Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- anonymous

Use the right endpoint and 6 rectangles to find the approximation of the area of the region between the x axis and between the interval [2, 5]
f(x) = 2x^2 - x -1. I'll post my work so far done.

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

Get your **free** account and access **expert** answers to this and **thousands** of other questions

- anonymous

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

\[\sum_{i = 1}^{6} f(x)\Delta x\]
Delta x is the width of each rectangle which is:
\[\frac{ 5 - 2 }{ 6 } = \frac{ 1 }{ 2 }\]
The counter for the right endpoint is:
\[i (\frac{ 1 }{ 2 }) = \frac{ i }{ 2 }\]
\[\sum_{i = 1}^{6} (2x^2 - x - 1)(\frac{ 1 }{ 2 })\]
since 1/2 is constant it can be in front
\[(\frac{ 1 }{ 2 })\sum_{i = 1}^{6} (2x^2 - x - 1)\]
\[(\frac{ 1 }{ 2 }) \sum_{i =1}^{6} 2x^2 - \sum_{i =1}^{6} x - \sum_{i =1}^{6} 1\]
\[(\frac{ 1 }{ 2 }) \sum_{i =1}^{6} 2(\frac{ i }{ 2 })^2 - \sum_{i =1}^{6} (\frac{ i }{ 2 }) - \sum_{i =1}^{6} 1\]
Is what I did until now correct?

- jim_thompson5910

Something seems off. Let me think

- anonymous

@jim_thompson5910 I'm supposed to use sigma (summation) notation not the definite integral method
I think I should have a parenthesis or bracket around all the sigma notations

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- jim_thompson5910

ok I see what went wrong
you start at a = 2
each term xi is defined as
xi = a + i*(delta x)
so plug in a = 2 and delta x = 0.5 to get
xi = a + i*(delta x)
xi = 2 + i*(0.5)
xi = 2 + 0.5i
so you should have
\[\Large \sum_{i = 1}^{6} f(x_i)\Delta x\]
\[\Large \sum_{i = 1}^{6} f(2+0.5i)0.5\]
\[\Large 0.5\sum_{i = 1}^{6}[2x_i^2 - x_i - 1]\]
\[\Large 0.5\sum_{i = 1}^{6}[ 2(2+0.5i)^2-(2+0.5i)-1]\]
hopefully that makes sense

- anonymous

@jim_thompson5910 Thank you. Looking it over. Didn't digest all of it yet.

- anonymous

I think I get it now. Will crunch the numbers.

- jim_thompson5910

tell me what you get

- anonymous

I didn't realize that xi is: a + i(delta x).
The problems we did in class a equaled 0.
So now I realize xi for this example is: 2 + i(delta x) so I need to replace each x with 2 + (i/2).
Please let me know if I'm still missing something

- jim_thompson5910

well in this case, a = 2 and b = 5
so maybe what your teacher did was shift all of f(x) 2 units to the left, so that a = 0 is the start of your interval

- anonymous

Correct. So I was thrown off with this one.

Looking for something else?

Not the answer you are looking for? Search for more explanations.