Mathematics Quick Revision Of Integral For CBSE-NCERT
Please Wait... While Loading Full Video

Class 12 Chapter 7 - Integrals

Quick Revision Of Integral For CBSE-NCERT

Integration as an Inverse Process of Differentiation

`\color{green} ✍️` If there is a function F such that `d/(dx) F(x) = f(x) , AA x ∈ I` (interval), then for any arbitrary real number C, (also called constant of integration)

`color{orange}{d/(dx) [F(x) +C ] = f(x) , x ∈ I"`

Symbolically, we write `color{orange} (∫ f (x) dx = F (x) + C.)`

Formulae of Integration(Must remember)

Derivatives Integrals (Anti Derivative )
(i) `d/(dx) ( (x^(n+1))/(n+1) ) = x^n` Particularly `d/(dx)x = 1` `int x^n dx = x^(n +1) /(n+1) + C , n != -1` ` intdx = x +C`
(ii) `color{blue} {d/(dx) (sinx) = cos x}` ; `color{green} { int cos x dx = sinx + C`
(iii) `color{blue} {d/(dx) ( - cos x ) = sinx}` ; `color{green} {int sinx dx = - cos x + C}`
(iv) `color{blue} {d/(dx) (tan x) = sec^2x}` ; `color{green} { int sec^2 x dx = tan x +C}`
(v) `color{blue} {d/(dx) (-cot x ) = cosec^2x}` ; ` color{green} { int cosec^x dx = - cot x +C} `
(vi) `color{blue} { d/(dx) (sec x ) = sec x tan x }` ; ` color{green} {int sec tan x dx = sec x + C}`
(vii) `color{blue} {d/(dx) (- cosec x ) = cosec x cot x }`; `color{green} {int cosec x cot x dx = - cosec x + C} `
(viii) `color{blue} {d/(dx) (sin^-1 x ) = 1/ sqrt (1 -x^2)}` `color{green} {int (dx)/sqrt(1 -x^2) = sin^(-1) x + C }`
(ix) `color{blue} {d/(dx) ( - cos^(-1) x ) = 1/ sqrt (1 -x^2)}` ` color{green} {int (dx)/sqrt(1 -x^2) = - cos^(-1) x + C}`
(x)` color{blue} {d/(dx) ( tan^(-1) x ) = 1/ (1 +x^2)}` `color{green} { int (dx)/sqrt(1 +x^2) = tan^(-1) x + C}`
(xi) `color{blue} { d/(dx) ( - cot ^(-1) x ) = 1/ (1 +x^2)}` `color{green} {int (dx)/sqrt(1 +x^2) = -cot^(-1) x + C}`
(xiii) `color{blue} {d/(dx) (sec^-1 x) = 1/(x sqrt(x^2 -1))}` ; `color{green} { int (dx)/(xsqrt(x^2 -1) ) = sec^-1 x +C}`
(xiii) `color{blue} {d/(dx) ( - cosec^-1 x) = 1/(x sqrt(x^2 -1))}` ; `color{green} { int (dx)/(xsqrt(x^2 -1) ) = -cosec^-1 x +C}`
(xiv) `color{blue} {g/(dx) (e^x) = e^x}` ; `color{green} {int e^x dx = e^x +C}`
(xv) `color{blue} {d/(dx) ( log | x | ) = 1/x}` ; ` color{green} {int 1/x dx = log |x| +C}`
(xvi) `color{blue} {d/(dx) ( a^x/ (log a) ) = a^x}` ; `color{green} {int a^x dx = a^x/(log a ) +C}`

Properties of indefinite integral

`"Property 1 :"`

`color{blue}"The process of differentiation and integration are inverses of each other"`

`color{orange} {=> d/(dx) ∫ f(x) dx = f(x)}` and ` color{orange} {∫ f' (x) dx = f(x) +C}` ,

where `C ` is any arbitrary constant.

`"Property 2 :"`

`color{blue}"Two indefinite integrals with the same derivative lead to the same family of curves and so they are equivalent. "`

`"Property 3 :"`

`color{blue}( ∫ [ f(x) +g(x) ] dx = ∫ f(x) dx + ∫ g(x) dx)`

`"Property 4 :"`

For any real number k, `color{blue}( ∫ k f(x) dx= k ∫ f(x) dx)`

`"Property 5 :"`

- By Properties (III) and (IV) can be generalised to a finite number of functions `color{green} {f_1, f_2, ..., f_n}` and the real numbers, `color{green} {k_1, k_2, ..., k_n}` giving

`color{orange} { ∫ [k_1 f_1 (x) + k_2 f_2 (x) + .....+ k_n f_n (x) ] dx}`

`color{orange} { = k_1 ∫ f_1 (x) dx + k_2 ∫ f_2(x) dx + .....+ k_n ∫ f_n (x) dx}` .

Comparison between differentiation and integration

Integration by substitution

Integration of perticular finction



Integration by Partial Fractions

The following Table indicates the types of simpler partial fractions that are to be associated with various kind of rational functions.


Integration by Parts

Definite integral as the limit of a sum

Evaluation of Definite Integrals by Substitution

To evaluate `color {red} {int_a^b f(x) dx}` , by substitution, the steps could be as follows:

1. Consider the integral without limits and substitute, y = f (x) or x = g(y) to reduce the given integral to a known form.

2. Integrate the new integrand with respect to the new variable without mentioning the constant of integration.

3. Resubstitute for the new variable and write the answer in terms of the original variable.

4. Find the values of answers obtained in (3) at the given limits of integral and find the difference of the values at the upper and lower limits.

Fundamental Theorem of Calculus

`\color{green} ✍️` `color{blue} text{Area function}`

`=>` We have defined `int_a^bf(x) dx` as the area of the region bounded by the curve `y = f(x)` the ordinates `x = a` and `x = b` and x-axis.

`=>` Let `x` be a given point in `[a, b].` Then `int_a^x f(x) dx` represents the area of the shaded region in Fig.

`=>` In other words, the area of this shaded region is a function of `x.` We denote this function of `x` by `A(x).`

`color {brown} {A(x) = int_a^x f(x) dx}`
 
SiteLock