Thread: Using Riemann Sums To Find Area

1. Using Riemann Sums To Find Area

The function is y = 3x - 4, and I am to find the area on the interval [2, 5] with 5 rectangles. I have done this twice already, and I deemed it futile to give it another chance. I found the width of each rectangle to be 3/5 and my left endpoint to be 2 + (i -1)3/5. I try to compute the inscribed rectangles with this information, but I can't seem to get it. It always ends up coming out greater than the actual area.

2. Re: Using Riemann Sums To Find Area

Originally Posted by Bashyboy
The function is y = 3x - 4, and I am to find the area on the interval [2, 5] with 5 rectangles. I have done this twice already, and I deemed it futile to give it another chance. I found the width of each rectangle to be 3/5 and my left endpoint to be 2 + (i -1)3/5. I try to compute the inscribed rectangles with this information, but I can't seem to get it. It always ends up coming out greater than the actual area.
Your setup looks okay. Remember that to find the height of each rectangle, you need to evaluate the function at the endpoint. It should look like

$\sum_{i=1}^5\left[3\left(2+\frac35(i-1)\right)-4\right]\left(\frac35\right)$

And after some simplifying,

$=\frac{3}{25}\sum_{i=1}^5(9i+1)=\frac3{25}\left(9 \sum_{i=1}^5i+\sum_{i=1}^51\right)=\frac{84}5$

3. Re: Using Riemann Sums To Find Area

If the problem only said "on the interval [2, 5] with 5 rectangles" there is a whole lot missing! For one thing is not necessary that all rectangle have same length base! If we assume that was intended, that each rectangle has base of length 3/5, then the "break points" are at 2, 13/5, 16/5, 19/5, 22/5, and 5. You then talk about the 'left endpoint" but, unless that was specified in the problem, you can take the height of the rectangle to be at calculated at any point within that rectangle.