`color{red}♦` 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}` .........................(1)

`\color{green} ✍️ color{blue} text{ First fundamental theorem of integral calculus}`

Theorem 1 : Let `f` be a continuous function on the closed interval `[a, b]` and let `A (x)` be the area function.

Then `color{green}{A′(x) = f (x),}` for all `x ∈ [a, b].`

`\color{green} ✍️ color{blue} text{ Second fundamental theorem of integral calculus}`

We state below an important theorem which enables us to evaluate definite integrals by making use of anti derivative.

Theorem 2 : Let `f` be continuous function defined on the closed interval `[a, b]` and `F` be an anti derivative of `f.`

Then, `color{brown} {int_a^b f(x) dx = [F(x)]_a^b = F(b) - F(a)}`

`color{green} text{Remarks}`

(i) Theorem 2 tells us that `int_a^b f(x) dx =` (value of the anti derivative F of f at the upper limit b – value of the same anti derivative at the lower limit a).

(iii) In `int_a^b f(x) dx` , the function `f` needs to be well defined and continuous in `[a, b].` For instance, the consideration of definite integral `int_(-2)^3 x (x^2 -1)^(1/2) dx` is erroneous since the function `f` expressed by `f (x) = x(x^2 -1)^(1/2)` is not defined in a portion `-1 < x < 1 ` of the closed interval [– 2, 3].

`text{Steps for calculating} color{brown} {int_a^bf(x)dx}`

(i) Find the indefinite integral `∫ f (x) dx .` Let this be `F(x)` There is no need to keep integration constant `C`

because if we consider `F(x) + C` instead of `F(x),` we get

`=>color{brown}{ int_a^b f(x) dx = [F(x) + C]_a^b = [F(b) +C ] - [F(a) +C] = F(b) - F(a)}`

Thus, the arbitrary constant disappears in evaluating the value of the definite integral.

(ii) Evaluate `F(b) – F(a) = [F ( x)]_a^b` , which is the value of `int_a^b f(x) dx`

`=>` 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}` .........................(1)

`\color{green} ✍️ color{blue} text{ First fundamental theorem of integral calculus}`

Theorem 1 : Let `f` be a continuous function on the closed interval `[a, b]` and let `A (x)` be the area function.

Then `color{green}{A′(x) = f (x),}` for all `x ∈ [a, b].`

`\color{green} ✍️ color{blue} text{ Second fundamental theorem of integral calculus}`

We state below an important theorem which enables us to evaluate definite integrals by making use of anti derivative.

Theorem 2 : Let `f` be continuous function defined on the closed interval `[a, b]` and `F` be an anti derivative of `f.`

Then, `color{brown} {int_a^b f(x) dx = [F(x)]_a^b = F(b) - F(a)}`

`color{green} text{Remarks}`

(i) Theorem 2 tells us that `int_a^b f(x) dx =` (value of the anti derivative F of f at the upper limit b – value of the same anti derivative at the lower limit a).

(iii) In `int_a^b f(x) dx` , the function `f` needs to be well defined and continuous in `[a, b].` For instance, the consideration of definite integral `int_(-2)^3 x (x^2 -1)^(1/2) dx` is erroneous since the function `f` expressed by `f (x) = x(x^2 -1)^(1/2)` is not defined in a portion `-1 < x < 1 ` of the closed interval [– 2, 3].

`text{Steps for calculating} color{brown} {int_a^bf(x)dx}`

(i) Find the indefinite integral `∫ f (x) dx .` Let this be `F(x)` There is no need to keep integration constant `C`

because if we consider `F(x) + C` instead of `F(x),` we get

`=>color{brown}{ int_a^b f(x) dx = [F(x) + C]_a^b = [F(b) +C ] - [F(a) +C] = F(b) - F(a)}`

Thus, the arbitrary constant disappears in evaluating the value of the definite integral.

(ii) Evaluate `F(b) – F(a) = [F ( x)]_a^b` , which is the value of `int_a^b f(x) dx`

Q 3175480366

Evaluate the following integrals:

(i) ` ∫_(2)^3 x^2 dx`

(ii) `∫_(4)^9 (sqrt x)/ ((30 -x^(3/2))^2 dx`

(iii) ` ∫_(1)^2 (x dx)/( (x+1) ( x+2) )`

(iv) `∫_(0)^(pi/4) sin^3 2t cos 2 t dt`

Class 12 Chapter 7 Example 27

(i) ` ∫_(2)^3 x^2 dx`

(ii) `∫_(4)^9 (sqrt x)/ ((30 -x^(3/2))^2 dx`

(iii) ` ∫_(1)^2 (x dx)/( (x+1) ( x+2) )`

(iv) `∫_(0)^(pi/4) sin^3 2t cos 2 t dt`

Class 12 Chapter 7 Example 27

(i) Let `I = ∫_(2)^3 x^2 dx` . Since ` ∫ x^2 dx = x^3/3 = F(x)`

Therefore, by the second fundamental theorem, we get

`I=F(3) - F(2) = 27/3 - 8/3 = 19/3`

(ii) Let `I = ∫_(4)^9 (sqrt x)/( (30- x^(3/2))^2) dx` . We first find the anti derivative of the integrand.

Put `30- x^3/2 = t`. Then `-3/2 sqrt x dx = dt` or `sqrt x dx = -2/3 dt`

Thus, ` ∫ (sqrt x)/( (30- x^(3/2))^2 ) dx = -2/3 ∫ (dt)/(t^2) = 2/3 [1/t] = 2/3 [1/(30- x^(3/2)) ] =F (x)`

Therefore, by the second fundamental theorem of calculus, we have

`I= F (9) -F(4) =2/3 [ 1/(30 - x^(3/2)) ]_(4)^9`

` = 2/3 [1/(30-27) - 1/(30-8)] = 2/3 [1/3 - 1/22] = 19/99`

(iii) Let `I = ∫_(1)^2 (x dx )/( (x+1) (x+2) )`

Using partial fraction, we get `x/( (x+1) (x+2) ) = (-1)/(x+1) +2/(x+2)`

So ` ∫ (x dx)/( (x+1) (x+2) ) = - log | x+1 | + 2 log | x+2 | = F(x)`

Therefore, by the second fundamental theorem of calculus, we have

`I = F(2) – F(1) = [– log 3 + 2 log 4] – [– log 2 + 2 log 3]`

` = -3 log 3 + log 2 + 2 log 4 = log (32/27)`

(iv) Let `I = ∫_(0)^(pi/4) sin^3 2t cos 2 t dt` Consider ` ∫ sin^3 2t cos 2 t dt`

Put `sin 2t = u` so that `2 cos 2t dt = du` or `cos 2t dt = 1/2 du`

So ` ∫ sin^3 2t cos 2t dt = 1/2 ∫ u^3 du`

`= 1/8 [ u^4 ] = 1/8 sin^4 2t = F (t )` say

Therefore, by the second fundamental theorem of integral calculus

`I= F (pi/4) -F (0) = 1/8 [ sin^4 (pi/2) - sin^4 0 ] = 1/8`