distribution of constants over integrals
$begingroup$
I'm seeing this as part of a proof for the reduction formula and I see this:
So am I correct for saying that if you multiply the $sin^{n-2}{x}$ by $(1 - sin^2{x})$, you get $sin^{n-2}{x} - sin^n{x}$ and so the $(n-1) int sin^{n-2}{x}cdot cos^2{x}$ becomes what is shown below? is that right? Generally, $int a - b = int a - int b$ right? That integral rule is visually intuitive right?
integration
asked Jan 5 at 1:45
Jwan622Jwan622
2,34611632
$endgroup$
add a comment |
$begingroup$
I'm seeing this as part of a proof for the reduction formula and I see this:
So am I correct for saying that if you multiply the $sin^{n-2}{x}$ by $(1 - sin^2{x})$, you get $sin^{n-2}{x} - sin^n{x}$ and so the $(n-1) int sin^{n-2}{x}cdot cos^2{x}$ becomes what is shown below? is that right? Generally, $int a - b = int a - int b$ right? That integral rule is visually intuitive right?
integration
asked Jan 5 at 1:45
Jwan622Jwan622
2,34611632
$endgroup$
add a comment |
$begingroup$
I'm seeing this as part of a proof for the reduction formula and I see this:
So am I correct for saying that if you multiply the $sin^{n-2}{x}$ by $(1 - sin^2{x})$, you get $sin^{n-2}{x} - sin^n{x}$ and so the $(n-1) int sin^{n-2}{x}cdot cos^2{x}$ becomes what is shown below? is that right? Generally, $int a - b = int a - int b$ right? That integral rule is visually intuitive right?
integration
asked Jan 5 at 1:45
Jwan622Jwan622
2,34611632
$endgroup$
I'm seeing this as part of a proof for the reduction formula and I see this:
So am I correct for saying that if you multiply the $sin^{n-2}{x}$ by $(1 - sin^2{x})$, you get $sin^{n-2}{x} - sin^n{x}$ and so the $(n-1) int sin^{n-2}{x}cdot cos^2{x}$ becomes what is shown below? is that right? Generally, $int a - b = int a - int b$ right? That integral rule is visually intuitive right?
integration
integration
asked Jan 5 at 1:45
Jwan622Jwan622
2,34611632
asked Jan 5 at 1:45
Jwan622Jwan622
2,34611632
asked Jan 5 at 1:45
Jwan622Jwan622
2,34611632
asked Jan 5 at 1:45
Jwan622Jwan622
2,34611632
asked Jan 5 at 1:45
Jwan622Jwan622
2,34611632
2,34611632
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Assuming that all of your functions are "nice", the integral operator is linear, and will split over sums and respects multiplication by constants. Formally, we have that
$$
int(af(x)+bg(x)),dx = aint f(x),dx + bint g(x),dx
$$
for all real numbers $a$ and $b$ and all functions $f$ and $g$ which have anti-derivatives.
answered Jan 5 at 1:51
ItsJustTranscendenceBroItsJustTranscendenceBro
1712
$endgroup$
$begingroup$
How does one define 'nice'?
$endgroup$
– user150203
Jan 5 at 1:57
$begingroup$
@DavidG I mentioned it in the final sentence; you need to be able to find anti-derivatives for both $f$ and $g$. The issue is that the class of functions which have anti-derivatives is difficult to characterize, so most of the time we pick a criterion that is way too powerful, like $f$ and $g$ are continuous. That would be enough to know that they have anti-derivatives, but it is not a necessary property.
$endgroup$
– ItsJustTranscendenceBro
Jan 5 at 1:59
$begingroup$
Cheers, appreciate the detail.
$endgroup$
– user150203
Jan 5 at 2:03
1
$begingroup$
@DavidG For more information for anyone who reads this in the future (and you as well if you don't already know), you can look at this post for more info.
$endgroup$
– ItsJustTranscendenceBro
Jan 5 at 2:04
$begingroup$
Awesome, I will check out. Thanks again.
$endgroup$
– user150203
Jan 5 at 2:10
add a comment |
$begingroup$
Yes, the fact that $int [a(x)-b(x)]dx$ is equal to $int a(x)dx-int b(x)dx$ is a well-known valid rule for manipulating integrals. This is also known as the "difference rule" for integrals. There is a good proof explaining why this rule is true on Paul's Online Notes.
In particular, your manipulation of these integrals above seems correct.
answered Jan 5 at 1:50
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Assuming that all of your functions are "nice", the integral operator is linear, and will split over sums and respects multiplication by constants. Formally, we have that
$$
int(af(x)+bg(x)),dx = aint f(x),dx + bint g(x),dx
$$
for all real numbers $a$ and $b$ and all functions $f$ and $g$ which have anti-derivatives.
answered Jan 5 at 1:51
ItsJustTranscendenceBroItsJustTranscendenceBro
1712
$endgroup$
$begingroup$
How does one define 'nice'?
$endgroup$
– user150203
Jan 5 at 1:57
$begingroup$
@DavidG I mentioned it in the final sentence; you need to be able to find anti-derivatives for both $f$ and $g$. The issue is that the class of functions which have anti-derivatives is difficult to characterize, so most of the time we pick a criterion that is way too powerful, like $f$ and $g$ are continuous. That would be enough to know that they have anti-derivatives, but it is not a necessary property.
$endgroup$
– ItsJustTranscendenceBro
Jan 5 at 1:59
$begingroup$
Cheers, appreciate the detail.
$endgroup$
– user150203
Jan 5 at 2:03
1
$begingroup$
@DavidG For more information for anyone who reads this in the future (and you as well if you don't already know), you can look at this post for more info.
$endgroup$
– ItsJustTranscendenceBro
Jan 5 at 2:04
$begingroup$
Awesome, I will check out. Thanks again.
$endgroup$
– user150203
Jan 5 at 2:10
add a comment |
$begingroup$
Assuming that all of your functions are "nice", the integral operator is linear, and will split over sums and respects multiplication by constants. Formally, we have that
$$
int(af(x)+bg(x)),dx = aint f(x),dx + bint g(x),dx
$$
for all real numbers $a$ and $b$ and all functions $f$ and $g$ which have anti-derivatives.
answered Jan 5 at 1:51
ItsJustTranscendenceBroItsJustTranscendenceBro
1712
$endgroup$
$begingroup$
How does one define 'nice'?
$endgroup$
– user150203
Jan 5 at 1:57
$begingroup$
@DavidG I mentioned it in the final sentence; you need to be able to find anti-derivatives for both $f$ and $g$. The issue is that the class of functions which have anti-derivatives is difficult to characterize, so most of the time we pick a criterion that is way too powerful, like $f$ and $g$ are continuous. That would be enough to know that they have anti-derivatives, but it is not a necessary property.
$endgroup$
– ItsJustTranscendenceBro
Jan 5 at 1:59
$begingroup$
Cheers, appreciate the detail.
$endgroup$
– user150203
Jan 5 at 2:03
1
$begingroup$
@DavidG For more information for anyone who reads this in the future (and you as well if you don't already know), you can look at this post for more info.
$endgroup$
– ItsJustTranscendenceBro
Jan 5 at 2:04
$begingroup$
Awesome, I will check out. Thanks again.
$endgroup$
– user150203
Jan 5 at 2:10
add a comment |
$begingroup$
Assuming that all of your functions are "nice", the integral operator is linear, and will split over sums and respects multiplication by constants. Formally, we have that
$$
int(af(x)+bg(x)),dx = aint f(x),dx + bint g(x),dx
$$
for all real numbers $a$ and $b$ and all functions $f$ and $g$ which have anti-derivatives.
answered Jan 5 at 1:51
ItsJustTranscendenceBroItsJustTranscendenceBro
1712
$endgroup$
Assuming that all of your functions are "nice", the integral operator is linear, and will split over sums and respects multiplication by constants. Formally, we have that
$$
int(af(x)+bg(x)),dx = aint f(x),dx + bint g(x),dx
$$
for all real numbers $a$ and $b$ and all functions $f$ and $g$ which have anti-derivatives.
answered Jan 5 at 1:51
ItsJustTranscendenceBroItsJustTranscendenceBro
1712
answered Jan 5 at 1:51
ItsJustTranscendenceBroItsJustTranscendenceBro
1712
answered Jan 5 at 1:51
ItsJustTranscendenceBroItsJustTranscendenceBro
1712
answered Jan 5 at 1:51
ItsJustTranscendenceBroItsJustTranscendenceBro
1712
1712
$begingroup$
How does one define 'nice'?
$endgroup$
– user150203
Jan 5 at 1:57
$begingroup$
@DavidG I mentioned it in the final sentence; you need to be able to find anti-derivatives for both $f$ and $g$. The issue is that the class of functions which have anti-derivatives is difficult to characterize, so most of the time we pick a criterion that is way too powerful, like $f$ and $g$ are continuous. That would be enough to know that they have anti-derivatives, but it is not a necessary property.
$endgroup$
– ItsJustTranscendenceBro
Jan 5 at 1:59
$begingroup$
Cheers, appreciate the detail.
$endgroup$
– user150203
Jan 5 at 2:03
1
$begingroup$
@DavidG For more information for anyone who reads this in the future (and you as well if you don't already know), you can look at this post for more info.
$endgroup$
– ItsJustTranscendenceBro
Jan 5 at 2:04
$begingroup$
Awesome, I will check out. Thanks again.
$endgroup$
– user150203
Jan 5 at 2:10
add a comment |
$begingroup$
How does one define 'nice'?
$endgroup$
– user150203
Jan 5 at 1:57
$begingroup$
@DavidG I mentioned it in the final sentence; you need to be able to find anti-derivatives for both $f$ and $g$. The issue is that the class of functions which have anti-derivatives is difficult to characterize, so most of the time we pick a criterion that is way too powerful, like $f$ and $g$ are continuous. That would be enough to know that they have anti-derivatives, but it is not a necessary property.
$endgroup$
– ItsJustTranscendenceBro
Jan 5 at 1:59
$begingroup$
Cheers, appreciate the detail.
$endgroup$
– user150203
Jan 5 at 2:03
1
$begingroup$
@DavidG For more information for anyone who reads this in the future (and you as well if you don't already know), you can look at this post for more info.
$endgroup$
– ItsJustTranscendenceBro
Jan 5 at 2:04
$begingroup$
Awesome, I will check out. Thanks again.
$endgroup$
– user150203
Jan 5 at 2:10
$begingroup$
How does one define 'nice'?
$endgroup$
– user150203
Jan 5 at 1:57
$begingroup$
How does one define 'nice'?
$endgroup$
– user150203
Jan 5 at 1:57
$begingroup$
@DavidG I mentioned it in the final sentence; you need to be able to find anti-derivatives for both $f$ and $g$. The issue is that the class of functions which have anti-derivatives is difficult to characterize, so most of the time we pick a criterion that is way too powerful, like $f$ and $g$ are continuous. That would be enough to know that they have anti-derivatives, but it is not a necessary property.
$endgroup$
– ItsJustTranscendenceBro
Jan 5 at 1:59
$begingroup$
@DavidG I mentioned it in the final sentence; you need to be able to find anti-derivatives for both $f$ and $g$. The issue is that the class of functions which have anti-derivatives is difficult to characterize, so most of the time we pick a criterion that is way too powerful, like $f$ and $g$ are continuous. That would be enough to know that they have anti-derivatives, but it is not a necessary property.
$endgroup$
– ItsJustTranscendenceBro
Jan 5 at 1:59
$begingroup$
Cheers, appreciate the detail.
$endgroup$
– user150203
Jan 5 at 2:03
$begingroup$
Cheers, appreciate the detail.
$endgroup$
– user150203
Jan 5 at 2:03
1
1
$begingroup$
@DavidG For more information for anyone who reads this in the future (and you as well if you don't already know), you can look at this post for more info.
$endgroup$
– ItsJustTranscendenceBro
Jan 5 at 2:04
$begingroup$
@DavidG For more information for anyone who reads this in the future (and you as well if you don't already know), you can look at this post for more info.
$endgroup$
– ItsJustTranscendenceBro
Jan 5 at 2:04
$begingroup$
Awesome, I will check out. Thanks again.
$endgroup$
– user150203
Jan 5 at 2:10
$begingroup$
Awesome, I will check out. Thanks again.
$endgroup$
– user150203
Jan 5 at 2:10
add a comment |
$begingroup$
Yes, the fact that $int [a(x)-b(x)]dx$ is equal to $int a(x)dx-int b(x)dx$ is a well-known valid rule for manipulating integrals. This is also known as the "difference rule" for integrals. There is a good proof explaining why this rule is true on Paul's Online Notes.
In particular, your manipulation of these integrals above seems correct.
answered Jan 5 at 1:50
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
add a comment |
$begingroup$
Yes, the fact that $int [a(x)-b(x)]dx$ is equal to $int a(x)dx-int b(x)dx$ is a well-known valid rule for manipulating integrals. This is also known as the "difference rule" for integrals. There is a good proof explaining why this rule is true on Paul's Online Notes.
In particular, your manipulation of these integrals above seems correct.
answered Jan 5 at 1:50
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
add a comment |
$begingroup$
Yes, the fact that $int [a(x)-b(x)]dx$ is equal to $int a(x)dx-int b(x)dx$ is a well-known valid rule for manipulating integrals. This is also known as the "difference rule" for integrals. There is a good proof explaining why this rule is true on Paul's Online Notes.
In particular, your manipulation of these integrals above seems correct.
answered Jan 5 at 1:50
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
Yes, the fact that $int [a(x)-b(x)]dx$ is equal to $int a(x)dx-int b(x)dx$ is a well-known valid rule for manipulating integrals. This is also known as the "difference rule" for integrals. There is a good proof explaining why this rule is true on Paul's Online Notes.
In particular, your manipulation of these integrals above seems correct.
answered Jan 5 at 1:50
Noble MushtakNoble Mushtak
15.3k1835
answered Jan 5 at 1:50
Noble MushtakNoble Mushtak
15.3k1835
answered Jan 5 at 1:50
Noble MushtakNoble Mushtak
15.3k1835
answered Jan 5 at 1:50
Noble MushtakNoble Mushtak
15.3k1835
15.3k1835
add a comment |
add a comment |
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