Respuesta :
The summand (R?) is missing, but we can always come up with another one.
Divide the interval [0, 1] into [tex]n[/tex] subintervals of equal length [tex]\dfrac{1-0}n=\dfrac1n[/tex]:
[tex][0,1]=\left[0,\dfrac1n\right]\cup\left[\dfrac1n,\dfrac2n\right]\cup\cdots\cup\left[1-\dfrac1n,1\right][/tex]
Let's consider a left-endpoint sum, so that we take values of [tex]f(\ell_i)={\ell_i}^3[/tex] where [tex]\ell_i[/tex] is given by the sequence
[tex]\ell_i=\dfrac{i-1}n[/tex]
with [tex]1\le i\le n[/tex]. Then the definite integral is equal to the Riemann sum
[tex]\displaystyle\int_0^1x^3\,\mathrm dx=\lim_{n\to\infty}\sum_{i=1}^n\left(\frac{i-1}n\right)^3\frac{1-0}n[/tex]
[tex]=\displaystyle\lim_{n\to\infty}\frac1{n^4}\sum_{i=1}^n(i-1)^3[/tex]
[tex]=\displaystyle\lim_{n\to\infty}\frac1{n^4}\sum_{i=0}^{n-1}i^3[/tex]
[tex]=\displaystyle\lim_{n\to\infty}\frac{n^2(n-1)^2}{4n^4}=\boxed{\frac14}[/tex]
The limit expression for the area under the curve y = x³ as n approaches infinity is; ¹/₄
What is the integral limit?
The given definition is area A of the region S that lies under the graph of the continuous function which is the limit of the sum of the areas of approximating rectangles.
The expression for the area under the curve y = x³ from 0 to 1 as a limit is;
[tex]\lim_{n \to \infty} \Sigma^{n} _{i = 1} (\frac{i}{n})^{3} * \frac{1}{n} }[/tex]
From the expression above, when we factor out 1/n⁴, we will get;
[tex]\lim_{n \to \infty} \frac{1}{n^{4} } \Sigma^{n} _{i = 1} \frac{i}{n} }[/tex]
This is further broken down to get;
[tex]\lim_{n \to \infty} \frac{1}{n^{4} } (\frac{n(n + 1)}{2} )^{2} } }[/tex]
This will be simplified to;
[tex]\lim_{n \to \infty} \frac{1}{n^{4} } \frac{(n^{4} + 2n^{3} + n^{2}) }{2}[/tex]
This would be simplified to;
[tex]\lim_{n \to \infty} \frac{1}{4} + \frac{1}{2n} + \frac{1}{4n^{2} }[/tex]
At limit of n approaches ∞, we have;
Limit = ¹/₄
Read more about integral limits at; https://brainly.com/question/14388434
