What is the definite integral from x=2 to x=4: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 - 1 what is the definite integral...

What is the definite integral from x=2 to x=4:

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 - 1 what is the definite integral from x=2 to x=4.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to use the continuity principle, hence, if the function `f(x)` is continuous over the interval `[1,4],` hence, the function is continuous at any point `c = 2 in [1,4]` , such that:

`int_1^4 f(x)dx = int_1^2 f(x)dx + int_2^4 f(x) dx`

The problem provides the information that `int_1^2 f(x)dx = 2 ` and `int_1^4 f(x)dx = -1 ` such that:

`-1 = 2 + int_2^4 f(x) dx => int_2^4 f(x) dx = -1 - 2`

`int_2^4 f(x) dx = -3`

Hence, evaluating `int_2^4 f(x) dx` , using continuity principle over an interval, yields `int_2^4 f(x) dx = -3` .

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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

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