# Maya's question at Yahoo! Answers regarding a Riemann sum and definite integral

#### MarkFL

Staff member
Here is the question:

I need some help with calculus 1 please?

Here's my problem:
Write a Riemann sum and then a definite integral representing the area of the region, using the strip shown in the figure below. Evaluate the integral exactly.

What is the approximate area of the strip with respect to y? (Use Delta y for Δy as necessary.)

In your definite integral what is the upper endpoint given that the lower endpoint is 0?

and finally, what is the result when you evaluate the definite integral?

10 points to best answer! Thanks!
I have posted a link there to this topic so the OP can see my work.

#### MarkFL

Staff member
Hello Maya,

I will choose to use the lower end of each strip to determine its width.

For an arbitrary strip, its area can be found as follows:

$$\displaystyle A_k=bh$$

where:

$$\displaystyle y_k=k\frac{y_n-y_0}{n}=k\frac{1-0}{n}=\frac{k}{n}$$

$$\displaystyle b=y_k-y_k^2=\frac{nk-k^2}{n^2}$$

$$\displaystyle h=\Delta y=y_{k+1}-y_{k}=\frac{1}{n}$$

and so we have:

$$\displaystyle A_k=\left(y_k-y_k^2 \right)\Delta y=\frac{nk-k^2}{n^3}$$

Now, summing the strips, we find:

$$\displaystyle A_n=\sum_{k=0}^{n-1}\left(\frac{nk-k^2}{n^3} \right)=\frac{1}{n^3}\sum_{k=0}^{n-1}\left(nk-k^2 \right)$$

Using the following identities:

$$\displaystyle \sum_{k=0}^{n-1}(k)=\frac{n(n-1)}{2}$$

$$\displaystyle \sum_{k=0}^{n-1}(k^2)=\frac{n(n-1)(2n-1)}{6}$$

we obtain:

$$\displaystyle A_n=\frac{1}{n^3}\left(\frac{n^2(n-1)}{2}-\frac{n(n-1)(2n-1)}{6} \right)$$

$$\displaystyle A_n=\frac{1}{n^3}\left(\frac{n\left(n^2-1 \right)}{6} \right)=\frac{n^2-1}{6n^2}$$

Thus, when we write the definite integral and evaluate it, we should find it is equal to:

$$\displaystyle A=\lim_{n\to\infty}A_n=\frac{1}{6}$$

Now, to represent the area as a definite integral, we may use:

$$\displaystyle A=\int_0^1 y-y^2\,dy=\left[\frac{y^2}{2}-\frac{y^3}{3} \right]_0^1=\frac{1}{6}\left[3y^2-2y^3 \right]_0^1=\frac{1}{6}((3-2)-(0-0))=\frac{1}{6}$$