Why don't we write the second constant
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While doing integration by parts we usually do not write the constant derived from the second part of the formula and instead we assign general constant for the whole integration.What is the reason for that?
calculus integration definite-integrals
asked Dec 24 '18 at 22:27
Arif RustamovArif Rustamov
367
$endgroup$
add a comment |
$begingroup$
While doing integration by parts we usually do not write the constant derived from the second part of the formula and instead we assign general constant for the whole integration.What is the reason for that?
calculus integration definite-integrals
asked Dec 24 '18 at 22:27
Arif RustamovArif Rustamov
367
$endgroup$
5
$begingroup$
We do—but as: $C_1+C_2=:C$
$endgroup$
– LoveTooNap29
Dec 24 '18 at 22:28
$begingroup$
Two constants added together gives another constant, so we generally combine them for simplicity
$endgroup$
– WaveX
Dec 24 '18 at 22:30
$begingroup$
So,let's say whenever we need to find that constant we will be able to find the sum of both of them,simultaneously?
$endgroup$
– Arif Rustamov
Dec 24 '18 at 22:30
7
$begingroup$
The sum of two arbitrary constants is another arbitrary constant.
$endgroup$
– Winther
Dec 24 '18 at 22:31
add a comment |
$begingroup$
While doing integration by parts we usually do not write the constant derived from the second part of the formula and instead we assign general constant for the whole integration.What is the reason for that?
calculus integration definite-integrals
asked Dec 24 '18 at 22:27
Arif RustamovArif Rustamov
367
$endgroup$
While doing integration by parts we usually do not write the constant derived from the second part of the formula and instead we assign general constant for the whole integration.What is the reason for that?
calculus integration definite-integrals
calculus integration definite-integrals
asked Dec 24 '18 at 22:27
Arif RustamovArif Rustamov
367
asked Dec 24 '18 at 22:27
Arif RustamovArif Rustamov
367
asked Dec 24 '18 at 22:27
Arif RustamovArif Rustamov
367
asked Dec 24 '18 at 22:27
Arif RustamovArif Rustamov
367
asked Dec 24 '18 at 22:27
Arif RustamovArif Rustamov
367
367
5
$begingroup$
We do—but as: $C_1+C_2=:C$
$endgroup$
– LoveTooNap29
Dec 24 '18 at 22:28
$begingroup$
Two constants added together gives another constant, so we generally combine them for simplicity
$endgroup$
– WaveX
Dec 24 '18 at 22:30
$begingroup$
So,let's say whenever we need to find that constant we will be able to find the sum of both of them,simultaneously?
$endgroup$
– Arif Rustamov
Dec 24 '18 at 22:30
7
$begingroup$
The sum of two arbitrary constants is another arbitrary constant.
$endgroup$
– Winther
Dec 24 '18 at 22:31
add a comment |
5
$begingroup$
We do—but as: $C_1+C_2=:C$
$endgroup$
– LoveTooNap29
Dec 24 '18 at 22:28
$begingroup$
Two constants added together gives another constant, so we generally combine them for simplicity
$endgroup$
– WaveX
Dec 24 '18 at 22:30
$begingroup$
So,let's say whenever we need to find that constant we will be able to find the sum of both of them,simultaneously?
$endgroup$
– Arif Rustamov
Dec 24 '18 at 22:30
7
$begingroup$
The sum of two arbitrary constants is another arbitrary constant.
$endgroup$
– Winther
Dec 24 '18 at 22:31
5
5
$begingroup$
We do—but as: $C_1+C_2=:C$
$endgroup$
– LoveTooNap29
Dec 24 '18 at 22:28
$begingroup$
We do—but as: $C_1+C_2=:C$
$endgroup$
– LoveTooNap29
Dec 24 '18 at 22:28
$begingroup$
Two constants added together gives another constant, so we generally combine them for simplicity
$endgroup$
– WaveX
Dec 24 '18 at 22:30
$begingroup$
Two constants added together gives another constant, so we generally combine them for simplicity
$endgroup$
– WaveX
Dec 24 '18 at 22:30
$begingroup$
So,let's say whenever we need to find that constant we will be able to find the sum of both of them,simultaneously?
$endgroup$
– Arif Rustamov
Dec 24 '18 at 22:30
$begingroup$
So,let's say whenever we need to find that constant we will be able to find the sum of both of them,simultaneously?
$endgroup$
– Arif Rustamov
Dec 24 '18 at 22:30
7
7
$begingroup$
The sum of two arbitrary constants is another arbitrary constant.
$endgroup$
– Winther
Dec 24 '18 at 22:31
$begingroup$
The sum of two arbitrary constants is another arbitrary constant.
$endgroup$
– Winther
Dec 24 '18 at 22:31
add a comment |
2 Answers
2
active
oldest
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$begingroup$
Let's consider what this constant means. When you write, for example, $int 2x dx = x^2 + C$, this essentially means that $2x$ has a lot of antiderivatives: $x^2$ is one, $x^2 + 1$ is another, $x^2 - 3.272$ is a third, and so on. You can put any real number where the $C$ is, and you'll get a proper antiderivative.
What would happen if you have two constants? Say, we have $x^2 + C_1 + C_2$. This means we can replace $C_1$ with any real number, and $C_2$ with any real number. Like, we can choose $C_1=3$ and $C_2 = -3.15$, and get $x^2 + 3 - 3.15$. But anything we can achieve by adding two real numbers, we can also achieve by adding only one - their sum (in our example $x^2 - 0.15$). So having two constants doesn't give us anything more than just having one constant.
answered Dec 25 '18 at 0:05
Todor MarkovTodor Markov
2,410412
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add a comment |
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I think the question is being misinterpreted. The formula for integration by parts is $$int f(x) g(x) , dx=f(x) int g(x) , dx-intleft(f'(x) int g(x) , dxright) , dx$$ Suppose we use anti-derivative $G(x)$ of $g(x) $ then the above formula can be written as $$int f(x) g(x) , dx=f(x) G(x) - int f'(x) G(x) , dx$$ If instead of $G(x) $ we use an anti-derivative $G(x) + C$ then we get $$int f(x) g(x) , dx=f(x) G(x) + Cf(x) - int f'(x) G(x) , dx-Cint f'(x) , dx$$ The effect of last term is cancelled by the second term $Cf(x) $ and at most a constant remains as expected and thus we don't really need to add a constant explicitly for the anti-derivative of $g(x) $.
answered Dec 25 '18 at 5:30
Paramanand SinghParamanand Singh
50.3k556163
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
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$begingroup$
Let's consider what this constant means. When you write, for example, $int 2x dx = x^2 + C$, this essentially means that $2x$ has a lot of antiderivatives: $x^2$ is one, $x^2 + 1$ is another, $x^2 - 3.272$ is a third, and so on. You can put any real number where the $C$ is, and you'll get a proper antiderivative.
What would happen if you have two constants? Say, we have $x^2 + C_1 + C_2$. This means we can replace $C_1$ with any real number, and $C_2$ with any real number. Like, we can choose $C_1=3$ and $C_2 = -3.15$, and get $x^2 + 3 - 3.15$. But anything we can achieve by adding two real numbers, we can also achieve by adding only one - their sum (in our example $x^2 - 0.15$). So having two constants doesn't give us anything more than just having one constant.
answered Dec 25 '18 at 0:05
Todor MarkovTodor Markov
2,410412
$endgroup$
add a comment |
$begingroup$
Let's consider what this constant means. When you write, for example, $int 2x dx = x^2 + C$, this essentially means that $2x$ has a lot of antiderivatives: $x^2$ is one, $x^2 + 1$ is another, $x^2 - 3.272$ is a third, and so on. You can put any real number where the $C$ is, and you'll get a proper antiderivative.
What would happen if you have two constants? Say, we have $x^2 + C_1 + C_2$. This means we can replace $C_1$ with any real number, and $C_2$ with any real number. Like, we can choose $C_1=3$ and $C_2 = -3.15$, and get $x^2 + 3 - 3.15$. But anything we can achieve by adding two real numbers, we can also achieve by adding only one - their sum (in our example $x^2 - 0.15$). So having two constants doesn't give us anything more than just having one constant.
answered Dec 25 '18 at 0:05
Todor MarkovTodor Markov
2,410412
$endgroup$
add a comment |
$begingroup$
Let's consider what this constant means. When you write, for example, $int 2x dx = x^2 + C$, this essentially means that $2x$ has a lot of antiderivatives: $x^2$ is one, $x^2 + 1$ is another, $x^2 - 3.272$ is a third, and so on. You can put any real number where the $C$ is, and you'll get a proper antiderivative.
What would happen if you have two constants? Say, we have $x^2 + C_1 + C_2$. This means we can replace $C_1$ with any real number, and $C_2$ with any real number. Like, we can choose $C_1=3$ and $C_2 = -3.15$, and get $x^2 + 3 - 3.15$. But anything we can achieve by adding two real numbers, we can also achieve by adding only one - their sum (in our example $x^2 - 0.15$). So having two constants doesn't give us anything more than just having one constant.
answered Dec 25 '18 at 0:05
Todor MarkovTodor Markov
2,410412
$endgroup$
Let's consider what this constant means. When you write, for example, $int 2x dx = x^2 + C$, this essentially means that $2x$ has a lot of antiderivatives: $x^2$ is one, $x^2 + 1$ is another, $x^2 - 3.272$ is a third, and so on. You can put any real number where the $C$ is, and you'll get a proper antiderivative.
What would happen if you have two constants? Say, we have $x^2 + C_1 + C_2$. This means we can replace $C_1$ with any real number, and $C_2$ with any real number. Like, we can choose $C_1=3$ and $C_2 = -3.15$, and get $x^2 + 3 - 3.15$. But anything we can achieve by adding two real numbers, we can also achieve by adding only one - their sum (in our example $x^2 - 0.15$). So having two constants doesn't give us anything more than just having one constant.
answered Dec 25 '18 at 0:05
Todor MarkovTodor Markov
2,410412
answered Dec 25 '18 at 0:05
Todor MarkovTodor Markov
2,410412
answered Dec 25 '18 at 0:05
Todor MarkovTodor Markov
2,410412
answered Dec 25 '18 at 0:05
Todor MarkovTodor Markov
2,410412
2,410412
add a comment |
add a comment |
$begingroup$
I think the question is being misinterpreted. The formula for integration by parts is $$int f(x) g(x) , dx=f(x) int g(x) , dx-intleft(f'(x) int g(x) , dxright) , dx$$ Suppose we use anti-derivative $G(x)$ of $g(x) $ then the above formula can be written as $$int f(x) g(x) , dx=f(x) G(x) - int f'(x) G(x) , dx$$ If instead of $G(x) $ we use an anti-derivative $G(x) + C$ then we get $$int f(x) g(x) , dx=f(x) G(x) + Cf(x) - int f'(x) G(x) , dx-Cint f'(x) , dx$$ The effect of last term is cancelled by the second term $Cf(x) $ and at most a constant remains as expected and thus we don't really need to add a constant explicitly for the anti-derivative of $g(x) $.
answered Dec 25 '18 at 5:30
Paramanand SinghParamanand Singh
50.3k556163
$endgroup$
add a comment |
$begingroup$
I think the question is being misinterpreted. The formula for integration by parts is $$int f(x) g(x) , dx=f(x) int g(x) , dx-intleft(f'(x) int g(x) , dxright) , dx$$ Suppose we use anti-derivative $G(x)$ of $g(x) $ then the above formula can be written as $$int f(x) g(x) , dx=f(x) G(x) - int f'(x) G(x) , dx$$ If instead of $G(x) $ we use an anti-derivative $G(x) + C$ then we get $$int f(x) g(x) , dx=f(x) G(x) + Cf(x) - int f'(x) G(x) , dx-Cint f'(x) , dx$$ The effect of last term is cancelled by the second term $Cf(x) $ and at most a constant remains as expected and thus we don't really need to add a constant explicitly for the anti-derivative of $g(x) $.
answered Dec 25 '18 at 5:30
Paramanand SinghParamanand Singh
50.3k556163
$endgroup$
add a comment |
$begingroup$
I think the question is being misinterpreted. The formula for integration by parts is $$int f(x) g(x) , dx=f(x) int g(x) , dx-intleft(f'(x) int g(x) , dxright) , dx$$ Suppose we use anti-derivative $G(x)$ of $g(x) $ then the above formula can be written as $$int f(x) g(x) , dx=f(x) G(x) - int f'(x) G(x) , dx$$ If instead of $G(x) $ we use an anti-derivative $G(x) + C$ then we get $$int f(x) g(x) , dx=f(x) G(x) + Cf(x) - int f'(x) G(x) , dx-Cint f'(x) , dx$$ The effect of last term is cancelled by the second term $Cf(x) $ and at most a constant remains as expected and thus we don't really need to add a constant explicitly for the anti-derivative of $g(x) $.
answered Dec 25 '18 at 5:30
Paramanand SinghParamanand Singh
50.3k556163
$endgroup$
I think the question is being misinterpreted. The formula for integration by parts is $$int f(x) g(x) , dx=f(x) int g(x) , dx-intleft(f'(x) int g(x) , dxright) , dx$$ Suppose we use anti-derivative $G(x)$ of $g(x) $ then the above formula can be written as $$int f(x) g(x) , dx=f(x) G(x) - int f'(x) G(x) , dx$$ If instead of $G(x) $ we use an anti-derivative $G(x) + C$ then we get $$int f(x) g(x) , dx=f(x) G(x) + Cf(x) - int f'(x) G(x) , dx-Cint f'(x) , dx$$ The effect of last term is cancelled by the second term $Cf(x) $ and at most a constant remains as expected and thus we don't really need to add a constant explicitly for the anti-derivative of $g(x) $.
answered Dec 25 '18 at 5:30
Paramanand SinghParamanand Singh
50.3k556163
answered Dec 25 '18 at 5:30
Paramanand SinghParamanand Singh
50.3k556163
answered Dec 25 '18 at 5:30
Paramanand SinghParamanand Singh
50.3k556163
answered Dec 25 '18 at 5:30
Paramanand SinghParamanand Singh
50.3k556163
50.3k556163
add a comment |
add a comment |
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5
$begingroup$
We do—but as: $C_1+C_2=:C$
$endgroup$
– LoveTooNap29
Dec 24 '18 at 22:28
$begingroup$
Two constants added together gives another constant, so we generally combine them for simplicity
$endgroup$
– WaveX
Dec 24 '18 at 22:30
$begingroup$
So,let's say whenever we need to find that constant we will be able to find the sum of both of them,simultaneously?
$endgroup$
– Arif Rustamov
Dec 24 '18 at 22:30
7
$begingroup$
The sum of two arbitrary constants is another arbitrary constant.
$endgroup$
– Winther
Dec 24 '18 at 22:31