If f(4) = 11, f ' is continuous, and 6f '(x) dx4 = 18, what is the value of f(6)?f(6) =

The problem involves the Fundamental Theorem of Calculus, which states that if a function f is continuous over the interval [a, b] and F is an antiderivative of f on [a, b], then the definite integral from a to b of f(x) dx is F(b) - F(a).

Given that f(4) = 11 and ∫6f '(x) dx from 4 to x = 18, we can rewrite the integral as 6∫f '(x) dx from 4 to x = 18.

Dividing both sides by 6 gives ∫f '(x) dx from 4 to x = 3.

By the Fundamental Theorem of Calculus, ∫f '(x) dx from 4 to x = f(x) - f(4).

So, f(x) - f(4) = 3.

Substituting the given value f(4) = 11, we get f(x) - 11 = 3.

Solving for f(x), we find f(x) = 14.

Since the question asks for f(6), we can say that f(6) = 14.