`\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`