If I had a point at (-2,-2) and I reflected it across y=1, where would the new point be located
1 answer:
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Look at the attached picture for the correct process.
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:
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:
We can also write n-th partial sum: