integral with a derivative function

Question

if g'(x) = f(x) with g(-11)=a express the value of the integral of -4 to 0 f(x+5)dx in terms of a

Answer 1

Given: g'(x) = f(x), g(-11) = a

Find (definite) integral (of f(x)) on -4 to 0

Find f(x+5)dx in terms of ‘a'

Ignoring f(x + 5) for now, and focusing on just f(x), and assuming Continuity requirements of f(x) on the open interval (-4, 0), then we know the definite integral ∫₋₄⁰f(x)dx exists, but we do not know what it is.

We are not given a continuous function f(x) to integrate over [-4, 0], therefore we do not know whether any resulting function g(x) would be continuous on [-4, 0].

Even if we were given f(x) to integrate, g(x) would only be considered valid on the closed interval of integration [-4, 0] without further information, and we do not even know what g(x) would look like in any neighborhood of x = -11, completely outside our interval… g(x) could look completely different in another interval of continuity, and still be piecewise differentiable (loosely speaking).

In the same way, back to f(x), without knowing f(x) outside of the interval (-4, 0) , whose range only measures 4, all values f(x + 5) would result from domain values ‘x+5’, which _all_ lie outside our known range of continuity for integrability, and may not even exist, and be undefined.

Lastly, it is important to remember: The “indefinite" Integral is one of an infinite number of “functions”, while the “definite” Integral is not a function, but a “number”, a value.

Even if we were given the “value” of the definite integral ∫₋₄⁰f(x)dx (which is just a number), we would still have no idea on what is the “function" g(x), anywhere specific on [-4, 0], much less the value of g(-11) outside that range.

So, insufficient information… some context is missing.

Sufficient information could include: Continuous function f(x) on the closed interval [-11, 5], then we could express g(x) as a function, determine *which* g(x) is our integral function [outside the range integration, but inside its interval of continuity], evaluate it at g(-11) to understand ‘a’, then apply this knowledge to the domain range [1, 5] of f(x) that results from f(x + 5).

Answer 2

Writing u=x+5, then du=dx, and -4<x<0 is equivalent to 1<u<5. The original integral is the same as the integral of f(u)du over the [1,5], which by the Newton-Leibniz formula, is equal to g(5)-g(1). The information of g(-11), however, is irrelevant. Although f(x) may have infinitely many anti-derivatives (aka primitive functions), they differ by a constant. There is not enough information to relate this integral with the value of g(-11).