Left endpoint approximation

[Hckyplayer8]

So the given is that the sum of the integral of f(x) = x² over the interval [0,1] is 1/3. Thus the combined area of the left-endpoint rectangles = 1/3.

So my job is to figure out how many rectangles were used to come up with that summation?

 

[MarkFL]

What you're going to want to do is come up with a formula in terms of the number of rectangles (usually \(n\)) and then take the limit as \(n\to\infty\).

 

[Steven G]

No no no. Just like before you approximate the area using N rectangles and then you compute the limit as N-->\(\displaystyle \infty\)

(b-a)/n = (1-0)/N = 1/N.

Use the formula [MATH]\left \{ \sum_{j=0}^{n-1} f \left (a + j * \dfrac{b-a}{n} \right ) * \dfrac{b-a}{n} \right \}.[/MATH]

 

[Hckyplayer8]

Yeesh. I hate this N rectangle problem.

I'll study the last thread that was about it and then I will try to apply it to this one and will get back to you all.

 

[Deleted member 4993]

No no no. Just like before you approximate the area using N rectangles and then you compute the limit as N-->\(\displaystyle \infty\)

(b-a)/n = (1-0)/N = 1/N.

Use the formula [MATH]\left \{ \sum_{j=0}^{n-1} f \left (a + j * \dfrac{b-a}{n} \right ) * \dfrac{b-a}{n} \right \}.[/MATH]

Jomo,

When you reply to a post - use the "reply" button inside the referenced post (bottom RHS corner). That will avoid the Khan-fusion regarding determination of "whose post are you responding to".

Your post above started with "No no no" - and I thought you are vehemently disagreeing with Mark's response (#2) just above your reply.

 

[Steven G]

No no no. Just like before you approximate the area using N rectangles and then you compute the limit as N-->\(\displaystyle \infty\)

Jomo,

When you reply to a post - use the "reply" button inside the referenced post (bottom RHS corner). That will avoid the Khan-fusion regarding determination of "whose post are you responding to".

Your post above started with "No no no" - and I thought you are vehemently disagreeing with Mark's response (#2) just above your reply.

I do it just to confuse you but I will try to be better. Thanks for informing me about this.

 

[MarkFL]

View attachment 14586

So the given is that the sum of the integral of f(x) = x² over the interval [0,1] is 1/3. Thus the combined area of the left-endpoint rectangles = 1/3.

So my job is to figure out how many rectangles were used to come up with that summation?

I would begin by plotting the area we are to find:



Now, since we are using a left sum, we are going to want to let:

[MATH]x_0=0[/MATH]
[MATH]x_k=x_{k-1}+\frac{1-0}{n}\implies x_k=\frac{k}{n}[/MATH] where \(0\le k\le n-1\)

[MATH]\int_0^1 x^2\,dx\approx A_n[/MATH]
[MATH]A_n=\sum_{k=0}^{n-1}\left(\frac{1}{n}\cdot x_k^2\right)=\frac{1}{n^3}\sum_{k=0}^{n-1}\left(k^2\right)=\frac{(n-1)(2n-1)}{6n^2}[/MATH]
Hence:

[MATH]\int_0^1 x^2\,dx=A_{\infty}=\lim_{n\to\infty}\left(\frac{(n-1)(2n-1)}{6n^2}\right)=\frac{1}{3}[/MATH]
Does that make sense?