Verified Answer :

LammettHash

\(\dfrac1{50}\left(\sqrt{\dfrac1{50}}+\sqrt{\dfrac2{50}}+\cdots+\sqrt{\dfrac{50}{50}}\right)=\displaystyle\sum_{n=1}^{50}\sqrt{\dfrac n{50}}\frac1{50}\)

describes a sum of the areas of 50 rectangles, each of width \(\dfrac1{50}\), and the \(n\)th rectangle has height \(\sqrt{\dfrac n{50}}\).

\(\sqrt{\dfrac n{50}}\) ranges from a small positive number to 1, which means the integration interval must be \([0,1]\). The sum is then the right-endpoint Riemann sum of \(\sqrt x\) over the interval with \(50\) equally spaced subintervals, so

\(\displaystyle\sum_{n=1}^{50}\sqrt{\dfrac n{50}}\frac1{50}\approx\int_0^1\sqrt x\,\mathrm dx\)

The sum itself evaluates to roughly 0.676, while the exact value of the integral is \(\dfrac23\approx0.667\).

okpalawalter8

The sum bears a value relatively close to that of the integral

\(0.676 \approx 0.667\)

Sum approximation

Question Parameters:

\(\frac{1}{50} (\sqrt{\frac{1}{50}} +\sqrt{\frac{1}{50}} + \sqrt{\frac{1}{50}} + ... + \sqrt{ \frac{1}{50}}\)

Generally, the equation for the expression  is mathematically given as

 \(\frac{1}{50} (\sqrt{\frac{1}{50}} +\sqrt{\frac{1}{50}} + \sqrt{\frac{1}{50}} + ... + \sqrt{ \frac{1}{50}}\)

This expression above readily shows an expression of 50 rectangles whose individual width is 1/50 and as progression continues you get a height of n/50

Hence, we decipher the integration interval to be [0,1]

Therefore, the sum of  right-endpoint Riemann

\(\sum_n=1^50 \sqrt{n}{50} \frac{1}{50}=\int^1_0 \sqrt{x} dx\)

Where

\(\sum_n=1^50 \sqrt{n}{50} \frac{1}{50} \approx 0.676\)

\(\int^1_0 \sqrt{x} dx \approx 0.667\)

Therefore, we can say that the sum bears a value relatively close to that of the integral

\(0.676 \approx 0.667\)

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