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Find a closed form expression for the nth right Riemann sum of this integral?

Find a closed form expression for the nth right Riemann sum of this integral?-example-1

1 Answer

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Partitioning the interval
[6,11] into
n equally-spaced subintervals gives
n rectangles of width
\Delta x=\frac{11-6}n=\frac5n and of heights determined by the right endpoints of each subinterval.

If
a=x_1=6, then
x_2=6+\frac5n,
x_3=6+2\frac5n, and so on, up to
b=x_(n+1)=6+n\frac5n=11. Because we're using the right endpoints, the approximation will consider
x_2,\ldots,x_(n+1)

The definite integral is then approximated by


\displaystyle\int_6^(11)(1-5x)\,\mathrm dx\approx\sum_(i=2)^(n+1)f(x_i)\Delta x=\sum_(i=2)^(n+1)\left(1-\left(6+(i-1)\frac5n\right)\frac5n

You have


\displaystyle\sum_(i=2)^(n+1)\left(1-5\left(6+(i-1)\frac5n\right)\frac5n=\sum_(i=2)^(n+1)\left(-29-\frac{25}n(i-1)\right)\frac5n

=\displaystyle-{145}n\sum_(i=2)^(n+1)1-(125)/(n^2)\sum_(i=2)^(n+1)(i-1)

=-145-(125(n^2+n))/(2n^2)

=-\frac{145}2-(125)/(2n)

To check that this is correct, let's make sure the sum converges to the exact value of the definite integral. As
n\to\infty, you have the sum converging to
-\frac{145}2.

Meanwhile,


\int_6^(11)(1-5x)\,\mathrm dx=\left[x-\frac52x^2\right]_(x=6)^(x=11)=-\frac{415}2

so we're done.
User Michael Borgwardt
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