Integration of $(x)d(x^2)$, is my solution mathematically correct?
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Just for my own understanding of how exactly integration works, are these steps correct:
$$begin{align}int x,d(x^2) qquad &implies x^2 = u \ & implies x= sqrt{u}end{align}$$
Thus, it becomes $$intsqrt{u},du = frac{2}{3}u^{3/2} implies {2over3}x^3$$
integration
edited Dec 6 at 2:03
Decaf-Math
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asked Dec 6 at 1:53
strateeg32
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up vote
1
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Just for my own understanding of how exactly integration works, are these steps correct:
$$begin{align}int x,d(x^2) qquad &implies x^2 = u \ & implies x= sqrt{u}end{align}$$
Thus, it becomes $$intsqrt{u},du = frac{2}{3}u^{3/2} implies {2over3}x^3$$
integration
edited Dec 6 at 2:03
Decaf-Math
3,154825
asked Dec 6 at 1:53
strateeg32
1384
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Just for my own understanding of how exactly integration works, are these steps correct:
$$begin{align}int x,d(x^2) qquad &implies x^2 = u \ & implies x= sqrt{u}end{align}$$
Thus, it becomes $$intsqrt{u},du = frac{2}{3}u^{3/2} implies {2over3}x^3$$
integration
edited Dec 6 at 2:03
Decaf-Math
3,154825
asked Dec 6 at 1:53
strateeg32
1384
Just for my own understanding of how exactly integration works, are these steps correct:
$$begin{align}int x,d(x^2) qquad &implies x^2 = u \ & implies x= sqrt{u}end{align}$$
Thus, it becomes $$intsqrt{u},du = frac{2}{3}u^{3/2} implies {2over3}x^3$$
integration
integration
edited Dec 6 at 2:03
Decaf-Math
3,154825
asked Dec 6 at 1:53
strateeg32
1384
edited Dec 6 at 2:03
Decaf-Math
3,154825
asked Dec 6 at 1:53
strateeg32
1384
edited Dec 6 at 2:03
Decaf-Math
3,154825
edited Dec 6 at 2:03
Decaf-Math
3,154825
edited Dec 6 at 2:03
Decaf-Math
3,154825
3,154825
asked Dec 6 at 1:53
strateeg32
1384
asked Dec 6 at 1:53
strateeg32
1384
asked Dec 6 at 1:53
strateeg32
1384
1384
add a comment |
add a comment |
2 Answers
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While fundamentally correct, aside from your lack of a $+C$ constant, notationally your work is a bit of a mess (before someone edited it for clarity's sake).
First, we start with
$$int x d(x^2)$$
From here, we make the substitution $u = x^2$. This gives $x = sqrt{u}$. Thus,
$$x d(x^2) = sqrt{u} du ;;; Rightarrow ;;; int xd(x^2) = int u^{1/2}du = frac{2}{3}u^{3/2} + C$$
Reutilizing our substitution, we finally get our answer:
$$frac{2}{3}u^{3/2} + C = frac{2}{3}x^3 +C$$
Your errors make it hard to follow your thought process, because $sqrt{u} du neq frac{2}{3}u^{3/2}$, when you assert they're equal in your last line.
As a note, this question was asked before: you might want to check more thoroughly next time. The original question.
answered Dec 6 at 2:01
Eevee Trainer
3,370225
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2
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Yes, except for an integration on the LHS of the last line. Also make sure to keep track of your limits. You would get the same expression when you use $d(x^2) = 2x dx$
answered Dec 6 at 2:03
zero
608
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
While fundamentally correct, aside from your lack of a $+C$ constant, notationally your work is a bit of a mess (before someone edited it for clarity's sake).
First, we start with
$$int x d(x^2)$$
From here, we make the substitution $u = x^2$. This gives $x = sqrt{u}$. Thus,
$$x d(x^2) = sqrt{u} du ;;; Rightarrow ;;; int xd(x^2) = int u^{1/2}du = frac{2}{3}u^{3/2} + C$$
Reutilizing our substitution, we finally get our answer:
$$frac{2}{3}u^{3/2} + C = frac{2}{3}x^3 +C$$
Your errors make it hard to follow your thought process, because $sqrt{u} du neq frac{2}{3}u^{3/2}$, when you assert they're equal in your last line.
As a note, this question was asked before: you might want to check more thoroughly next time. The original question.
answered Dec 6 at 2:01
Eevee Trainer
3,370225
add a comment |
up vote
1
down vote
accepted
While fundamentally correct, aside from your lack of a $+C$ constant, notationally your work is a bit of a mess (before someone edited it for clarity's sake).
First, we start with
$$int x d(x^2)$$
From here, we make the substitution $u = x^2$. This gives $x = sqrt{u}$. Thus,
$$x d(x^2) = sqrt{u} du ;;; Rightarrow ;;; int xd(x^2) = int u^{1/2}du = frac{2}{3}u^{3/2} + C$$
Reutilizing our substitution, we finally get our answer:
$$frac{2}{3}u^{3/2} + C = frac{2}{3}x^3 +C$$
Your errors make it hard to follow your thought process, because $sqrt{u} du neq frac{2}{3}u^{3/2}$, when you assert they're equal in your last line.
As a note, this question was asked before: you might want to check more thoroughly next time. The original question.
answered Dec 6 at 2:01
Eevee Trainer
3,370225
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
While fundamentally correct, aside from your lack of a $+C$ constant, notationally your work is a bit of a mess (before someone edited it for clarity's sake).
First, we start with
$$int x d(x^2)$$
From here, we make the substitution $u = x^2$. This gives $x = sqrt{u}$. Thus,
$$x d(x^2) = sqrt{u} du ;;; Rightarrow ;;; int xd(x^2) = int u^{1/2}du = frac{2}{3}u^{3/2} + C$$
Reutilizing our substitution, we finally get our answer:
$$frac{2}{3}u^{3/2} + C = frac{2}{3}x^3 +C$$
Your errors make it hard to follow your thought process, because $sqrt{u} du neq frac{2}{3}u^{3/2}$, when you assert they're equal in your last line.
As a note, this question was asked before: you might want to check more thoroughly next time. The original question.
answered Dec 6 at 2:01
Eevee Trainer
3,370225
While fundamentally correct, aside from your lack of a $+C$ constant, notationally your work is a bit of a mess (before someone edited it for clarity's sake).
First, we start with
$$int x d(x^2)$$
From here, we make the substitution $u = x^2$. This gives $x = sqrt{u}$. Thus,
$$x d(x^2) = sqrt{u} du ;;; Rightarrow ;;; int xd(x^2) = int u^{1/2}du = frac{2}{3}u^{3/2} + C$$
Reutilizing our substitution, we finally get our answer:
$$frac{2}{3}u^{3/2} + C = frac{2}{3}x^3 +C$$
Your errors make it hard to follow your thought process, because $sqrt{u} du neq frac{2}{3}u^{3/2}$, when you assert they're equal in your last line.
As a note, this question was asked before: you might want to check more thoroughly next time. The original question.
answered Dec 6 at 2:01
Eevee Trainer
3,370225
answered Dec 6 at 2:01
Eevee Trainer
3,370225
answered Dec 6 at 2:01
Eevee Trainer
3,370225
answered Dec 6 at 2:01
Eevee Trainer
3,370225
3,370225
add a comment |
add a comment |
up vote
2
down vote
Yes, except for an integration on the LHS of the last line. Also make sure to keep track of your limits. You would get the same expression when you use $d(x^2) = 2x dx$
answered Dec 6 at 2:03
zero
608
add a comment |
up vote
2
down vote
Yes, except for an integration on the LHS of the last line. Also make sure to keep track of your limits. You would get the same expression when you use $d(x^2) = 2x dx$
answered Dec 6 at 2:03
zero
608
add a comment |
up vote
2
down vote
up vote
2
down vote
Yes, except for an integration on the LHS of the last line. Also make sure to keep track of your limits. You would get the same expression when you use $d(x^2) = 2x dx$
answered Dec 6 at 2:03
zero
608
Yes, except for an integration on the LHS of the last line. Also make sure to keep track of your limits. You would get the same expression when you use $d(x^2) = 2x dx$
answered Dec 6 at 2:03
zero
608
answered Dec 6 at 2:03
zero
608
answered Dec 6 at 2:03
zero
608
answered Dec 6 at 2:03
zero
608
608
add a comment |
add a comment |
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