# How would I write the Riemann sum needed to find the area under the curve given by the function #f(x)=5x^2+3x+2# over the interval [2,6]?

=#sum20/n{(2^2 +2^2+2^2...)+(16/n +32/n+48/n...)+(16/n^2 +64/n^2+144/n^2+..)} +sum12/n{(2+2+2...)+4/nsum (1+2+3+...)}+ 4/n *2n #

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To find the Riemann sum for the area under the curve of (f(x) = 5x^2 + 3x + 2) over the interval ([2,6]), you would divide the interval into (n) subintervals, each of width (\Delta x). Then, choose sample points within each subinterval and evaluate the function at those points. The Riemann sum is the sum of the areas of rectangles formed by the function values at the sample points and the width of the subintervals.

The formula for the Riemann sum is:

[ \sum_{i=1}^{n} f(x_i^*) \Delta x ]

where (x_i^*) is a sample point in the (i)th subinterval, and (\Delta x) is the width of each subinterval, given by (\frac{b-a}{n}) for the interval ([a, b]).

In this case, (a = 2), (b = 6), and (f(x) = 5x^2 + 3x + 2). Therefore, (\Delta x = \frac{6 - 2}{n}).

The sample points (x_i^*) can be chosen arbitrarily within each subinterval.

The Riemann sum can be expressed as:

[ \sum_{i=1}^{n} (5(x_i^*)^2 + 3x_i^* + 2) \Delta x ]

Substituting (\Delta x) and the sample points, (x_i^*), as desired, will give you the specific Riemann sum approximation.

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