Question

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.

Answer 1

\[\sum_{i = 1}^{6} f(x)\Delta x\]

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?

Answer 2

Something seems off. Let me think

Answer 3

@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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

tell me what you get

Answer 8

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

Answer 9

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

Answer 10

Correct. So I was thrown off with this one.