Answer:
[tex]\displaystyle\int_0^5{2x\sqrt{1+x}}\,dx[/tex]
Step-by-step explanation:
When computing the right Riemann sum corresponding to the integral from "a" to "b", the k-th value of x is
x[k] = a + k·∆x
as k goes from 1 to n, where the interval width (∆x) is (b-a)/n.
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Here, we have a=0, b=5, ∆x = 5/n and x[k] = 5k/n. So, the integral ...
[tex]\displaystyle\int_0^5{2x\sqrt{1+x}}\,dx[/tex]
becomes the sum ...
[tex]\displaystyle\Delta x\cdot\sum\limits_{k=1}^{n}{2x_k\sqrt{1+x_k}\\\\=\dfrac{5}{n}\sum\limits_{k=1}^{n}{2\dfrac{5k}{n}}\sqrt{1+\dfrac{5k}{n}}[/tex]
This matches the given sum, so the above integral is the answer to the question.
