Converting from Riemann sum to definite integral
Multi tool use
$begingroup$
Can someone please explain how to convert this into a definite integral in the form
$ int_a^b f(x)dx $
And please explain how you get a
and b
and the rest.
$$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{a-1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n}(a-1))^{2}} bigg)$$
One question: I understand that this:
$$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n})^{2}} bigg) = frac{1}{x^2}$$
What happens to the $$frac{a-1}{n}$$ and $${(a-1)}$$
When:
$$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{a-1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n}(a-1))^{2}} bigg) = int_1^infty frac{1}{x^2}dx $$
calculus integration definite-integrals
$endgroup$
add a comment |
$begingroup$
Can someone please explain how to convert this into a definite integral in the form
$ int_a^b f(x)dx $
And please explain how you get a
and b
and the rest.
$$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{a-1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n}(a-1))^{2}} bigg)$$
One question: I understand that this:
$$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n})^{2}} bigg) = frac{1}{x^2}$$
What happens to the $$frac{a-1}{n}$$ and $${(a-1)}$$
When:
$$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{a-1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n}(a-1))^{2}} bigg) = int_1^infty frac{1}{x^2}dx $$
calculus integration definite-integrals
$endgroup$
add a comment |
$begingroup$
Can someone please explain how to convert this into a definite integral in the form
$ int_a^b f(x)dx $
And please explain how you get a
and b
and the rest.
$$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{a-1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n}(a-1))^{2}} bigg)$$
One question: I understand that this:
$$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n})^{2}} bigg) = frac{1}{x^2}$$
What happens to the $$frac{a-1}{n}$$ and $${(a-1)}$$
When:
$$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{a-1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n}(a-1))^{2}} bigg) = int_1^infty frac{1}{x^2}dx $$
calculus integration definite-integrals
$endgroup$
Can someone please explain how to convert this into a definite integral in the form
$ int_a^b f(x)dx $
And please explain how you get a
and b
and the rest.
$$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{a-1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n}(a-1))^{2}} bigg)$$
One question: I understand that this:
$$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n})^{2}} bigg) = frac{1}{x^2}$$
What happens to the $$frac{a-1}{n}$$ and $${(a-1)}$$
When:
$$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{a-1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n}(a-1))^{2}} bigg) = int_1^infty frac{1}{x^2}dx $$
calculus integration definite-integrals
calculus integration definite-integrals
edited Mar 22 '14 at 9:23
M.E.
asked Mar 22 '14 at 8:19
M.E.M.E.
1771313
1771313
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The point is to guess. First of all, forget about the limit $ato infty$. When you want to approximate $int_a^b f(x)dx$ by sums of area of rectangles, you divide the intervals $[a, b]$ into $n$ equals parts with width $frac{b-a}{n}$. Thus one would guess the integral would be something like
$$int_{1+ c}^{a+c} f(x) dx$$
for some unknown $c$. Now for $f$, we put in $k=0$ to see that the summation term is
$$frac{1}{(1)^2},$$
and put in $k=n-1$ we have
$$frac{1}{(1+ frac{n-1}{n}(a-1))^2}sim frac{1}{(1+ (a-1))^2} = frac{1}{(a)^2} .$$
when $n$ is very big. Thus one good guess would be
$$int_1^a frac{1}{x^2} dx .$$
Now you take care of the limit $ato infty$ to see that it should be
$$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{a-1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n}(a-1))^{2}} bigg) = int_1^infty frac{1}{x^2}dx = -frac{1}{x}bigg|_1^infty = 1$$
$endgroup$
$begingroup$
Great answer. Thank you very much! But I have one question.Why isn't this correct for the third step: $$frac{1}{(1+ frac{n-1}{n}(a-1))^2}sim frac{1}{(2+ (a-1))^2} = frac{1}{(a+1)^2} $$? @John
$endgroup$
– M.E.
Mar 22 '14 at 8:50
$begingroup$
@Andy: When $n$ is big, $frac{n-1}{n} sim 1$, so it should be $1/a^2$.
$endgroup$
– user99914
Mar 22 '14 at 9:03
$begingroup$
you're right, when n is big, $$frac{n-1}{n} sim 1$$ but what about the other 1 that is$$ 1 + frac{n-1}{n}$$? And since $$frac{n-1}{n} sim 1 $$, $$1+frac{n-1}{n} = 2$$ Right? By the way sorry if my questions annoy you. I am just a bit confused.
$endgroup$
– M.E.
Mar 22 '14 at 9:07
$begingroup$
@Andy: In your equation I do not know why there is a $2$. It should be $$frac{1}{(1+ frac{n-1}{n}(a-1))^2}simfrac{1}{(1 + (a-1))^2} = frac{1}{a^2}$$
$endgroup$
– user99914
Mar 22 '14 at 9:08
1
$begingroup$
@Andy The $frac{n-1}{n}$ is being multiplied to $a-1$ first and then you add 1.
$endgroup$
– mathematician
Mar 22 '14 at 9:13
|
show 1 more comment
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1 Answer
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1 Answer
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$begingroup$
The point is to guess. First of all, forget about the limit $ato infty$. When you want to approximate $int_a^b f(x)dx$ by sums of area of rectangles, you divide the intervals $[a, b]$ into $n$ equals parts with width $frac{b-a}{n}$. Thus one would guess the integral would be something like
$$int_{1+ c}^{a+c} f(x) dx$$
for some unknown $c$. Now for $f$, we put in $k=0$ to see that the summation term is
$$frac{1}{(1)^2},$$
and put in $k=n-1$ we have
$$frac{1}{(1+ frac{n-1}{n}(a-1))^2}sim frac{1}{(1+ (a-1))^2} = frac{1}{(a)^2} .$$
when $n$ is very big. Thus one good guess would be
$$int_1^a frac{1}{x^2} dx .$$
Now you take care of the limit $ato infty$ to see that it should be
$$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{a-1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n}(a-1))^{2}} bigg) = int_1^infty frac{1}{x^2}dx = -frac{1}{x}bigg|_1^infty = 1$$
$endgroup$
$begingroup$
Great answer. Thank you very much! But I have one question.Why isn't this correct for the third step: $$frac{1}{(1+ frac{n-1}{n}(a-1))^2}sim frac{1}{(2+ (a-1))^2} = frac{1}{(a+1)^2} $$? @John
$endgroup$
– M.E.
Mar 22 '14 at 8:50
$begingroup$
@Andy: When $n$ is big, $frac{n-1}{n} sim 1$, so it should be $1/a^2$.
$endgroup$
– user99914
Mar 22 '14 at 9:03
$begingroup$
you're right, when n is big, $$frac{n-1}{n} sim 1$$ but what about the other 1 that is$$ 1 + frac{n-1}{n}$$? And since $$frac{n-1}{n} sim 1 $$, $$1+frac{n-1}{n} = 2$$ Right? By the way sorry if my questions annoy you. I am just a bit confused.
$endgroup$
– M.E.
Mar 22 '14 at 9:07
$begingroup$
@Andy: In your equation I do not know why there is a $2$. It should be $$frac{1}{(1+ frac{n-1}{n}(a-1))^2}simfrac{1}{(1 + (a-1))^2} = frac{1}{a^2}$$
$endgroup$
– user99914
Mar 22 '14 at 9:08
1
$begingroup$
@Andy The $frac{n-1}{n}$ is being multiplied to $a-1$ first and then you add 1.
$endgroup$
– mathematician
Mar 22 '14 at 9:13
|
show 1 more comment
$begingroup$
The point is to guess. First of all, forget about the limit $ato infty$. When you want to approximate $int_a^b f(x)dx$ by sums of area of rectangles, you divide the intervals $[a, b]$ into $n$ equals parts with width $frac{b-a}{n}$. Thus one would guess the integral would be something like
$$int_{1+ c}^{a+c} f(x) dx$$
for some unknown $c$. Now for $f$, we put in $k=0$ to see that the summation term is
$$frac{1}{(1)^2},$$
and put in $k=n-1$ we have
$$frac{1}{(1+ frac{n-1}{n}(a-1))^2}sim frac{1}{(1+ (a-1))^2} = frac{1}{(a)^2} .$$
when $n$ is very big. Thus one good guess would be
$$int_1^a frac{1}{x^2} dx .$$
Now you take care of the limit $ato infty$ to see that it should be
$$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{a-1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n}(a-1))^{2}} bigg) = int_1^infty frac{1}{x^2}dx = -frac{1}{x}bigg|_1^infty = 1$$
$endgroup$
$begingroup$
Great answer. Thank you very much! But I have one question.Why isn't this correct for the third step: $$frac{1}{(1+ frac{n-1}{n}(a-1))^2}sim frac{1}{(2+ (a-1))^2} = frac{1}{(a+1)^2} $$? @John
$endgroup$
– M.E.
Mar 22 '14 at 8:50
$begingroup$
@Andy: When $n$ is big, $frac{n-1}{n} sim 1$, so it should be $1/a^2$.
$endgroup$
– user99914
Mar 22 '14 at 9:03
$begingroup$
you're right, when n is big, $$frac{n-1}{n} sim 1$$ but what about the other 1 that is$$ 1 + frac{n-1}{n}$$? And since $$frac{n-1}{n} sim 1 $$, $$1+frac{n-1}{n} = 2$$ Right? By the way sorry if my questions annoy you. I am just a bit confused.
$endgroup$
– M.E.
Mar 22 '14 at 9:07
$begingroup$
@Andy: In your equation I do not know why there is a $2$. It should be $$frac{1}{(1+ frac{n-1}{n}(a-1))^2}simfrac{1}{(1 + (a-1))^2} = frac{1}{a^2}$$
$endgroup$
– user99914
Mar 22 '14 at 9:08
1
$begingroup$
@Andy The $frac{n-1}{n}$ is being multiplied to $a-1$ first and then you add 1.
$endgroup$
– mathematician
Mar 22 '14 at 9:13
|
show 1 more comment
$begingroup$
The point is to guess. First of all, forget about the limit $ato infty$. When you want to approximate $int_a^b f(x)dx$ by sums of area of rectangles, you divide the intervals $[a, b]$ into $n$ equals parts with width $frac{b-a}{n}$. Thus one would guess the integral would be something like
$$int_{1+ c}^{a+c} f(x) dx$$
for some unknown $c$. Now for $f$, we put in $k=0$ to see that the summation term is
$$frac{1}{(1)^2},$$
and put in $k=n-1$ we have
$$frac{1}{(1+ frac{n-1}{n}(a-1))^2}sim frac{1}{(1+ (a-1))^2} = frac{1}{(a)^2} .$$
when $n$ is very big. Thus one good guess would be
$$int_1^a frac{1}{x^2} dx .$$
Now you take care of the limit $ato infty$ to see that it should be
$$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{a-1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n}(a-1))^{2}} bigg) = int_1^infty frac{1}{x^2}dx = -frac{1}{x}bigg|_1^infty = 1$$
$endgroup$
The point is to guess. First of all, forget about the limit $ato infty$. When you want to approximate $int_a^b f(x)dx$ by sums of area of rectangles, you divide the intervals $[a, b]$ into $n$ equals parts with width $frac{b-a}{n}$. Thus one would guess the integral would be something like
$$int_{1+ c}^{a+c} f(x) dx$$
for some unknown $c$. Now for $f$, we put in $k=0$ to see that the summation term is
$$frac{1}{(1)^2},$$
and put in $k=n-1$ we have
$$frac{1}{(1+ frac{n-1}{n}(a-1))^2}sim frac{1}{(1+ (a-1))^2} = frac{1}{(a)^2} .$$
when $n$ is very big. Thus one good guess would be
$$int_1^a frac{1}{x^2} dx .$$
Now you take care of the limit $ato infty$ to see that it should be
$$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{a-1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n}(a-1))^{2}} bigg) = int_1^infty frac{1}{x^2}dx = -frac{1}{x}bigg|_1^infty = 1$$
answered Mar 22 '14 at 8:36
user99914
$begingroup$
Great answer. Thank you very much! But I have one question.Why isn't this correct for the third step: $$frac{1}{(1+ frac{n-1}{n}(a-1))^2}sim frac{1}{(2+ (a-1))^2} = frac{1}{(a+1)^2} $$? @John
$endgroup$
– M.E.
Mar 22 '14 at 8:50
$begingroup$
@Andy: When $n$ is big, $frac{n-1}{n} sim 1$, so it should be $1/a^2$.
$endgroup$
– user99914
Mar 22 '14 at 9:03
$begingroup$
you're right, when n is big, $$frac{n-1}{n} sim 1$$ but what about the other 1 that is$$ 1 + frac{n-1}{n}$$? And since $$frac{n-1}{n} sim 1 $$, $$1+frac{n-1}{n} = 2$$ Right? By the way sorry if my questions annoy you. I am just a bit confused.
$endgroup$
– M.E.
Mar 22 '14 at 9:07
$begingroup$
@Andy: In your equation I do not know why there is a $2$. It should be $$frac{1}{(1+ frac{n-1}{n}(a-1))^2}simfrac{1}{(1 + (a-1))^2} = frac{1}{a^2}$$
$endgroup$
– user99914
Mar 22 '14 at 9:08
1
$begingroup$
@Andy The $frac{n-1}{n}$ is being multiplied to $a-1$ first and then you add 1.
$endgroup$
– mathematician
Mar 22 '14 at 9:13
|
show 1 more comment
$begingroup$
Great answer. Thank you very much! But I have one question.Why isn't this correct for the third step: $$frac{1}{(1+ frac{n-1}{n}(a-1))^2}sim frac{1}{(2+ (a-1))^2} = frac{1}{(a+1)^2} $$? @John
$endgroup$
– M.E.
Mar 22 '14 at 8:50
$begingroup$
@Andy: When $n$ is big, $frac{n-1}{n} sim 1$, so it should be $1/a^2$.
$endgroup$
– user99914
Mar 22 '14 at 9:03
$begingroup$
you're right, when n is big, $$frac{n-1}{n} sim 1$$ but what about the other 1 that is$$ 1 + frac{n-1}{n}$$? And since $$frac{n-1}{n} sim 1 $$, $$1+frac{n-1}{n} = 2$$ Right? By the way sorry if my questions annoy you. I am just a bit confused.
$endgroup$
– M.E.
Mar 22 '14 at 9:07
$begingroup$
@Andy: In your equation I do not know why there is a $2$. It should be $$frac{1}{(1+ frac{n-1}{n}(a-1))^2}simfrac{1}{(1 + (a-1))^2} = frac{1}{a^2}$$
$endgroup$
– user99914
Mar 22 '14 at 9:08
1
$begingroup$
@Andy The $frac{n-1}{n}$ is being multiplied to $a-1$ first and then you add 1.
$endgroup$
– mathematician
Mar 22 '14 at 9:13
$begingroup$
Great answer. Thank you very much! But I have one question.Why isn't this correct for the third step: $$frac{1}{(1+ frac{n-1}{n}(a-1))^2}sim frac{1}{(2+ (a-1))^2} = frac{1}{(a+1)^2} $$? @John
$endgroup$
– M.E.
Mar 22 '14 at 8:50
$begingroup$
Great answer. Thank you very much! But I have one question.Why isn't this correct for the third step: $$frac{1}{(1+ frac{n-1}{n}(a-1))^2}sim frac{1}{(2+ (a-1))^2} = frac{1}{(a+1)^2} $$? @John
$endgroup$
– M.E.
Mar 22 '14 at 8:50
$begingroup$
@Andy: When $n$ is big, $frac{n-1}{n} sim 1$, so it should be $1/a^2$.
$endgroup$
– user99914
Mar 22 '14 at 9:03
$begingroup$
@Andy: When $n$ is big, $frac{n-1}{n} sim 1$, so it should be $1/a^2$.
$endgroup$
– user99914
Mar 22 '14 at 9:03
$begingroup$
you're right, when n is big, $$frac{n-1}{n} sim 1$$ but what about the other 1 that is$$ 1 + frac{n-1}{n}$$? And since $$frac{n-1}{n} sim 1 $$, $$1+frac{n-1}{n} = 2$$ Right? By the way sorry if my questions annoy you. I am just a bit confused.
$endgroup$
– M.E.
Mar 22 '14 at 9:07
$begingroup$
you're right, when n is big, $$frac{n-1}{n} sim 1$$ but what about the other 1 that is$$ 1 + frac{n-1}{n}$$? And since $$frac{n-1}{n} sim 1 $$, $$1+frac{n-1}{n} = 2$$ Right? By the way sorry if my questions annoy you. I am just a bit confused.
$endgroup$
– M.E.
Mar 22 '14 at 9:07
$begingroup$
@Andy: In your equation I do not know why there is a $2$. It should be $$frac{1}{(1+ frac{n-1}{n}(a-1))^2}simfrac{1}{(1 + (a-1))^2} = frac{1}{a^2}$$
$endgroup$
– user99914
Mar 22 '14 at 9:08
$begingroup$
@Andy: In your equation I do not know why there is a $2$. It should be $$frac{1}{(1+ frac{n-1}{n}(a-1))^2}simfrac{1}{(1 + (a-1))^2} = frac{1}{a^2}$$
$endgroup$
– user99914
Mar 22 '14 at 9:08
1
1
$begingroup$
@Andy The $frac{n-1}{n}$ is being multiplied to $a-1$ first and then you add 1.
$endgroup$
– mathematician
Mar 22 '14 at 9:13
$begingroup$
@Andy The $frac{n-1}{n}$ is being multiplied to $a-1$ first and then you add 1.
$endgroup$
– mathematician
Mar 22 '14 at 9:13
|
show 1 more comment
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