The answer is most likely A.
The integration interval [a, b] is split up into n subintervals of equal length (so each subinterval has width (b - a)/n, same as the coefficient of the sum of y terms) and approximated by the area of n rectangles with base (b - a)/n and height y.
n subintervals require n + 1 points, with
x₀ = a
x₁ = a + (b - a)/n
x₂ = a + 2(b - a)/n
and so on up to the last point x = b. The right endpoints are x₁, x₂, … etc. and the height of each rectangle are the corresponding y 's at these endpoints. Then you get the formula as given in the photo.
• "Average rate of change" isn't really relevant here. The AROC of a function G(x) continuous* over an interval [a, b] is equal to the slope of the secant line through x = a and x = b, i.e. the value of the difference quotient
(G(b) - G(a) ) / (b - a)
If G(x) happens to be the antiderivative of a function g(x), then this is the same as the average value of g(x) on the same interval,
[tex]g_{\rm ave}=\dfrac{G(b)-G(a)}{b-a}=\dfrac1{b-a}\displaystyle\int_a^b g(x)\,\mathrm dx[/tex]
(* I'm actually not totally sure that continuity is necessary for the AROC to exist; I've asked this question before without getting a particularly satisfying answer.)
• "Trapezoidal rule" doesn't apply here. Split up [a, b] into n subintervals of equal width (b - a)/n. Over the first subinterval, the area of a trapezoid with "bases" y₀ and y₁ and "height" (b - a)/n is
(y₀ + y₁) (b - a)/n
but y₀ is clearly missing in the sum, and also the next term in the sum would be
(y₁ + y₂) (b - a)/n
the sum of these two areas would reduce to
(b - a)/n = (y₀ + 2 y₁ + y₂)
which would mean all the terms in-between would need to be doubled as well to get
[tex]\displaystyle\int_a^b f(x)\,\mathrm dx\approx\frac{b-a}n\left(y_0+2y_1+2y_2+\cdots+2y_{n-1}+y_n\right)[/tex]
