Question

Could somebody help me with this? (use Riemann sums)
1) Estimate the area under f(x)= x^2 + 3x from 0 to 5 using 20 right rectangles.

Answer 1

oh god.. 20 rectangles is a pain.. the width of each is 0.25 since the total length is from 0 to 5

Answer 2

And since it's right hand, you use right endpoints, so:
0.25*f(0.25) + 0.25*f(0.5) + 0.25*f(0.75) + ... + 0.25*f(5)

Answer 3

Let\[\Delta x = \frac{|5 - 0|}{20} = 0.25\]The Riemann Sum\[\S = \sum_{i = 1}^{20}f(\Delta x_{i}) \Delta x_{i}\]\[=\sum_{i = 1}^{20}(\Delta x_{i})^{3} + 3(\Delta x_{i})^{2}\]\[=\sum_{i = 1}^{20}(\Delta x_i)^{3} + 3\sum_{i = 1}^{20} (\Delta x_i)^{2}\]Plug the delta x and you do the sum. Its kinda long and im lazy >.>

Answer 4

this is what I did:
(1/4)[((.25)^2 + 3(.25))+((.50)^2 + 3(.50))+((.75)^2 + 3(.75))+((1)^2 + 3(1))+((1.25)^2 + 3(1.25))+((1.50)^2 + 3(1.50))+((1.75)^2 + 3(1.75))+((2)^2 + 3(2))+((2.25)^2 + 3(2.25))+((2.5)^2 + 3(2.5))+((2.75)^2 + 3(2.75))+((3)^2 + 3(3))+((3.25)^2 + 3(3.25))+((3.5)^2 + 3(3.5))+((3.75)^2 + 3(3.75))+((4)^2 + 3(4))+((4.25)^2 + 3(4.25))+((4.5)^2 + 3(4.5))+((4.75)^2 + 3(4.75))+((5)^2 + 3(5))+((5.25)^2 + 3(5.25))+((5.5)^2 + 3(5.5))+((5.75)^2 + 3(5.75))+(6^2 + 3(6))]

Answer 5

that's basically right, it's just tedious long arithmetic.. dumb question really

Answer 6

ohh wait why did u go all the way to 6?

Answer 7

stop at f(5), since u have the interval from 0 to 5

Answer 8

ahhh.... I think that's where I went wrong! Guess I got carried away while I was doing this!

Answer 9

LOL

Answer 10

maybe if the delta x = 1 its manageable. You will just use the sum of n cubes/squares thing.

Answer 11

Okay, I got 84, which isn't too far off from the actual answer of 79. Thanks everyone!