Respuesta :
Answer:
Step-by-step explanation:
a)
[tex]A=\int\limits^a_b {f(x)} \, dx =\lim_{n \to \infty} \sum\limits^n_{i=1}{f(x_i)}\, \bigtriangleup x\\\\a=0;b=1\rightarrow \bigtriangleup x=\frac{1-0}{n}=\frac{1}{n}\\\\x_0=0;x_1=\frac{1}{n};x_2=\frac{2}{n};x_3=\frac{3}{n};..;x_i=\frac{i}{n}\\\\f(x)=x^3\\\\f(x_i)=[\frac{i}{n}]^3=\frac{i^3}{n^3}[/tex]
Then
[tex]A= \lim_{n \to \infty} {\sum\limits^n_{i=1}(\frac{i^3}{n^3})*\frac{1}{n}}[/tex]
b)
[tex]A= \lim_{n \to \infty} [\frac{1}{n}*\sum\limits^n_{i=1}\frac{i^3}{n^3}]\\\\\\A= \lim_{n \to \infty} [\frac{1}{n}*\frac{1}{n^3}\sum\limits^n_{i=1} i^3]\\\\\\A= \lim_{n \to \infty} [\frac{1}{n^4}*[\frac{n(n+1)}{2}]^2]\\\\\\A= \lim_{n \to \infty} [\frac{1}{n^4}*\frac{n^2(n+1)^2}{4}]=\lim_{n \to \infty} \frac{(n+1)^2}{4n^2}\\\\\\A=\frac{1}{4}*\lim_{n \to \infty} (\frac{n+1}{n})^2=\frac{1}{4}*1^2=\frac{1}{4}[/tex]
The area can be expressed as a limit as:
[tex]A = \lim_{n \to \infty} \sum_i \frac{1}{n} (\frac{i}{n} )^3[/tex]
And using L'Hopital's rule, we solve:
A = 1/4.
How to find the expression for the area?
a) The integral for the area is:
[tex]A = \int\limits^1_0 {x^3} \, dx[/tex]
To write as a limit, we define Δx = (1 - 0)/n = 1/n, and take the limit when n tends to infinity and write:
[tex]A = \lim_{n \to \infty} \sum_i \Delta x*f(i*\Delta x)[/tex]
Replacing Δx we have:
[tex]A = \lim_{n \to \infty} \sum_i \frac{1}{n} (\frac{i}{n} )^3[/tex]
Where i goes between 0 and n.
b) Using the given relation, we can write:
[tex]A = \lim_{n \to \infty} \sum_i \frac{1}{n} (\frac{i}{n} )^3\\\\A = \lim_{n \to \infty} \frac{1}{n^4} \sum_i i^3\\\\A = \lim_{n \to \infty} \frac{1}{n^4}(\frac{n*(n + 1)}{2})^2 \\[/tex]
Now, using the L'hophital rule we get:
[tex]A = \lim_{n \to \infty} \frac{1}{n^4}(\frac{n*(n + 1)}{2})^2 \\\\A = \frac{4}{4*4} = \frac{1}{4}[/tex]
If you want to learn more about limits, you can read:
https://brainly.com/question/21348158