Mixing
Equality between 2 summations unclear
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I'm learning about integrals and in particular upper and lower Riemann sums. Now in an example I came about the following:
$$L(f, P_n) = sum_i = 1^n(x_i - 1)^2Delta x = fraca^3n^3sum_i=1^n(i - 1)^2 = ...$$
I don't understand how this equality comes about, in particular $fraca^3n^3$ seem to appear out of thin air at the moment.
Any hints, definitions or theorems I shoud consult to get a feeling for this?
calculus algebra-precalculus summation
asked Jul 28 at 10:29
Learner
325
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up vote
0
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favorite
I'm learning about integrals and in particular upper and lower Riemann sums. Now in an example I came about the following:
$$L(f, P_n) = sum_i = 1^n(x_i - 1)^2Delta x = fraca^3n^3sum_i=1^n(i - 1)^2 = ...$$
I don't understand how this equality comes about, in particular $fraca^3n^3$ seem to appear out of thin air at the moment.
Any hints, definitions or theorems I shoud consult to get a feeling for this?
calculus algebra-precalculus summation
asked Jul 28 at 10:29
Learner
325
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm learning about integrals and in particular upper and lower Riemann sums. Now in an example I came about the following:
$$L(f, P_n) = sum_i = 1^n(x_i - 1)^2Delta x = fraca^3n^3sum_i=1^n(i - 1)^2 = ...$$
I don't understand how this equality comes about, in particular $fraca^3n^3$ seem to appear out of thin air at the moment.
Any hints, definitions or theorems I shoud consult to get a feeling for this?
calculus algebra-precalculus summation
asked Jul 28 at 10:29
Learner
325
I'm learning about integrals and in particular upper and lower Riemann sums. Now in an example I came about the following:
$$L(f, P_n) = sum_i = 1^n(x_i - 1)^2Delta x = fraca^3n^3sum_i=1^n(i - 1)^2 = ...$$
I don't understand how this equality comes about, in particular $fraca^3n^3$ seem to appear out of thin air at the moment.
Any hints, definitions or theorems I shoud consult to get a feeling for this?
calculus algebra-precalculus summation
asked Jul 28 at 10:29
Learner
325
asked Jul 28 at 10:29
Learner
325
asked Jul 28 at 10:29
Learner
325
asked Jul 28 at 10:29
Learner
325
325
add a comment |Â
add a comment |Â
1 Answer
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As you are integrating $x^2$ with respect to $x$ from $x=0$ to $x=a$, you divide the area into $n$ strips of width $Delta x=frac an$ unit each. Your $x_i-1$ is consequently $x_i-1=(i-1)Delta x =fracan(i-1)$, so you have $(x_i-1)^2=left(fracan(i-1)right)^2$. Simplifying, you have
$$(x_i - 1)^2Delta x = fraca^3n^3(i - 1)^2 $$
answered Jul 28 at 10:36
Karn Watcharasupat
3,7992426
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
As you are integrating $x^2$ with respect to $x$ from $x=0$ to $x=a$, you divide the area into $n$ strips of width $Delta x=frac an$ unit each. Your $x_i-1$ is consequently $x_i-1=(i-1)Delta x =fracan(i-1)$, so you have $(x_i-1)^2=left(fracan(i-1)right)^2$. Simplifying, you have
$$(x_i - 1)^2Delta x = fraca^3n^3(i - 1)^2 $$
answered Jul 28 at 10:36
Karn Watcharasupat
3,7992426
add a comment |Â
up vote
2
down vote
accepted
As you are integrating $x^2$ with respect to $x$ from $x=0$ to $x=a$, you divide the area into $n$ strips of width $Delta x=frac an$ unit each. Your $x_i-1$ is consequently $x_i-1=(i-1)Delta x =fracan(i-1)$, so you have $(x_i-1)^2=left(fracan(i-1)right)^2$. Simplifying, you have
$$(x_i - 1)^2Delta x = fraca^3n^3(i - 1)^2 $$
answered Jul 28 at 10:36
Karn Watcharasupat
3,7992426
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
As you are integrating $x^2$ with respect to $x$ from $x=0$ to $x=a$, you divide the area into $n$ strips of width $Delta x=frac an$ unit each. Your $x_i-1$ is consequently $x_i-1=(i-1)Delta x =fracan(i-1)$, so you have $(x_i-1)^2=left(fracan(i-1)right)^2$. Simplifying, you have
$$(x_i - 1)^2Delta x = fraca^3n^3(i - 1)^2 $$
answered Jul 28 at 10:36
Karn Watcharasupat
3,7992426
As you are integrating $x^2$ with respect to $x$ from $x=0$ to $x=a$, you divide the area into $n$ strips of width $Delta x=frac an$ unit each. Your $x_i-1$ is consequently $x_i-1=(i-1)Delta x =fracan(i-1)$, so you have $(x_i-1)^2=left(fracan(i-1)right)^2$. Simplifying, you have
$$(x_i - 1)^2Delta x = fraca^3n^3(i - 1)^2 $$
answered Jul 28 at 10:36
Karn Watcharasupat
3,7992426
answered Jul 28 at 10:36
Karn Watcharasupat
3,7992426
answered Jul 28 at 10:36
Karn Watcharasupat
3,7992426
answered Jul 28 at 10:36
Karn Watcharasupat
3,7992426
3,7992426
add a comment |Â
add a comment |Â
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