`color {red} star` Introduction

`color {red} star` Integration as an Inverse Process of Differentiation

`color {red} star` Formulae Of integration

`color {red} star`Geometrical interpretation of indefinite integral

`color {red} star` Properties of indefinite integral

`color {red} star` Integration as an Inverse Process of Differentiation

`color {red} star` Formulae Of integration

`color {red} star`Geometrical interpretation of indefinite integral

`color {red} star` Properties of indefinite integral

`\color{green} ✍️` Process of finding anti derivatives is called integration .

`\color{green} ✍️` There are two forms of the integrals, e.g., indefinite and definite integrals, which together constitute the Integral Calculus.

`\color{green} ✍️` There are two forms of the integrals, e.g., indefinite and definite integrals, which together constitute the Integral Calculus.

`\color{green} ✍️` Integration is the inverse process of differentiation.

`\color{green} ✍️` Instead of differentiating a function, we are given the derivative of a function and asked to find its primitive, i.e., the original function. Such a process is called integration or anti 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"`

`\color{green} ✍️` A new symbol, namely, ∫ f (x) dx which will represent the entire class of"`

`\color{green}" anti derivatives read as the indefinite integral of f with respect to x

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

Notation Given that ` (dy)/(dx)= f(x)` , we write `y = ∫ f (x) dx .`

- we mention below the following symbols/terms/phrases with their meanings as given in the Table.

`\color{green} ✍️` Instead of differentiating a function, we are given the derivative of a function and asked to find its primitive, i.e., the original function. Such a process is called integration or anti 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"`

`\color{green} ✍️` A new symbol, namely, ∫ f (x) dx which will represent the entire class of"`

`\color{green}" anti derivatives read as the indefinite integral of f with respect to x

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

Notation Given that ` (dy)/(dx)= f(x)` , we write `y = ∫ f (x) dx .`

- we mention below the following symbols/terms/phrases with their meanings as given in the Table.

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}` |

● Let f (x) = 2x, Then `color{orange}(∫ f (x) dx = x^2 + C)` . For different values of C, we get different integrals. But these integrals are very similar geometrically.

● `y = x^2 + C,` where C is arbitrary constant, represents a family of integrals. By assigning different values to `C,` we get different members of the family.

● Clearly, for `C = 0,` we obtain `y = x^2 ,` a parabola with its vertex on the origin. The curve `y = x^2 + 1` for `C = 1` is obtained by shifting the parabola `y = x^2` one unit along y-axis in positive direction. For `C = – 1, y = x^2 – 1` is obtained by shifting the parabola `y = x^2

` one unit along y-axis in the negative direction and so on.

● So `color{orange}{∫ f (x) dx = x^2 + C}" represents a family of curves."` The different values of C will correspond to different members of this family and these members can be obtained by shifting any one of the curves parallel to itself.

● `"This is the geometrical interpretation of indefinite integral."`

● `y = x^2 + C,` where C is arbitrary constant, represents a family of integrals. By assigning different values to `C,` we get different members of the family.

● Clearly, for `C = 0,` we obtain `y = x^2 ,` a parabola with its vertex on the origin. The curve `y = x^2 + 1` for `C = 1` is obtained by shifting the parabola `y = x^2` one unit along y-axis in positive direction. For `C = – 1, y = x^2 – 1` is obtained by shifting the parabola `y = x^2

` one unit along y-axis in the negative direction and so on.

● So `color{orange}{∫ f (x) dx = x^2 + C}" represents a family of curves."` The different values of C will correspond to different members of this family and these members can be obtained by shifting any one of the curves parallel to itself.

● `"This is the geometrical interpretation 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.

`color{red}{ul{" Proof :"` Let F be any anti derivative of f, i.e.,

`color{orange} {d/(dx) F(x) = f(x)}`

Then ` color{orange} {∫ f (x) dx = F(x) +C}`

`therefore` `color{orange} {d/(dx) ∫ f(x) dx = d/(dx) (F(x) +C )}`

` color{orange} { = d/(dx) F(x) = f(x)}`

Similarly, we note that

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

where C is arbitrary constant called constant of integration.

`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.

`color{red}{ul{" Proof :"` Let F be any anti derivative of f, i.e.,

`color{orange} {d/(dx) F(x) = f(x)}`

Then ` color{orange} {∫ f (x) dx = F(x) +C}`

`therefore` `color{orange} {d/(dx) ∫ f(x) dx = d/(dx) (F(x) +C )}`

` color{orange} { = d/(dx) F(x) = f(x)}`

Similarly, we note that

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

where C is arbitrary constant called constant of integration.

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

`color{red}{ul{" Proof :"` Let f and g be two functions such that

`color{orange} {d/(dx) ∫ f(x) dx = d/(dx) ∫ g (x) dx}`

or `color{orange} {d/(dx) [ ∫ f(x) dx- ∫ g(x) dx ] = 0}`

Hence ` color{orange} {∫ f(x) dx - ∫ g (x) dx = C}` , where C is any real number

or ` color{orange} {∫ f (x) dx = ∫ g (x) dx + C}`

So the families of curves ` color{green} { ∫ f(x) dx +C_1 ,C_1 ∈ R }`

and ` color{green} { ∫ g(x) dx +C_2, C_2 ∈ R }`are identical

Hence, in this sense , `color{orange} {∫ f (x) dx}` and ` color{orange} {∫ g (x) dx} ` are equivalent.

`color{red}{ul{" Proof :"` Let f and g be two functions such that

`color{orange} {d/(dx) ∫ f(x) dx = d/(dx) ∫ g (x) dx}`

or `color{orange} {d/(dx) [ ∫ f(x) dx- ∫ g(x) dx ] = 0}`

Hence ` color{orange} {∫ f(x) dx - ∫ g (x) dx = C}` , where C is any real number

or ` color{orange} {∫ f (x) dx = ∫ g (x) dx + C}`

So the families of curves ` color{green} { ∫ f(x) dx +C_1 ,C_1 ∈ R }`

and ` color{green} { ∫ g(x) dx +C_2, C_2 ∈ R }`are identical

Hence, in this sense , `color{orange} {∫ f (x) dx}` and ` color{orange} {∫ g (x) dx} ` are equivalent.

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

`color{red}{ul{" Proof :"` By Property (I), we have

`color{orange} {d/(dx) [ ∫ [ f(x) + g(x)] dx ] = f(x) + g(x)}` ........(1)

On the otherhand, we find that

`color{orange} {d/(dx) [ ∫ f(x) dx + ∫ g(x) dx ] = d/(dx) ∫ f(x) dx + d/(dx) ∫ g(x) dx}`

`color{orange} {= f(x) + g(x)}` ........(2)

Thus, in view of Property (II), it follows by (1) and (2) that

`color{orange} { ∫ ( f(x) +g(x) ) dx = ∫ f(x) dx + ∫ g(x) dx}`

`color{red}{ul{" Proof :"` By Property (I), we have

`color{orange} {d/(dx) [ ∫ [ f(x) + g(x)] dx ] = f(x) + g(x)}` ........(1)

On the otherhand, we find that

`color{orange} {d/(dx) [ ∫ f(x) dx + ∫ g(x) dx ] = d/(dx) ∫ f(x) dx + d/(dx) ∫ g(x) dx}`

`color{orange} {= f(x) + g(x)}` ........(2)

Thus, in view of Property (II), it follows by (1) and (2) that

`color{orange} { ∫ ( f(x) +g(x) ) dx = ∫ f(x) dx + ∫ g(x) dx}`

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

Proof By the Property (I), `color{orange} {d/(dx) ∫ k f(x) dx = k f(x)}`

Also `color{orange} {d/(dx) [ k ∫ f(x) dx ] = k d/(dx) ∫ f(x) dx = k f(x)}`

Therefore, using the Property (II), we have ` color{orange} {∫ f(x) dx = k ∫ f(x) dx}` .

Proof By the Property (I), `color{orange} {d/(dx) ∫ k f(x) dx = k f(x)}`

Also `color{orange} {d/(dx) [ k ∫ f(x) dx ] = k d/(dx) ∫ f(x) dx = k f(x)}`

Therefore, using the Property (II), we have ` color{orange} {∫ f(x) dx = k ∫ f(x) dx}` .

- 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}` .

`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}` .

Q 3175278166

Write an anti derivative for each of the following functions using the

method of inspection:

(i) `cos 2x `

(ii) `3x^2 + 4x^3`

(iii) `1/x , x ≠ 0 `

Class 12 Chapter 7 Example 1

method of inspection:

(i) `cos 2x `

(ii) `3x^2 + 4x^3`

(iii) `1/x , x ≠ 0 `

Class 12 Chapter 7 Example 1

(i) We look for a function whose derivative is cos 2x. Recall that

`d/(dx) sin 2x = 2 cos 2x`

or `cos 2x = 1/2 d/(dx) (sin 2x ) = d/(dx) (1/2 sin 2x )`

Therefore, an anti derivative of cos 2x is 1/2 sin 2x .

(ii) We look for a function whose derivative is `3x^2 + 4x^3`. Note that

`d/(dx) (x^3 + x^4 ) = 3x^2 + 4x^3`

Therefore, an anti derivative of `3x^2 + 4x^3` is `x^3 + x^4`.

(iii) We know that

`d/(dx) ( log x ) =1/x , x > 0` and `d/(dx) [ log (-x) ] = 1/(-x) (-1) = 1/x , x < 0`

Combining above, we get `d/(dx) ( log |x| ) = 1/x , x ≠ 0`

Therefore, ` int 1/x dx = log |x|` is one of the anti derivatives of 1/x .

Q 3115278169

Find the following integrals:

(i) `int (x^3-1)/(x^2) dx`

(ii) `int (x^2/3 +1) dx`

(iii) `int ( x^3/2 + 2 e^x -1/x ) dx`

Class 12 Chapter 7 Example 2

(i) `int (x^3-1)/(x^2) dx`

(ii) `int (x^2/3 +1) dx`

(iii) `int ( x^3/2 + 2 e^x -1/x ) dx`

Class 12 Chapter 7 Example 2

(i) We have

` int (x^3-1)/( x^2) dx = int x dx - int x^(-2) dx` (by Property V)

`= ( ( x^(1+1) )/(1+1) + C_1 ) - ( ( x^(-2+1) )/(-2+1) + C_2 ) ; C_1 , C_2` are constants of integration

`= x^2/2 + C_1 - ( x^(-1))/( -1) -C_2 = x^2/2 +1/x +C_1 -C_2`

`= (x^2)/2 +1/x +C` , where `C = C_1 – C_2` is another constant of integration

(ii) We have

`int (x^2/3 +1) dx = int x^2/3 dx+ int dx`

` = (x^(2/3 +1) )/(2/3 +1) + x + C= 3/5 x^5/3 + x+ C`

(iii) We have

`int (x^3/2 + 2 e^x - 1/x) dx = int x^3/2 dx + int 2 e^x dx - int 1/x dx`

`= (x^(3/2+1) )/( 3/2+1) + 2 e^x - log | x| + C`

`= 2/5 x^5/2 + 2 e^x -log |x| + C`

Q 3115378260

Find the following integrals:

(i) `int (sin x + cos x) dx`

(ii) `int cosec x (cosec x + cot x) dx`

(iii) `int (1- sin x )/( cos^2 x) dx`

Class 12 Chapter 7 Example 3

(i) `int (sin x + cos x) dx`

(ii) `int cosec x (cosec x + cot x) dx`

(iii) `int (1- sin x )/( cos^2 x) dx`

Class 12 Chapter 7 Example 3

(i) We have

∫(sin x + cos x) dx = ∫sin x dx + ∫cos x dx

= – cos x + sin x + C

(ii) We have

∫(cosec x (cosec x + cot x) dx = ∫cosec2 x dx + ∫cosec x cot x dx

= – cot x – cosec x + C

(iii) We have

` int (1- sin x )/(cos^2 x) = int 2/(cos^2 x) dx - int (sin x)/(cos^2 x) dx`

` = int sec^2 x dx - int tan x sec x dx`

`= tan x – sec x + C`

Q 3125378261

Find the anti derivative F of f defined by `f (x) = 4x^3 – 6` , where F (0) = 3

Class 12 Chapter 7 Example 4

Class 12 Chapter 7 Example 4

One anti derivative of f (x) is `x^4 – 6x` since

`d/(dx) ( x^4 -6x ) = 4x^3 - 6`

Therefore, the anti derivative F is given by

`F(x) = x^4 – 6x + C`, where C is constant.

Given that F(0) = 3, which gives,

3 = 0 – 6 × 0 + C or C = 3

Hence, the required anti derivative is the unique function F defined by

`F(x) = x^4 – 6x + 3`.

(i) We see that if F is an anti derivative of f, then so is `F + C,` where `C` is an constant. Thus, if we know one anti derivative `F` of a function `f,` we can write down an infinite number of anti derivatives of `f` by adding any constant to F expressed by `F(x) + C, C ∈ R.` In applications, it is often necessary to satisfy an additional condition which then determines a specific value of `C` giving unique anti derivative of the given function.

(ii) Sometimes, F is not expressible in terms of elementary functions viz., polynomial, logarithmic, exponential, trigonometric functions and their inverses etc. We are therefore blocked for finding `∫f (x) dx.` For example, it is not possible to find `∫e^((-x)^2) dx` by inspection since we can not find a function whose derivative is `e^((-x)^2).`

(iii) When the variable of integration is denoted by a variable other than `x,` the integral formulae are modified accordingly. For instance

`int y^4 dy = y^5/5 + c`

(ii) Sometimes, F is not expressible in terms of elementary functions viz., polynomial, logarithmic, exponential, trigonometric functions and their inverses etc. We are therefore blocked for finding `∫f (x) dx.` For example, it is not possible to find `∫e^((-x)^2) dx` by inspection since we can not find a function whose derivative is `e^((-x)^2).`

(iii) When the variable of integration is denoted by a variable other than `x,` the integral formulae are modified accordingly. For instance

`int y^4 dy = y^5/5 + c`