a) use definition 2 to find an expression for the area under the curve y=x^3 from 0 to 1 as a limit.
(b) the following formula for the sum of the cubes of the first n integers is proved in Appendix E. useit to evaluate the limit in part (a).
1^3 + 2^3 +3^3+.....n^3 = [n(n+1)/2]^2

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

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