# Question #89f06

Then teach the underlying concepts
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Jim H Share
Jan 26, 2018

#### Explanation:

Using ${x}_{i} =$right endpoints, we have

${\int}_{a}^{b} f \left(x\right) \mathrm{dx} = {\lim}_{n \rightarrow \infty} {\sum}_{i = 1}^{n} \left[f \left(a + i \frac{b - a}{n}\right) \frac{b - a}{n}\right]$

$= {\lim}_{n \rightarrow \infty} {\sum}_{i = 1}^{n} {i}^{27} / {n}^{27} \cdot \frac{1}{n}$

$= {\lim}_{n \rightarrow \infty} {\sum}_{i = 1}^{n} {\left(\frac{i}{n}\right)}^{27} \cdot \frac{1}{n}$

$f \left(x\right) = {x}^{27}$, $a = 0$ and $b = 1$, so

${\int}_{0}^{1} {x}^{27} \mathrm{dx}$

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

1

### This answer has been featured!

Featured answers represent the very best answers the Socratic community can create.

Jim H Share
Jan 26, 2018

#### Explanation:

Using ${x}_{i} =$right endpoints, we have

${\int}_{a}^{b} f \left(x\right) \mathrm{dx} = {\lim}_{n \rightarrow \infty} {\sum}_{i = 1}^{n} \left[f \left(a + i \frac{b - a}{n}\right) \frac{b - a}{n}\right]$

So ${\lim}_{n \rightarrow \infty} {\sum}_{i = 1}^{n} 2 {e}^{\sqrt{\frac{i}{n}}} \frac{1}{n}$

has $f \left(a + i \frac{b - a}{n}\right) = 2 {e}^{\sqrt{\frac{i}{n}}}$ and $\frac{b - a}{n} - \frac{1}{n}$

So $\frac{b - a}{n} = \frac{1}{n}$ and $a + i \frac{b - a}{n} = \frac{i}{n} = i \left(\frac{1}{n}\right)$, so $a = 0$.

Observe that when $i = n$, we have $a + i \frac{b - a}{n} = b$.
In this case we get $b = \frac{n}{n} = 1$

We have

${\int}_{a}^{b} f \left(x\right) \mathrm{dx} = {\lim}_{n \rightarrow \infty} {\sum}_{i = 1}^{n} \left[f \left(a + i \frac{b - a}{n}\right) \frac{b - a}{n}\right]$

${\lim}_{n \rightarrow \infty} {\sum}_{i = 1}^{n} 2 {e}^{\sqrt{0 + \frac{i}{n}}} \frac{1}{n}$

So $f \left(x\right) = 2 {e}^{\sqrt{x}}$

and we get

${\lim}_{n \rightarrow \infty} {\sum}_{i = 1}^{n} 2 {e}^{\sqrt{\frac{i}{n}}} \frac{1}{n} = {\int}_{0}^{1} 2 {e}^{\sqrt{x}} \mathrm{dx}$

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