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Question

Express the limit as an integral (or multiple of an integral) and evaluate.

limN5Nj=1N4+5j/N\lim _ { N \rightarrow \infty } \frac { 5 } { N } \sum _ { j = 1 } ^ { N } \sqrt { 4 + 5 j / N }

Solution

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Answered 1 year ago
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Recall that the right-endpoint limit definition of the definite integral is given as:

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

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

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

limN5NΔxj=1N4a+5jNjΔxf(xj)\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:

f(x)=xa=4Δx=5N=94Nb=9\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:

limN5Nj=1N4+5jN=49xdx\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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