CALCULUS: For the function whose values are given in the table below, the integral from 0 to 6 of f(x)dx is approximated by a Reimann Sum using the value at the left endpoint value of each of three intervals with width 2. x 0 1 2 3 4 5 6

f(x) 0 0.25 0.48 0.68 0.84 0.95 1

The approximation is:

A. 2.64

B. 3.64

C. 3.72

The table gives values of f(x) at various x in the interval [0, 6], and you have to use them to approximate the definite integral,

∫₀⁶ f(x) dx

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 f(x) these endpoints, so x ∈ {0, 2, 4} and f(x) ∈ {0, 0.48, 0.84}

• approximate the area under f(x) on [0, 6] with the sum of the areas of rectangles with dimensions 2 × f(x) (that is, width = 2 and height = f(x) for each rectangle)

So the Riemann sum is

∫₀⁶ f(x) dx ≈ 2 ∑ f(x)

for x ∈ {0, 2, 4}, which is about

∫₀⁶ f(x) dx ≈ 2 (0 + 0.48 + 0.84) = 2.64

making the answer A.

x+y=90

x=30-y=x

step-by-step explanation:

if you understand you will get the answer

(a) the volume of the cone is 20/3 in3

hence, option a is correct.