Areas And Distances (Intro. to Definite Integral)

[in the rye]

Hey everyone,

Today in my Calculus 1 lecture we covered Areas and Distances, which serves as a prequel to the definite integral in my book. I am confused on some notation the book uses, and I cannot seem to find a clear explanation anywhere that I look.

n
∑ f(x_i) ΔX ≅ A
i=1

First, let me explain this how I understand it, then correct me where I am wrong.

I understand this this is a simple way of saying that the approximate area under your line. And I know what the summation means.

My confusion is over the x_i, and the start/end point. I know that f(x_i) is defining the height of your rectangle based on the x value you chose in your x sub-interval. However, I'm confused with the relation of i=1 and n to this point. Say we have f(x) = x². If we used this formula with left rectangles, one of our 'i's would have to be at 0. Does this mean that we alter the formula to say(?):

n
∑ f(x_i-1) ΔX ≅ A
i=1

or

n
∑ f(x_i) ΔX ≅ A
i=0

For some reason this is just confusing the hell out of me. My book really doesn't clarify this enough, and i know in the future this will be important for Cal 2, so I want to get a handle on it now. A tutor said told me you would have to change it to one of these formulas, but to me, that doesn't make any sense. Why wouldn't the formula just remain the same, but have f(x₁) = 0?

It makes no sense to me why you would write it as either of the two methods the tutor told me because it would mean you're creating an interval that doesn't exist. Interval 0 doesn't exist, where in my mind interval 1 would be f(0) = 0 giving your sub interval area as ΔX(0)².

 

[Mark44]

Hey everyone,

Today in my Calculus 1 lecture we covered Areas and Distances, which serves as a prequel to the definite integral in my book. I am confused on some notation the book uses, and I cannot seem to find a clear explanation anywhere that I look.

n
∑ f(x_i) ΔX ≅ A
i=1

First, let me explain this how I understand it, then correct me where I am wrong.

I understand this this is a simple way of saying that the approximate area under your line. And I know what the summation means.

My confusion is over the x_i, and the start/end point. I know that f(x_i) is defining the height of your rectangle based on the x value you chose in your x sub-interval. However, I'm confused with the relation of i=1 and n to this point. Say we have f(x) = x². If we used this formula with left rectangles, one of our 'i's would have to be at 0. Does this mean that we alter the formula to say(?):

n
∑ f(x_i-1) ΔX ≅ A
i=1

or

n
∑ f(x_i) ΔX ≅ A
i=0

For some reason this is just confusing the hell out of me. My book really doesn't clarify this enough, and i know in the future this will be important for Cal 2, so I want to get a handle on it now. A tutor said told me you would have to change it to one of these formulas, but to me, that doesn't make any sense. Why wouldn't the formula just remain the same, but have f(x₁) = 0?

In the first summation you show, it is implied that some interval [a, b] is divided up into n subintervals. $x_1$ is some point in the first subinterval, $x_2$ is some point in the second subinterval, and so on, with one $x_i$ in each subinterval.

 

[in the rye]

In the first summation you show, it is implied that some interval [a, b] is divided up into n subintervals. $x_1$ is some point in the first subinterval, $x_2$ is some point in the second subinterval, and so on, with one $x_i$ in each subinterval.

So I am correct in thinking that x₀ doesn't exist using the summation formula, correct? I edited the ending of my comment, which may expand my confusion

 

[Mark44]

So I am correct in thinking that x₀ doesn't exist using the summation formula, correct? I edited the ending of my comment, which may expand my confusion

Let me correct your first summation:
$\sum_{i = 1}^n f(x_i)\Delta x$
Here $x_i$ is some point in the i-th subinterval.

 

[in the rye]

Let me correct your first summation:
$\sum_{i = 1}^n f(x_i)\Delta x$
Here $x_i$ is some point in the i-th subinterval.

Right, for some reason I have difficulty applying this, though. So for simplicity-sake let's say we use y = x² over the interval [0, 6], with only 2 left rectangles.I would have that my A = 3[f(0) + f(3)]. This would mean that x₁ = 0 and x₂ = 3in the summation formula, correct? Giving:$\sum_{i = 1}^2 f(x_i)\Delta x$

NOT

$\sum_{i = 0}^1 f(x_i)\Delta x$

Where $Δx$ = 3

 

[Mark44]

Right, for some reason I have difficulty applying this, though. So for simplicity-sake let's say we use y = x², over the interval [0, 6], with only 2 left rectangles.I would have that my A = 3[f(0) + f(3)]. This would mean that x₁ = 0 and x₂ = 3 in the summation formula, correct? Giving:$\sum_{i = 1}^2 f(x_i)\Delta x$

NOT

$\sum_{i = 0}^1 f(x_i)\Delta x$

Where $Δx$ = 3

Yes.