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, f' is continuous, and ∫ from 4 to x of 6f'(x) dx = 18, we can use the Fundamental Theorem of Calculus to find f(6).

First, note that the integral of 6f'(x) from 4 to x is 6 times the integral of f'(x) from 4 to x. This is because the constant multiple rule of integration allows us to factor out constants.

So, 6 ∫ from 4 to x of f'(x) dx = 18 implies that ∫ from 4 to x of f'(x) dx = 18/6 = 3.

By the Fundamental Theorem of Calculus, ∫ from 4 to x of f'(x) dx = 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 get f(x) = 14.

Therefore, the value of f(6) is 14.