We will use the right Riemann sum. We can break this integral in two parts.

We take the interval and we divide it n times:

The area of the i-th rectangle in the right Riemann sum is:

For the first part of our integral we have:

For the second part we have:

We can now put it all together:
![\sum_{i=1}^{i=n} [(\Delta x)^4 i^3-6(\Delta x)^2i]\\\sum_{i=1}^{i=n}[ (\frac{3}{n})^4 i^3-6(\frac{3}{n})^2i]\\ \sum_{i=1}^{i=n}(\frac{3}{n})^2i[(\frac{3}{n})^2 i^2-6]](https://tex.z-dn.net/?f=%5Csum_%7Bi%3D1%7D%5E%7Bi%3Dn%7D%20%5B%28%5CDelta%20x%29%5E4%20i%5E3-6%28%5CDelta%20x%29%5E2i%5D%5C%5C%5Csum_%7Bi%3D1%7D%5E%7Bi%3Dn%7D%5B%20%28%5Cfrac%7B3%7D%7Bn%7D%29%5E4%20i%5E3-6%28%5Cfrac%7B3%7D%7Bn%7D%29%5E2i%5D%5C%5C%0A%5Csum_%7Bi%3D1%7D%5E%7Bi%3Dn%7D%28%5Cfrac%7B3%7D%7Bn%7D%29%5E2i%5B%28%5Cfrac%7B3%7D%7Bn%7D%29%5E2%20i%5E2-6%5D)
We can also write n-th partial sum:
20% of 52
Set up a proportion.
20/100 = 52/1
Divide
52*0.2 = 10.40
Final answer: 10.4
It’s d because they need that one
C. Since the data is going down by -8, and it starts at +45