Splitting up the interval of integration into [tex]n[/tex] subintervals gives the partition
[tex]\left[0,\dfrac1n\right],\left[\dfrac1n,\dfrac2n\right],\ldots,\left[\dfrac{n-1}n,1\right][/tex]
Each subinterval has length [tex]\dfrac{1-0}n=\dfrac1n[/tex]. The right endpoints of each subinterval follow the sequence
[tex]r_i=\dfrac in[/tex]
with [tex]i=1,2,3,\ldots,n[/tex]. Then the left-endpoint Riemann sum that approximates the definite integral is
[tex]\displaystyle\sum_{i=1}^n\frac{{r_i}^3}n[/tex]
and taking the limit as [tex]n\to\infty[/tex] gives the area exactly. We have
[tex]\displaystyle\lim_{n\to\infty}\frac1n\sum_{i=1}^n\left(\frac in\right)^3=\lim_{n\to\infty}\frac{n^2(n+1)^2}{4n^3}=\boxed{\frac14}[/tex]
The limit of the area under the curve y = x3 from 0 to 1 as a limit is 1/4
First, we would have to split the interval of integration:
[tex][0, 1/n], [1/n, 2/n]. . .[n-1/n, 1][/tex]
Each subinterval has length [tex]1-0/n= 1/n[/tex]
Hence, the right endpoints of each subinterval follow the sequence:
[tex]ri= i/n[/tex]
With i= 1, 2, 3,. . .n
Then the left-endpoint Riemann sum that approximates the definite integral is
Σ i=1 r^3/n
and taking the limit as n-->∞ gives the exact area.
Hence, the final answer is:
= 1/n Σ, limn-->∞, i=1 (i/n)^3
= 1/4
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