How do you use the midpoint rule to approximate the integral #3x8x^2dx# from the interval [1,4] with #n=3#?
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To use the midpoint rule to approximate the integral (3x  8x^2 , dx) from the interval ([1, 4]) with (n = 3), follow these steps:

Determine the width of each subinterval: ( \Delta x = \frac{b  a}{n} = \frac{4  (1)}{3} = \frac{5}{3} ).

Identify the midpoints of each subinterval: ( x_1 = 1 + \frac{\Delta x}{2} = 1 + \frac{5/3}{2} = \frac{1}{3} ), ( x_2 = \frac{1}{3} + \frac{5}{3} = \frac{4}{3} ), ( x_3 = \frac{4}{3} + \frac{5}{3} = 3 ).

Evaluate the function at each midpoint: ( f(x_1) = 3(1/3)  8(1/3)^2 ), ( f(x_2) = 3(4/3)  8(4/3)^2 ), ( f(x_3) = 3(3)  8(3)^2 ).

Calculate the sum of these function values multiplied by the width of each subinterval: ( \text{Approximation} = \Delta x \left[ f(x_1) + f(x_2) + f(x_3) \right] ).

Simplify the expression to find the numerical approximation of the integral.
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