Mixing
Solving Indefinite integral without FTC
$begingroup$
While I was watching some physics lectures, I saw a professor write down the $int r*dr$. The writing multiplication sign (normally just implied) prompted me to attempt to solve this integral without the use of the fundamental theorem. So I wrote as follows:
$r(Delta r) +(r+Delta r)(Delta r)+(r+2Delta r)(Delta r)...$
However, if I take this $Delta r$ to $0$, I will end up with 0. What am I doing wrong?
Thanks :)
calculus nonstandard-analysis
asked Jan 28 at 3:40
Dude156Dude156
572316
$endgroup$
add a comment |
$begingroup$
While I was watching some physics lectures, I saw a professor write down the $int r*dr$. The writing multiplication sign (normally just implied) prompted me to attempt to solve this integral without the use of the fundamental theorem. So I wrote as follows:
$r(Delta r) +(r+Delta r)(Delta r)+(r+2Delta r)(Delta r)...$
However, if I take this $Delta r$ to $0$, I will end up with 0. What am I doing wrong?
Thanks :)
calculus nonstandard-analysis
asked Jan 28 at 3:40
Dude156Dude156
572316
$endgroup$
add a comment |
$begingroup$
While I was watching some physics lectures, I saw a professor write down the $int r*dr$. The writing multiplication sign (normally just implied) prompted me to attempt to solve this integral without the use of the fundamental theorem. So I wrote as follows:
$r(Delta r) +(r+Delta r)(Delta r)+(r+2Delta r)(Delta r)...$
However, if I take this $Delta r$ to $0$, I will end up with 0. What am I doing wrong?
Thanks :)
calculus nonstandard-analysis
asked Jan 28 at 3:40
Dude156Dude156
572316
$endgroup$
While I was watching some physics lectures, I saw a professor write down the $int r*dr$. The writing multiplication sign (normally just implied) prompted me to attempt to solve this integral without the use of the fundamental theorem. So I wrote as follows:
$r(Delta r) +(r+Delta r)(Delta r)+(r+2Delta r)(Delta r)...$
However, if I take this $Delta r$ to $0$, I will end up with 0. What am I doing wrong?
Thanks :)
calculus nonstandard-analysis
calculus nonstandard-analysis
asked Jan 28 at 3:40
Dude156Dude156
572316
asked Jan 28 at 3:40
Dude156Dude156
572316
asked Jan 28 at 3:40
Dude156Dude156
572316
asked Jan 28 at 3:40
Dude156Dude156
572316
asked Jan 28 at 3:40
Dude156Dude156
572316
572316
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
You need to be more specific, group terms where warranted, and take a limit.
Let's add bounds to the integral.
$$int_a^b r dr$$
Now, say we are going to divide the interval into $n$ chunks, so that each interval is of width $Delta r = (b-a)/n$.
Starting at $r = a$, our Riemann approximation becomes...
$$a left( frac{b-a}{n} right) + left( a + frac{b-a}{n} right) left( frac{b-a}{n} right) + left( a + 2 cdot frac{b-a}{n} right) left( frac{b-a}{n} right) + cdots + left( a + (n-1)cdot frac{b-a}{n} right) left( frac{b-a}{n} right)$$
Simplfy the expression.
$$left[ a + left( a + frac{b-a}{n} right) + left( a + 2 cdot frac{b-a}{n} right) + cdots + left( a + (n-1)cdot frac{b-a}{n} right) right] left( frac{b-a}{n} right)$$
$$left[ na + left(frac{b-a}{n} right) + left(2 cdot frac{b-a}{n} right) + cdots + left((n-1)cdot frac{b-a}{n} right) right] left( frac{b-a}{n} right)$$
Use the fact that $1 + 2 + cdots + (n-1) = frac{n(n-1)}{2}$.
$$left[ na + left( frac{n(n-1)}{2} right) left(frac{b-a}{n} right) right] left( frac{b-a}{n} right)$$
Continue simplifying...
$$na left( frac{b-a}{n} right) + left( frac{n(n-1)}{2} right) left(frac{b-a}{n} right)^2 $$
$$a(b-a) + frac{(n-1)(b-a)^2}{2n}$$
$$a(b-a) + left( frac{1}{2} - frac{1}{2n}right)(b^2 - 2ab + a^2)$$
$$ab - a^2 + frac{a^2}{2} - ab + frac{b^2}{2} - frac{b^2 - 2ab + a^2}{2n}$$
$$frac{b^2}{2} -frac{a^2}{2} - frac{b^2 - 2ab + a^2}{2n}$$
Now, take the limit as $Delta r rightarrow infty$.
$$lim_{n rightarrow infty} left[ frac{b^2}{2} -frac{a^2}{2} - frac{b^2 - 2ab + a^2}{2n} right] = frac{b^2}{2} -frac{a^2}{2}$$
Thus,
$$int_a^b r dr = frac{b^2}{2} - frac{a^2}{2},$$
which implies the indefinite integral takes the following form.
$$int r dr = frac{r^2}{2} + C$$
answered Jan 28 at 4:01
Trevor KafkaTrevor Kafka
5309
$endgroup$
$begingroup$
Yes, here you took a definite integral using the reimann sum method. I know about this technique, however I'm looking at something that doesn't require an introduction of bounds. Is there any way to do that?
$endgroup$
– Dude156
Jan 28 at 4:10
1
$begingroup$
You can't do a Riemann approximation without bounds, no. The indefinite integral function is defined in terms of the definite integral. $$int f(x) dx equiv left( int_0^x f(u)du right) + C$$
$endgroup$
– Trevor Kafka
Jan 28 at 4:20
$begingroup$
ahh, now I understand what you mean by "be more specific". Thanks man!
$endgroup$
– Dude156
Jan 28 at 4:38
add a comment |
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1 Answer
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$begingroup$
You need to be more specific, group terms where warranted, and take a limit.
Let's add bounds to the integral.
$$int_a^b r dr$$
Now, say we are going to divide the interval into $n$ chunks, so that each interval is of width $Delta r = (b-a)/n$.
Starting at $r = a$, our Riemann approximation becomes...
$$a left( frac{b-a}{n} right) + left( a + frac{b-a}{n} right) left( frac{b-a}{n} right) + left( a + 2 cdot frac{b-a}{n} right) left( frac{b-a}{n} right) + cdots + left( a + (n-1)cdot frac{b-a}{n} right) left( frac{b-a}{n} right)$$
Simplfy the expression.
$$left[ a + left( a + frac{b-a}{n} right) + left( a + 2 cdot frac{b-a}{n} right) + cdots + left( a + (n-1)cdot frac{b-a}{n} right) right] left( frac{b-a}{n} right)$$
$$left[ na + left(frac{b-a}{n} right) + left(2 cdot frac{b-a}{n} right) + cdots + left((n-1)cdot frac{b-a}{n} right) right] left( frac{b-a}{n} right)$$
Use the fact that $1 + 2 + cdots + (n-1) = frac{n(n-1)}{2}$.
$$left[ na + left( frac{n(n-1)}{2} right) left(frac{b-a}{n} right) right] left( frac{b-a}{n} right)$$
Continue simplifying...
$$na left( frac{b-a}{n} right) + left( frac{n(n-1)}{2} right) left(frac{b-a}{n} right)^2 $$
$$a(b-a) + frac{(n-1)(b-a)^2}{2n}$$
$$a(b-a) + left( frac{1}{2} - frac{1}{2n}right)(b^2 - 2ab + a^2)$$
$$ab - a^2 + frac{a^2}{2} - ab + frac{b^2}{2} - frac{b^2 - 2ab + a^2}{2n}$$
$$frac{b^2}{2} -frac{a^2}{2} - frac{b^2 - 2ab + a^2}{2n}$$
Now, take the limit as $Delta r rightarrow infty$.
$$lim_{n rightarrow infty} left[ frac{b^2}{2} -frac{a^2}{2} - frac{b^2 - 2ab + a^2}{2n} right] = frac{b^2}{2} -frac{a^2}{2}$$
Thus,
$$int_a^b r dr = frac{b^2}{2} - frac{a^2}{2},$$
which implies the indefinite integral takes the following form.
$$int r dr = frac{r^2}{2} + C$$
answered Jan 28 at 4:01
Trevor KafkaTrevor Kafka
5309
$endgroup$
$begingroup$
Yes, here you took a definite integral using the reimann sum method. I know about this technique, however I'm looking at something that doesn't require an introduction of bounds. Is there any way to do that?
$endgroup$
– Dude156
Jan 28 at 4:10
1
$begingroup$
You can't do a Riemann approximation without bounds, no. The indefinite integral function is defined in terms of the definite integral. $$int f(x) dx equiv left( int_0^x f(u)du right) + C$$
$endgroup$
– Trevor Kafka
Jan 28 at 4:20
$begingroup$
ahh, now I understand what you mean by "be more specific". Thanks man!
$endgroup$
– Dude156
Jan 28 at 4:38
add a comment |
$begingroup$
You need to be more specific, group terms where warranted, and take a limit.
Let's add bounds to the integral.
$$int_a^b r dr$$
Now, say we are going to divide the interval into $n$ chunks, so that each interval is of width $Delta r = (b-a)/n$.
Starting at $r = a$, our Riemann approximation becomes...
$$a left( frac{b-a}{n} right) + left( a + frac{b-a}{n} right) left( frac{b-a}{n} right) + left( a + 2 cdot frac{b-a}{n} right) left( frac{b-a}{n} right) + cdots + left( a + (n-1)cdot frac{b-a}{n} right) left( frac{b-a}{n} right)$$
Simplfy the expression.
$$left[ a + left( a + frac{b-a}{n} right) + left( a + 2 cdot frac{b-a}{n} right) + cdots + left( a + (n-1)cdot frac{b-a}{n} right) right] left( frac{b-a}{n} right)$$
$$left[ na + left(frac{b-a}{n} right) + left(2 cdot frac{b-a}{n} right) + cdots + left((n-1)cdot frac{b-a}{n} right) right] left( frac{b-a}{n} right)$$
Use the fact that $1 + 2 + cdots + (n-1) = frac{n(n-1)}{2}$.
$$left[ na + left( frac{n(n-1)}{2} right) left(frac{b-a}{n} right) right] left( frac{b-a}{n} right)$$
Continue simplifying...
$$na left( frac{b-a}{n} right) + left( frac{n(n-1)}{2} right) left(frac{b-a}{n} right)^2 $$
$$a(b-a) + frac{(n-1)(b-a)^2}{2n}$$
$$a(b-a) + left( frac{1}{2} - frac{1}{2n}right)(b^2 - 2ab + a^2)$$
$$ab - a^2 + frac{a^2}{2} - ab + frac{b^2}{2} - frac{b^2 - 2ab + a^2}{2n}$$
$$frac{b^2}{2} -frac{a^2}{2} - frac{b^2 - 2ab + a^2}{2n}$$
Now, take the limit as $Delta r rightarrow infty$.
$$lim_{n rightarrow infty} left[ frac{b^2}{2} -frac{a^2}{2} - frac{b^2 - 2ab + a^2}{2n} right] = frac{b^2}{2} -frac{a^2}{2}$$
Thus,
$$int_a^b r dr = frac{b^2}{2} - frac{a^2}{2},$$
which implies the indefinite integral takes the following form.
$$int r dr = frac{r^2}{2} + C$$
answered Jan 28 at 4:01
Trevor KafkaTrevor Kafka
5309
$endgroup$
$begingroup$
Yes, here you took a definite integral using the reimann sum method. I know about this technique, however I'm looking at something that doesn't require an introduction of bounds. Is there any way to do that?
$endgroup$
– Dude156
Jan 28 at 4:10
1
$begingroup$
You can't do a Riemann approximation without bounds, no. The indefinite integral function is defined in terms of the definite integral. $$int f(x) dx equiv left( int_0^x f(u)du right) + C$$
$endgroup$
– Trevor Kafka
Jan 28 at 4:20
$begingroup$
ahh, now I understand what you mean by "be more specific". Thanks man!
$endgroup$
– Dude156
Jan 28 at 4:38
add a comment |
$begingroup$
You need to be more specific, group terms where warranted, and take a limit.
Let's add bounds to the integral.
$$int_a^b r dr$$
Now, say we are going to divide the interval into $n$ chunks, so that each interval is of width $Delta r = (b-a)/n$.
Starting at $r = a$, our Riemann approximation becomes...
$$a left( frac{b-a}{n} right) + left( a + frac{b-a}{n} right) left( frac{b-a}{n} right) + left( a + 2 cdot frac{b-a}{n} right) left( frac{b-a}{n} right) + cdots + left( a + (n-1)cdot frac{b-a}{n} right) left( frac{b-a}{n} right)$$
Simplfy the expression.
$$left[ a + left( a + frac{b-a}{n} right) + left( a + 2 cdot frac{b-a}{n} right) + cdots + left( a + (n-1)cdot frac{b-a}{n} right) right] left( frac{b-a}{n} right)$$
$$left[ na + left(frac{b-a}{n} right) + left(2 cdot frac{b-a}{n} right) + cdots + left((n-1)cdot frac{b-a}{n} right) right] left( frac{b-a}{n} right)$$
Use the fact that $1 + 2 + cdots + (n-1) = frac{n(n-1)}{2}$.
$$left[ na + left( frac{n(n-1)}{2} right) left(frac{b-a}{n} right) right] left( frac{b-a}{n} right)$$
Continue simplifying...
$$na left( frac{b-a}{n} right) + left( frac{n(n-1)}{2} right) left(frac{b-a}{n} right)^2 $$
$$a(b-a) + frac{(n-1)(b-a)^2}{2n}$$
$$a(b-a) + left( frac{1}{2} - frac{1}{2n}right)(b^2 - 2ab + a^2)$$
$$ab - a^2 + frac{a^2}{2} - ab + frac{b^2}{2} - frac{b^2 - 2ab + a^2}{2n}$$
$$frac{b^2}{2} -frac{a^2}{2} - frac{b^2 - 2ab + a^2}{2n}$$
Now, take the limit as $Delta r rightarrow infty$.
$$lim_{n rightarrow infty} left[ frac{b^2}{2} -frac{a^2}{2} - frac{b^2 - 2ab + a^2}{2n} right] = frac{b^2}{2} -frac{a^2}{2}$$
Thus,
$$int_a^b r dr = frac{b^2}{2} - frac{a^2}{2},$$
which implies the indefinite integral takes the following form.
$$int r dr = frac{r^2}{2} + C$$
answered Jan 28 at 4:01
Trevor KafkaTrevor Kafka
5309
$endgroup$
You need to be more specific, group terms where warranted, and take a limit.
Let's add bounds to the integral.
$$int_a^b r dr$$
Now, say we are going to divide the interval into $n$ chunks, so that each interval is of width $Delta r = (b-a)/n$.
Starting at $r = a$, our Riemann approximation becomes...
$$a left( frac{b-a}{n} right) + left( a + frac{b-a}{n} right) left( frac{b-a}{n} right) + left( a + 2 cdot frac{b-a}{n} right) left( frac{b-a}{n} right) + cdots + left( a + (n-1)cdot frac{b-a}{n} right) left( frac{b-a}{n} right)$$
Simplfy the expression.
$$left[ a + left( a + frac{b-a}{n} right) + left( a + 2 cdot frac{b-a}{n} right) + cdots + left( a + (n-1)cdot frac{b-a}{n} right) right] left( frac{b-a}{n} right)$$
$$left[ na + left(frac{b-a}{n} right) + left(2 cdot frac{b-a}{n} right) + cdots + left((n-1)cdot frac{b-a}{n} right) right] left( frac{b-a}{n} right)$$
Use the fact that $1 + 2 + cdots + (n-1) = frac{n(n-1)}{2}$.
$$left[ na + left( frac{n(n-1)}{2} right) left(frac{b-a}{n} right) right] left( frac{b-a}{n} right)$$
Continue simplifying...
$$na left( frac{b-a}{n} right) + left( frac{n(n-1)}{2} right) left(frac{b-a}{n} right)^2 $$
$$a(b-a) + frac{(n-1)(b-a)^2}{2n}$$
$$a(b-a) + left( frac{1}{2} - frac{1}{2n}right)(b^2 - 2ab + a^2)$$
$$ab - a^2 + frac{a^2}{2} - ab + frac{b^2}{2} - frac{b^2 - 2ab + a^2}{2n}$$
$$frac{b^2}{2} -frac{a^2}{2} - frac{b^2 - 2ab + a^2}{2n}$$
Now, take the limit as $Delta r rightarrow infty$.
$$lim_{n rightarrow infty} left[ frac{b^2}{2} -frac{a^2}{2} - frac{b^2 - 2ab + a^2}{2n} right] = frac{b^2}{2} -frac{a^2}{2}$$
Thus,
$$int_a^b r dr = frac{b^2}{2} - frac{a^2}{2},$$
which implies the indefinite integral takes the following form.
$$int r dr = frac{r^2}{2} + C$$
answered Jan 28 at 4:01
Trevor KafkaTrevor Kafka
5309
answered Jan 28 at 4:01
Trevor KafkaTrevor Kafka
5309
answered Jan 28 at 4:01
Trevor KafkaTrevor Kafka
5309
answered Jan 28 at 4:01
Trevor KafkaTrevor Kafka
5309
5309
$begingroup$
Yes, here you took a definite integral using the reimann sum method. I know about this technique, however I'm looking at something that doesn't require an introduction of bounds. Is there any way to do that?
$endgroup$
– Dude156
Jan 28 at 4:10
1
$begingroup$
You can't do a Riemann approximation without bounds, no. The indefinite integral function is defined in terms of the definite integral. $$int f(x) dx equiv left( int_0^x f(u)du right) + C$$
$endgroup$
– Trevor Kafka
Jan 28 at 4:20
$begingroup$
ahh, now I understand what you mean by "be more specific". Thanks man!
$endgroup$
– Dude156
Jan 28 at 4:38
add a comment |
$begingroup$
Yes, here you took a definite integral using the reimann sum method. I know about this technique, however I'm looking at something that doesn't require an introduction of bounds. Is there any way to do that?
$endgroup$
– Dude156
Jan 28 at 4:10
1
$begingroup$
You can't do a Riemann approximation without bounds, no. The indefinite integral function is defined in terms of the definite integral. $$int f(x) dx equiv left( int_0^x f(u)du right) + C$$
$endgroup$
– Trevor Kafka
Jan 28 at 4:20
$begingroup$
ahh, now I understand what you mean by "be more specific". Thanks man!
$endgroup$
– Dude156
Jan 28 at 4:38
$begingroup$
Yes, here you took a definite integral using the reimann sum method. I know about this technique, however I'm looking at something that doesn't require an introduction of bounds. Is there any way to do that?
$endgroup$
– Dude156
Jan 28 at 4:10
$begingroup$
Yes, here you took a definite integral using the reimann sum method. I know about this technique, however I'm looking at something that doesn't require an introduction of bounds. Is there any way to do that?
$endgroup$
– Dude156
Jan 28 at 4:10
1
1
$begingroup$
You can't do a Riemann approximation without bounds, no. The indefinite integral function is defined in terms of the definite integral. $$int f(x) dx equiv left( int_0^x f(u)du right) + C$$
$endgroup$
– Trevor Kafka
Jan 28 at 4:20
$begingroup$
You can't do a Riemann approximation without bounds, no. The indefinite integral function is defined in terms of the definite integral. $$int f(x) dx equiv left( int_0^x f(u)du right) + C$$
$endgroup$
– Trevor Kafka
Jan 28 at 4:20
$begingroup$
ahh, now I understand what you mean by "be more specific". Thanks man!
$endgroup$
– Dude156
Jan 28 at 4:38
$begingroup$
ahh, now I understand what you mean by "be more specific". Thanks man!
$endgroup$
– Dude156
Jan 28 at 4:38
add a comment |
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