Answer:
[tex]\sum \limits ^{n}_{k=1} 4 \Big [ 1 + \dfrac{3k}{n} \Big] \Big [ \dfrac{3}{n} \Big ][/tex]
Step-by-step explanation:
Given the function:
f(x) = 4x; we are to determine the expression given the Reimman sum formula for the given function f(x) = 4x over the interval [1,4]
Since;
[tex]\Delta x = \dfrac{4-1}{x} = \dfrac{3}{x} \\ \\ x_i = a+ \Delta x_i[/tex]
where;
a = 1 and Δ = 4
∴
[tex]x_i = 1+ \dfrac{3}{x}i[/tex]
For i = k
[tex]x_k = 1+ \dfrac{3}{x}k[/tex]
However;
[tex]y(x_i) = 4x \\ \\ y(x_i) = 4(1 +\dfrac{3i}{x})[/tex]
Thus, the formula for the Reinmann sum is:
[tex]\sum \limits ^{n}_{k=1} \Big [ 4 \Big [ 1 + \dfrac{3i}{x} \Big] \Big ] \Delta x \\ \\ \\ \sum \limits ^{n}_{k=1} \Big [ 4 \Big [ 1 + \dfrac{3k}{x} \Big] \Big ] \dfrac{3}{x}[/tex]
Since we are taking the limit as n → ∞
[tex]\sum \limits ^{n}_{k=1} 4 \Big [ 1 + \dfrac{3k}{n} \Big] \Big [ \dfrac{3}{n} \Big ][/tex]
