Question

How do you calculate the left and right Riemann sum for the given function over the interval [0, ln2], using n=40 for `e^x`?

Answer 1

Left Riemann sum `= ln(2)/(40(2^(1/40)-1)) ~~ 0.99136068`

Right Riemann sum `= (2^(1/40)ln(2))/(40(2^(1/40)-1)) ~~ 1.00868936`

Explanation:

If `a_1, a_2, a_3,...` is a geometric series with initial term `a` and common ratio `r`, then the general term is:

`a_k = a r^(k-1)`

and the sum is given by:

`sum_(k=1)^n a_k = (a(r^n-1))/(r-1)`

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Given an exponential function `f(x)` and a sequence of equally spaced values of `x` at which we sample `f(x)`, the values at those points will be in geometric progression. So we can calculate the sum using the above formula.

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Dividing the interval `[0, ln(2)]` into `n` equal subintervals, the value of `f(x) = e^x` at the left hand end of each subinterval is:

`a_k = e^(1/n ((k-1) ln(2))) = 2^((k-1)/n)`

i.e. a geometric sequence with initial term `a=2^0 = 1` and common ratio `r = 2^(1/n)`

The left Riemann sum is:

`ln(2)/n sum_(k=1)^n 2^((k-1)/n) = ln(2)/n (2^(n/n) - 1)/(2^(1/n)-1) = ln(2)/(n(2^(1/n)-1))`

The right Rieman sum is given by multiplying this by `2^(1/n)` since every term is `2^(1/n)` times larger..

`(2^(1/n) ln(2))/(n(2^(1/n)-1)`

So with `n=40`, we find

Left Riemann sum `= ln(2)/(40(2^(1/40)-1)) ~~ 0.99136068`

Right Riemann sum `= (2^(1/40)ln(2))/(40(2^(1/40)-1)) ~~ 1.00868936`