# Help me check my HW (area under a curve)

• kingkong11
It shouldn't depend on whether the function is increasing or decreasing. However, the function itself may affect the accuracy of the approximation. In summary, the conversation discusses using the right endpoint formula to find the area under the curve of a function and the mistake made in applying the formula. The correct setup for the formula is also mentioned, as well as the potential effects of the function on the accuracy of the approximation.

## 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 $$\sum_{k=1}^n K^2 = \frac{n(n+1)(2n+1)}{6}$$

## 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:
$$\lim_{n\rightarrow\infty} \sum_{k=1}^n f(\frac{k}{n})*\frac{1}{n} = \left[4-(\frac{k^2}{n^2})\right]*\frac{1}{n}$$
$$= \frac{4}{n} - \frac{K^2}{n^3}$$
$$\lim_{n\rightarrow\infty} 4 - \frac{1}{n^3}(\frac{(n(n+1)(2n+1)}{6} )$$
$$= 4 - \frac{1}{3} = \frac{11}{3}$$

kingkong11 said:

## 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 $$\sum_{k=1}^n K^2 = \frac{n(n+1)(2n+1)}{6}$$

## 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:
$$\lim_{n\rightarrow\infty} \sum_{k=1}^n f(\frac{k}{n})*\frac{1}{n} = \left[4-(\frac{k^2}{n^2})\right]*\frac{1}{n}$$
$$= \frac{4}{n} - \frac{K^2}{n^3}$$
$$\lim_{n\rightarrow\infty} 4 - \frac{1}{n^3}(\frac{(n(n+1)(2n+1)}{6} )$$
$$= 4 - \frac{1}{3} = \frac{11}{3}$$

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)

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

physicsman2 said:
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

kingkong11 said:
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.

Last edited:
physicsman2 said:
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?

kingkong11 said:
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.