Help me check my HW (area under a curve)

[kingkong11]

Homework Statement

I need to use the right endpoint formula to find the area under the curve f(x) = 4-x² over the interval [1,2]
I integrated this problem and got 5/3. But doing it the long way got me 11/3. Can someone where I made my mistake on applying the right endpoint formula, thanks!

Homework Equations

right endpoint formula = f(\frac{b-a}{n}*k) and <br /> \sum_{k=1}^n K^2 = \frac{n(n+1)(2n+1)}{6}

The Attempt at a Solution

Before I use the right endpoint formula I tried integration and got:
\int_{1}^{2} 4-x^2 dx = \left[ 4x-\frac{x^3}{3}\right]_{1}^{2} = \frac{5}{3}

here is what I got using the right endpoint formula:
\lim_{n\rightarrow\infty} \sum_{k=1}^n f(\frac{k}{n})*\frac{1}{n}<br /> = \left[4-(\frac{k^2}{n^2})\right]*\frac{1}{n}
= \frac{4}{n} - \frac{K^2}{n^3}
\lim_{n\rightarrow\infty} 4 - \frac{1}{n^3}(\frac{(n(n+1)(2n+1)}{6} )
= 4 - \frac{1}{3} = \frac{11}{3}

 

[physicsman2]

Homework Statement

I need to use the right endpoint formula to find the area under the curve f(x) = 4-x² over the interval [1,2]
I integrated this problem and got 5/3. But doing it the long way got me 11/3. Can someone where I made my mistake on applying the right endpoint formula, thanks!

Homework Equations

right endpoint formula = f(\frac{b-a}{n}*k) and <br /> \sum_{k=1}^n K^2 = \frac{n(n+1)(2n+1)}{6}

The Attempt at a Solution

Before I use the right endpoint formula I tried integration and got:
\int_{1}^{2} 4-x^2 dx = \left[ 4x-\frac{x^3}{3}\right]_{1}^{2} = \frac{5}{3}

here is what I got using the right endpoint formula:
\lim_{n\rightarrow\infty} \sum_{k=1}^n f(\frac{k}{n})*\frac{1}{n}<br /> = \left[4-(\frac{k^2}{n^2})\right]*\frac{1}{n}
= \frac{4}{n} - \frac{K^2}{n^3}
\lim_{n\rightarrow\infty} 4 - \frac{1}{n^3}(\frac{(n(n+1)(2n+1)}{6} )
= 4 - \frac{1}{3} = \frac{11}{3}

You made a mistake using the right endpoint, mainly the set up of the problem the way I see it.

It should be something like:

lim as n --> infinity

summation from k = 1 to n of

[4 - (1 + k(1/n))^2] x (1/n)

 

[hgfalling]

Hint: try evaluating \int_0^1 f(x) dx

 

[kingkong11]

You made a mistake using the right endpoint, mainly the set up of the problem the way I see it.

It should be something like:

lim as n --> infinity

summation from k = 1 to n of

[4 - (1 + k(1/n))^2] x (1/n)

I looked at my setup again, but I still don't see the mistake. Where did you get the 1 from? [4 - (1 + k(1/n))^2] can you explain a little more please?
Thanks

 

[physicsman2]

I looked at my setup again, but I still don't see the mistake. Where did you get the 1 from? [4 - (1 + k(1/n))^2] can you explain a little more please?
Thanks

what you substitute for x should be in the form a + k(delta x) since it's a right endpoint approximation (at least the way I learned it). where a is the lower bound of integration or first number in the interval, in this case 1.

Edit: I get 5/3 my way.

 

[kingkong11]

what you substitute for x should be in the form a + k(delta x) since it's a right endpoint approximation (at least the way I learned it). where a is the lower bound of integration or first number in the interval, in this case 1.

Edit: I get 5/3 my way.

Yea, the formula I used is off that's why I was not able to figure out which part I made a mistake on.

While on the subject of formula, do I switch formula when I have an increasing function?

 

[physicsman2]

While on the subject of formula, do I switch formula when I have an increasing function?

What do you mean?

The formula depends on where you are approximating, i.e. right endpoint, left endpoint or midpoint.