Answer:
![\sum \limits ^(n)_(k=1) 4 \Big [ 1 + (3k)/(n) \Big] \Big [ (3)/(n) \Big ]](https://img.qammunity.org/2022/formulas/mathematics/college/cgb05jj1uzj2wgmx722zmxzwtt9634ldhn.png)
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;
![\Delta x = (4-1)/(x) = (3)/(x) \\ \\ x_i = a+ \Delta x_i](https://img.qammunity.org/2022/formulas/mathematics/college/re3r0n903mlfw2fm3xwu0cbdg3e9wxrh1q.png)
where;
a = 1 and Δ = 4
∴
![x_i = 1+ (3)/(x)i](https://img.qammunity.org/2022/formulas/mathematics/college/axmdf9c38q2v4qy6enkrg1nmwgp72v68jh.png)
For i = k
![x_k = 1+ (3)/(x)k](https://img.qammunity.org/2022/formulas/mathematics/college/koqt5lmbvzcbhygqzujnz4whpq6d6zddsy.png)
However;
![y(x_i) = 4x \\ \\ y(x_i) = 4(1 +(3i)/(x))](https://img.qammunity.org/2022/formulas/mathematics/college/xbj2y9ch85at9377bezfu3kkydrcz2eg08.png)
Thus, the formula for the Reinmann sum is:
![\sum \limits ^(n)_(k=1) \Big [ 4 \Big [ 1 + (3i)/(x) \Big] \Big ] \Delta x \\ \\ \\ \sum \limits ^(n)_(k=1) \Big [ 4 \Big [ 1 + (3k)/(x) \Big] \Big ] (3)/(x)](https://img.qammunity.org/2022/formulas/mathematics/college/ywligsrnsak5b7zng5g35362mnai6dd08a.png)
Since we are taking the limit as n → ∞
![\sum \limits ^(n)_(k=1) 4 \Big [ 1 + (3k)/(n) \Big] \Big [ (3)/(n) \Big ]](https://img.qammunity.org/2022/formulas/mathematics/college/cgb05jj1uzj2wgmx722zmxzwtt9634ldhn.png)