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Question

Express the limit as an integral (or multiple of an integral) and evaluate.$\lim _ { N \rightarrow \infty } \frac { 5 } { N } \sum _ { j = 1 } ^ { N } \sqrt { 4 + 5 j / N }$

Solution

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Recall that the right-endpoint limit definition of the definite integral is given as:

$\color{#c34632}\int_{a}^{b}f(x)dx=\lim\limits_{N\to\infty} \sum_{j=1}^{N}f(x_j)\Delta x$

Here, $\color{#c34632}\Delta x=\dfrac{b-a}{N}$ and $\color{#c34632}x_j=a+j\Delta x$.

For our given limit, we can label the following parts:

$\lim\limits_{N\to\infty} \underbrace{\dfrac{5}{N}}_{\color{#c34632}\Delta x} \sum_{j=1}^{N}\overbrace{\sqrt{\underbrace{4}_{{\color{#c34632}a}}+\underbrace{\dfrac{5j}{N}}_{{\color{#c34632}j\Delta x}}}}^{{\color{#c34632}f(x_j)}}$

So, that means:

$\begin{gather*} f(x)=\sqrt{x}\\ a=4\\ \Delta x =\dfrac{5}{N}=\dfrac{9-4}{N} \Rightarrow b=9 \end{gather*}$

Therefore, we get our definite integral to be:

$\lim\limits_{N\to\infty}\dfrac{5}{N}\sum_{j=1}^{N}\sqrt{4+\dfrac{5j}{N}}=\boxed{\color{#4257b2}\int_{4}^{9} \sqrt{x}dx}$

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