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Area function, First fundamental theorem of integral calculus and Second fundamental theorem of integral calculus

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**Area function:**

`int_a^b f(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 light shaded region in above fig. [Here it is assumed that f(x) > 0 for x ∈ [a, b], the assertion made below is equally true for other functions as well]. The area of this shaded region depends upon the value of x.

In other words, the area of this shaded region is a function of x. We denote this function of x by A(x). We call the function A(x) as Area function and is given by

A(x) = `int_a^x f(x) dx` ....(1)

Based on this definition, the two basic fundamental theorems have been given.

**1) First fundamental theorem of integral calculus:****Theorem:** Let f be a continuous function on the closed interval [a, b] and let A (x) be the area function. Then A′(x) = f (x), for all x ∈ [a, b].**2) Second fundamental theorem of integral calculus:****Theorem:** Let f be continuous function defined on the closed interval [a, b] and F be an anti derivative of f. Then `int_a^b f(x) dx = [F(x)]_a^b = F(b) -F(a)`

**Remarks:**

(i) In words, the Theorem 2 tells us that `∫_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).

(ii) This theorem is very useful, because it gives us a method of calculating the definite integral more easily, without calculating the limit of a sum.

(iii) The crucial operation in evaluating a definite integral is that of finding a function whose derivative is equal to the integrand. This strengthens the relationship between differentiation and integration.

(iv) In `∫_a^b f(x) dx` , the function f needs to be well defined and continuous in [a, b].

**Steps for calculating** `int_a^b f(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

`∫_a^b f(x) dx = [F(x) + C]_a^b = [F(b) + C] - [F(a) + 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 `∫_a^b f(x) dx` .

Video link : https://youtu.be/pFtNF7h2AOM