Given that 2<=f(x)<=3 for -5<=x<=1, use a property of definite integrals to find an interval in which the value of interval sign{-5}^{1} f(x)dx lies.

since 2 <= f(x) <= 3, and the width of the interval is 6,

2*6 <= ∫f(x) dx <= 3*6

To find an interval in which the value of the definite integral of f(x) dx lies, we can use the property of definite integrals that states:

If g(x) ≤ f(x) ≤ h(x) for every x in the interval [a, b], then ∫\[a]^{b} g(x) dx ≤ ∫\[a]^{b} f(x) dx ≤ ∫\[a]^{b} h(x) dx.

In this case, we know that 2 ≤ f(x) ≤ 3 for -5 ≤ x ≤ 1.

By applying the property of definite integrals, we have:

∫\[-5]^{1} 2 dx ≤ ∫\[-5]^{1} f(x) dx ≤ ∫\[-5]^{1} 3 dx.

Integrating each inequality separately, we obtain:

2x |_[-5]^[1] ≤ ∫\[-5]^{1} f(x) dx ≤ 3x |_[-5]^[1].

Evaluating the definite integrals on the left and right sides, we get:

2(1) - 2(-5) ≤ ∫\[-5]^{1} f(x) dx ≤ 3(1) - 3(-5).

Simplifying, we have:

12 ≤ ∫\[-5]^{1} f(x) dx ≤ 18.

Therefore, we can conclude that the value of the definite integral ∫\[-5]^{1} f(x) dx lies between 12 and 18, or in other words, the interval for the value of this definite integral is [12, 18].