# Given that definite integral of f(x) from x = 1 to x = 2 is 2 and definite integral of f(x) from x = 1 to x = 4 is - 1find integral of f(x) from x = 2 to x = 4 .

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If f(x) is defined and continuous over the interval [a, b], except maybe at a finite number of points, we'll write Int f(x)dx from a to b as:

Int f(x)dx (a->b) = Int f(x)dx (a->c) + Int f(x)dx(c->b)

We'll put the endpoints of the interval [a,b] as:

a = 1 and b = 4.

Int f(x)dx (1->4) = Int f(x)dx (1->2) + Int f(x)dx(2->4)

We'll subtract both sides by Int f(x)dx (1->2) and we'll use the symmetric property:

Int f(x)dx(2->4) = Int f(x)dx (1->4) - Int f(x)dx (1->2)

We'll substitute Int f(x)dx (1->4) = -1 and

Int f(x)dx (1->2) = 2 and we'll get:

Int f(x)dx(2->4) = -1 - 2 = -3

**So, the definite integral of the function f(x), from x = 2 to x = 4 is:**

**Int f(x)dx(2->4) = -1 - 2 = -3**