The answer is most likely A.
The integration interval [<em>a</em>, <em>b</em>] is split up into <em>n</em> subintervals of equal length (so each subinterval has width (<em>b</em> - <em>a</em>)/<em>n</em>, same as the coefficient of the sum of <em>y</em> terms) and approximated by the area of <em>n</em> rectangles with base (<em>b</em> - <em>a</em>)/<em>n</em> and height <em>y</em>.
<em>n</em> subintervals require <em>n</em> + 1 points, with
<em>x</em>₀ = <em>a</em>
<em>x</em>₁ = <em>a</em> + (<em>b</em> - <em>a</em>)/<em>n</em>
<em>x</em>₂ = <em>a</em> + 2(<em>b</em> - <em>a</em>)/<em>n</em>
and so on up to the last point <em>x</em> = <em>b</em>. The right endpoints are <em>x</em>₁, <em>x</em>₂, … etc. and the height of each rectangle are the corresponding <em>y </em>'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 <em>G(x)</em> continuous* over an interval [<em>a</em>, <em>b</em>] is equal to the slope of the secant line through <em>x</em> = <em>a</em> and <em>x</em> = <em>b</em>, i.e. the value of the difference quotient
(<em>G(b)</em> - <em>G(a)</em> ) / (<em>b</em> - <em>a</em>)
If <em>G(x)</em> happens to be the antiderivative of a function <em>g(x)</em>, then this is the same as the average value of <em>g(x)</em> on the same interval,
(* 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 [<em>a</em>, <em>b</em>] into <em>n</em> subintervals of equal width (<em>b</em> - <em>a</em>)/<em>n</em>. Over the first subinterval, the area of a trapezoid with "bases" <em>y</em>₀ and <em>y</em>₁ and "height" (<em>b</em> - <em>a</em>)/<em>n</em> is
(<em>y</em>₀ + <em>y</em>₁) (<em>b</em> - <em>a</em>)/<em>n</em>
but <em>y</em>₀ is clearly missing in the sum, and also the next term in the sum would be
(<em>y</em>₁ + <em>y</em>₂) (<em>b</em> - <em>a</em>)/<em>n</em>
the sum of these two areas would reduce to
(<em>b</em> - <em>a</em>)/<em>n</em> = (<em>y</em>₀ + <u>2</u> <em>y</em>₁ + <em>y</em>₂)
which would mean all the terms in-between would need to be doubled as well to get
