measure are some parts of differential calculus. Integral calculus On the contrary, finding a function given its derivative comes under integral calculus. Integration as an inverse process of differentiation: Ex: Let 3x2 d/dx(x3) = 3x2 â‡’x2 = 1/3 . d/dx (x3) â‡’ 2x2 = 2/3 . d/dx (x3) â‡’ 2x2 = d/dx (2x3/3) A function that could possibly have the given function as a derivative is called an 'anti-derivative' or 'primitive' of that function. Finding an anti-derivative of a function without actually integrating it is known as integration by the method of inspection. Ex: d/dx(x2 + 4) = 2x; d/dx (x2 + Ï€/2) = 2x; d/dx (x2 + log p) = 2x; Anti-derivative or integral of (2x) = (x2 + 4) or (x2 + Ï€/2) or (x2  + log p) d/dx (x2  + C) = 2x, where C is a constant. Anti-derivative or integral of (2x) = (x2  + C) If d/dx [F(x)] = g(x), âˆ€ x âˆˆ I then, for any arbitrary real number C, d/dx [F(x) + C] = g(x), âˆ€ x âˆˆ I F(x) + C, C âˆˆ R denotes a family of integrals of g. Functions with the same derivatives differ by a constant. Let g and h be two functions with the same derivatives on an interval I. âˆ´ g ' (x) = h ' (x) Consider: f(x) = g(x) - h(x), âˆ€ x âˆˆ I Differentiating both sides with respect to x, we get f '(x) = g '(x) - h '(x), âˆ€ x âˆˆ I f ' (x) = 0, âˆ€ x âˆˆ I ['.' g '(x) = h '(x)] â‡’ f(x) = C (Constant) â‡’ g(x) - h(x) = C (Constant) Mathematical notation to represent the integral of a function: This symbol represents the entire class of anti-derivatives. The function f of x is called the integrand. The x in (f of x) is called the variable of integration. DX means that the integration is with respect to x. If dy/dx = f(x), then y = âˆ« f(x)dx + C Some terms and phrases, and their meanings associated with integration.                Terms and Phrases                          Meaning      Integrate       Find the value of the integral      Integration       The process of finding the integral      Constant of integration       An arbitrary constant C. Derivatives $\frac{\text{d}}{\text{dx}}\text{}\left(\frac{{\text{x}}^{\text{n + 1}}}{\text{n + 1}}\right)={\text{x}}^{\text{n}}$ For n=0, $\frac{\text{d}}{\text{dx}}{\text{(x) = 1}}^{\text{}}$   = $\frac{{\text{x}}^{\text{n + 1}}}{\text{n + 1}}$ +C For n=0,  $âˆ«\text{1 dx = x + C}$   $\frac{\text{d}}{\text{dx}}{\text{(sin x) = cos x}}^{\text{}}$  $âˆ«\text{cos x dx = sin x + C}$   $\frac{\text{d}}{\text{dx}}\text{}\frac{\text{- 1}}{\text{cos x}}\text{= sin x}$  $âˆ«\text{sin x dx = - cos x + C}$  $\text{}\frac{\text{d}}{\text{dx}}\text{(tan x) =}{\text{sec}}^{\text{2}}\text{x}$  $âˆ«{\text{sec}}^{\text{2}}\text{x dx = tan x + C}$  $\frac{\text{d}}{\text{dx}}\text{}\frac{\text{-1}}{\text{cot x}}\text{=}{\text{cosec}}^{\text{2}}\text{x}$  $âˆ«{\text{cosec}}^{\text{2}}\text{x dx = - cot x + C}$ $\frac{\text{d}}{\text{dx}}\text{(sec x) = sec x tan x}$   $âˆ«\text{sec x tan x dx = sec x + C}$  $\frac{\text{d}}{\text{dx}}\text{(- cosec x) = cosec x cot x}$   $âˆ«\text{cosec x cot x dx = - cosec x + C}$  $\frac{\text{d}}{\text{dx}}\text{}\frac{\text{1}}{\text{(}{\text{sin}}^{\text{-1}}\text{x)}}\text{=}\frac{\text{1}}{\sqrt{\text{1 -}{\text{x}}^{\text{2}}}}$  $âˆ«\frac{\text{1}}{\sqrt{\text{1 -}{\text{x}}^{\text{2}}}}\text{dx =}{\text{sin}}^{\text{-1}}\text{x + C}$  $\frac{\text{d}}{\text{dx}}\text{}\frac{\text{1}}{\text{(}{\text{- cos}}^{\text{-1}}\text{x)}}\text{=}\frac{\text{1}}{\sqrt{\text{1 -}{\text{x}}^{\text{2}}}}$    $\frac{\text{d}}{\text{dx}}\text{(}{\text{tan}}^{\text{-1}}\text{x) =}\frac{\text{1}}{\text{1 +}{\text{x}}^{\text{2}}}$  $âˆ«\frac{\text{1}}{\text{1 +}{\text{x}}^{\text{2}}}\text{dx =}{\text{tan}}^{\text{-1}}\text{x + C}$  $\frac{\text{d}}{\text{dx}}\text{(}{\text{- cot}}^{\text{-1}}\text{x) =}\frac{\text{1}}{\text{1 +}{\text{x}}^{\text{2}}}$   $\frac{\text{d}}{\text{dx}}\text{(}{\text{sec}}^{\text{-1}}\text{x) =}\frac{\text{1}}{\text{x}\sqrt{{\text{x}}^{\text{2}}\text{- 1}}}$  $âˆ«\frac{\text{1}}{\text{x}\sqrt{{\text{x}}^{\text{2}}\text{- 1}}}\text{dx =}{\text{sec}}^{\text{-1}}\text{x + C}$   $\frac{\text{d}}{\text{dx}}\text{}\frac{\text{-1}}{\text{}{\text{cosec}}^{\text{-1}}\text{x}}\text{=}\frac{\text{1}}{\text{x}\sqrt{{\text{x}}^{\text{2}}\text{- 1}}}$     $\frac{\text{d}}{\text{dx}}\text{(}{\text{e}}^{\text{x}}\text{) =}{\text{e}}^{\text{x}}$  $âˆ«{\text{e}}^{\text{x}}\text{dx =}{\text{e}}^{\text{x}}\text{+ C}$   $\frac{\text{d}}{\text{dx}}\text{(log |x|) =}\frac{\text{1}}{\text{x}}$  $âˆ«\frac{\text{1}}{\text{x}}\text{dx = log |x| + C}$  $\frac{\text{d}}{\text{dx}}\text{(}\frac{{\text{a}}^{\text{x}}}{\text{log a}}\text{) =}{\text{a}}^{\text{x}}$  $âˆ«{\text{a}}^{\text{x}}\text{dx =}\frac{{\text{a}}^{\text{x}}\text{}}{\text{log a}}\text{+ C}$ Geometrical interpretation of indefinite integrals f(x) = 3xÂ² d/dx(x3) = 3x2 âˆ« 3x2 dx = x3 + C y = x3 + C represents a family of integrals. For C = 0, we obtain y = x3 . The curve y = x3 passes through the origin. For C = 1, we obtain, y = x3 + 1 For C= 2, we obtain, y = x3 + 2. For C = â€”1, we obtain, y = x3 -1. For C = â€”2, we obtain, y = x3 -2. The line 'x =1' intersects all these curves at different points. âˆ« f(x)dx = G(x) + C = y represents a family of curves.

#### Summary

The slope of a tangent to the curve y = f(x) at the point (xo,yo) is given by dy/dx|(xo,yo) Differential calculus To find derivatives, defining tangent lines to the graphs of the functions, calculating the slopes of lines and rate measure are some parts of differential calculus. Integral calculus On the contrary, finding a function given its derivative comes under integral calculus. Integration as an inverse process of differentiation: Ex: Let 3x2 d/dx(x3) = 3x2 â‡’x2 = 1/3 . d/dx (x3) â‡’ 2x2 = 2/3 . d/dx (x3) â‡’ 2x2 = d/dx (2x3/3) A function that could possibly have the given function as a derivative is called an 'anti-derivative' or 'primitive' of that function. Finding an anti-derivative of a function without actually integrating it is known as integration by the method of inspection. Ex: d/dx(x2 + 4) = 2x; d/dx (x2 + Ï€/2) = 2x; d/dx (x2 + log p) = 2x; Anti-derivative or integral of (2x) = (x2 + 4) or (x2 + Ï€/2) or (x2  + log p) d/dx (x2  + C) = 2x, where C is a constant. Anti-derivative or integral of (2x) = (x2  + C) If d/dx [F(x)] = g(x), âˆ€ x âˆˆ I then, for any arbitrary real number C, d/dx [F(x) + C] = g(x), âˆ€ x âˆˆ I F(x) + C, C âˆˆ R denotes a family of integrals of g. Functions with the same derivatives differ by a constant. Let g and h be two functions with the same derivatives on an interval I. âˆ´ g ' (x) = h ' (x) Consider: f(x) = g(x) - h(x), âˆ€ x âˆˆ I Differentiating both sides with respect to x, we get f '(x) = g '(x) - h '(x), âˆ€ x âˆˆ I f ' (x) = 0, âˆ€ x âˆˆ I ['.' g '(x) = h '(x)] â‡’ f(x) = C (Constant) â‡’ g(x) - h(x) = C (Constant) Mathematical notation to represent the integral of a function: This symbol represents the entire class of anti-derivatives. The function f of x is called the integrand. The x in (f of x) is called the variable of integration. DX means that the integration is with respect to x. If dy/dx = f(x), then y = âˆ« f(x)dx + C Some terms and phrases, and their meanings associated with integration.                Terms and Phrases                          Meaning      Integrate       Find the value of the integral      Integration       The process of finding the integral      Constant of integration       An arbitrary constant C. Derivatives $\frac{\text{d}}{\text{dx}}\text{}\left(\frac{{\text{x}}^{\text{n + 1}}}{\text{n + 1}}\right)={\text{x}}^{\text{n}}$ For n=0, $\frac{\text{d}}{\text{dx}}{\text{(x) = 1}}^{\text{}}$   = $\frac{{\text{x}}^{\text{n + 1}}}{\text{n + 1}}$ +C For n=0,  $âˆ«\text{1 dx = x + C}$   $\frac{\text{d}}{\text{dx}}{\text{(sin x) = cos x}}^{\text{}}$  $âˆ«\text{cos x dx = sin x + C}$   $\frac{\text{d}}{\text{dx}}\text{}\frac{\text{- 1}}{\text{cos x}}\text{= sin x}$  $âˆ«\text{sin x dx = - cos x + C}$  $\text{}\frac{\text{d}}{\text{dx}}\text{(tan x) =}{\text{sec}}^{\text{2}}\text{x}$  $âˆ«{\text{sec}}^{\text{2}}\text{x dx = tan x + C}$  $\frac{\text{d}}{\text{dx}}\text{}\frac{\text{-1}}{\text{cot x}}\text{=}{\text{cosec}}^{\text{2}}\text{x}$  $âˆ«{\text{cosec}}^{\text{2}}\text{x dx = - cot x + C}$ $\frac{\text{d}}{\text{dx}}\text{(sec x) = sec x tan x}$   $âˆ«\text{sec x tan x dx = sec x + C}$  $\frac{\text{d}}{\text{dx}}\text{(- cosec x) = cosec x cot x}$   $âˆ«\text{cosec x cot x dx = - cosec x + C}$  $\frac{\text{d}}{\text{dx}}\text{}\frac{\text{1}}{\text{(}{\text{sin}}^{\text{-1}}\text{x)}}\text{=}\frac{\text{1}}{\sqrt{\text{1 -}{\text{x}}^{\text{2}}}}$  $âˆ«\frac{\text{1}}{\sqrt{\text{1 -}{\text{x}}^{\text{2}}}}\text{dx =}{\text{sin}}^{\text{-1}}\text{x + C}$  $\frac{\text{d}}{\text{dx}}\text{}\frac{\text{1}}{\text{(}{\text{- cos}}^{\text{-1}}\text{x)}}\text{=}\frac{\text{1}}{\sqrt{\text{1 -}{\text{x}}^{\text{2}}}}$    $\frac{\text{d}}{\text{dx}}\text{(}{\text{tan}}^{\text{-1}}\text{x) =}\frac{\text{1}}{\text{1 +}{\text{x}}^{\text{2}}}$  $âˆ«\frac{\text{1}}{\text{1 +}{\text{x}}^{\text{2}}}\text{dx =}{\text{tan}}^{\text{-1}}\text{x + C}$  $\frac{\text{d}}{\text{dx}}\text{(}{\text{- cot}}^{\text{-1}}\text{x) =}\frac{\text{1}}{\text{1 +}{\text{x}}^{\text{2}}}$   $\frac{\text{d}}{\text{dx}}\text{(}{\text{sec}}^{\text{-1}}\text{x) =}\frac{\text{1}}{\text{x}\sqrt{{\text{x}}^{\text{2}}\text{- 1}}}$  $âˆ«\frac{\text{1}}{\text{x}\sqrt{{\text{x}}^{\text{2}}\text{- 1}}}\text{dx =}{\text{sec}}^{\text{-1}}\text{x + C}$   $\frac{\text{d}}{\text{dx}}\text{}\frac{\text{-1}}{\text{}{\text{cosec}}^{\text{-1}}\text{x}}\text{=}\frac{\text{1}}{\text{x}\sqrt{{\text{x}}^{\text{2}}\text{- 1}}}$     $\frac{\text{d}}{\text{dx}}\text{(}{\text{e}}^{\text{x}}\text{) =}{\text{e}}^{\text{x}}$  $âˆ«{\text{e}}^{\text{x}}\text{dx =}{\text{e}}^{\text{x}}\text{+ C}$   $\frac{\text{d}}{\text{dx}}\text{(log |x|) =}\frac{\text{1}}{\text{x}}$  $âˆ«\frac{\text{1}}{\text{x}}\text{dx = log |x| + C}$  $\frac{\text{d}}{\text{dx}}\text{(}\frac{{\text{a}}^{\text{x}}}{\text{log a}}\text{) =}{\text{a}}^{\text{x}}$  $âˆ«{\text{a}}^{\text{x}}\text{dx =}\frac{{\text{a}}^{\text{x}}\text{}}{\text{log a}}\text{+ C}$ Geometrical interpretation of indefinite integrals f(x) = 3xÂ² d/dx(x3) = 3x2 âˆ« 3x2 dx = x3 + C y = x3 + C represents a family of integrals. For C = 0, we obtain y = x3 . The curve y = x3 passes through the origin. For C = 1, we obtain, y = x3 + 1 For C= 2, we obtain, y = x3 + 2. For C = â€”1, we obtain, y = x3 -1. For C = â€”2, we obtain, y = x3 -2. The line 'x =1' intersects all these curves at different points. âˆ« f(x)dx = G(x) + C = y represents a family of curves.

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