To solve this integral, we can apply the Fundamental Theorem of Calculus, which states that if a function f is continuous on the interval [a, b] and F is an antiderivative of f on the interval [a, b], then ∫ f(x) dx from a to b equals F(b) - F(a).
Let's start by defining the function under the integral, which is f = 16x^2 - 49.
Next, we need to find its antiderivative. An antiderivative of a function f is a differentiable function F whose derivative is equal to f. In this case, the antiderivative of 16x^2 is (16/3)x^3 and the antiderivative of -49 is -49x. Therefore, the antiderivative F of f is F= (16/3)x^3 - 49x.
Now we can apply the Fundamental Theorem of Calculus by substituting the lower limit (4/7) into F and then subtracting this value from F evaluated at the upper limit (x). This gives us (16/3)x^3 - 49x + 27.0048590864917.
Next, let's analyze the trigonometric substitution options. We notice that the given integral and its antiderivative contain a quadratic function in x, thus, we need to choose a trigonometric function that simplifies the given function when squared. In this particular case, the secant function fits the bill. Hence, the substitution in this context would be B. x = 4 / 7 * sec(θ).
The next part of the problem requires us to find dx in terms of θ and dθ. Knowing that x = a * sec(θ), we take the derivative of both sides with respect to θ to get dx/dθ = a * sec(θ)*tan(θ). Hence, dx = a * sec(θ)*tan(θ)*dθ.
Finally, to evaluate the integral, we substitute x = 4/7*sec(θ) and dx = 4/7*sec(θ)*tan(θ)*dθ into the integral and calculate its value. This gives us the numerical result 5.33333333333333*x^3 - 49.0*x + 27.0048590864917.