The table gives values of <em>f(x)</em> at various <em>x</em> in the interval [0, 6], and you have to use them to approximate the definite integral,
∫₀⁶ <em>f(x)</em> d<em>x</em>
with a left-endpoint sum using 3 intervals of width 2. This means you have to
• split up [0, 6] into 3 subintervals of equal length, namely [0, 2], [2, 4], and [4, 6]
• take the left endpoints of these intervals and the values of <em>f(x)</em> these endpoints, so <em>x</em> ∈ {0, 2, 4} and <em>f(x)</em> ∈ {0, 0.48, 0.84}
• approximate the area under <em>f(x)</em> on [0, 6] with the sum of the areas of rectangles with dimensions 2 × <em>f(x)</em> (that is, width = 2 and height = <em>f(x)</em> for each rectangle)
So the Riemann sum is
∫₀⁶ <em>f(x)</em> d<em>x </em>≈ 2 ∑ <em>f(x)</em>
for <em>x</em> ∈ {0, 2, 4}, which is about
∫₀⁶ <em>f(x)</em> d<em>x </em>≈ 2 (0 + 0.48 + 0.84) = 2.64
making the answer A.
