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[tex]\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}[/tex]

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

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

[tex]\displaystyle\sum_{n=1}^{50}\sqrt{\dfrac n{50}}\frac1{50}\approx\int_0^1\sqrt x\,\mathrm dx[/tex]

The sum itself evaluates to roughly 0.676, while the exact value of the integral is [tex]\dfrac23\approx0.667[/tex].
okpalawalter8

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

[tex]0.676 \approx 0.667[/tex]

Sum approximation

Question Parameters:

[tex]\frac{1}{50} (\sqrt{\frac{1}{50}} +\sqrt{\frac{1}{50}} + \sqrt{\frac{1}{50}} + ... + \sqrt{ \frac{1}{50}}[/tex]

Generally, the equation for the expression  is mathematically given as

 [tex]\frac{1}{50} (\sqrt{\frac{1}{50}} +\sqrt{\frac{1}{50}} + \sqrt{\frac{1}{50}} + ... + \sqrt{ \frac{1}{50}}[/tex]

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

[tex]\sum_n=1^50 \sqrt{n}{50} \frac{1}{50}=\int^1_0 \sqrt{x} dx[/tex]

Where

[tex]\sum_n=1^50 \sqrt{n}{50} \frac{1}{50} \approx 0.676[/tex]

[tex]\int^1_0 \sqrt{x} dx \approx 0.667[/tex]

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

[tex]0.676 \approx 0.667[/tex]

For more information on Arithmetic visit

https://brainly.com/question/22568180

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