When change of variable makes an empty interval
Please consider the following case:
$$I = int^1_{-1}x^2dx$$
$$u(x) = x^2 rightarrow du = 2x,dx$$
$$u(-1) = 1, u(1) = 1$$
So
$$I = int^1_1frac{u}{2sqrt u} du = 0$$
Obviously the problem here is to only consider the positive root of u. I don't know how to consider both roots. This example is trivial but I have another example where such substitution would be really helpful:
$$I = int^1_{-1}frac{x^2(1 - x^2)^frac{3}{2}}{3} - frac{x^2(1 - x^2)^frac{5}{2}}{5} - frac{x^4(1 - x^2)^frac{3}{2}}{3} dx$$
I don't want you to solve it for me using another method, I know how to use an integral solver online. My question is how to properly do the change of variable.
integration definite-integrals change-of-variable
asked Dec 9 at 17:12
Winter
492421
add a comment |
Please consider the following case:
$$I = int^1_{-1}x^2dx$$
$$u(x) = x^2 rightarrow du = 2x,dx$$
$$u(-1) = 1, u(1) = 1$$
So
$$I = int^1_1frac{u}{2sqrt u} du = 0$$
Obviously the problem here is to only consider the positive root of u. I don't know how to consider both roots. This example is trivial but I have another example where such substitution would be really helpful:
$$I = int^1_{-1}frac{x^2(1 - x^2)^frac{3}{2}}{3} - frac{x^2(1 - x^2)^frac{5}{2}}{5} - frac{x^4(1 - x^2)^frac{3}{2}}{3} dx$$
I don't want you to solve it for me using another method, I know how to use an integral solver online. My question is how to properly do the change of variable.
integration definite-integrals change-of-variable
asked Dec 9 at 17:12
Winter
492421
The derivative of $x^2$ isn't 2
– orange
Dec 9 at 19:06
@orange Thanks, corrected
– Winter
Dec 9 at 19:42
related: $u$-substitution always evaluates to $0$
– Dando18
Dec 10 at 2:45
add a comment |
Please consider the following case:
$$I = int^1_{-1}x^2dx$$
$$u(x) = x^2 rightarrow du = 2x,dx$$
$$u(-1) = 1, u(1) = 1$$
So
$$I = int^1_1frac{u}{2sqrt u} du = 0$$
Obviously the problem here is to only consider the positive root of u. I don't know how to consider both roots. This example is trivial but I have another example where such substitution would be really helpful:
$$I = int^1_{-1}frac{x^2(1 - x^2)^frac{3}{2}}{3} - frac{x^2(1 - x^2)^frac{5}{2}}{5} - frac{x^4(1 - x^2)^frac{3}{2}}{3} dx$$
I don't want you to solve it for me using another method, I know how to use an integral solver online. My question is how to properly do the change of variable.
integration definite-integrals change-of-variable
asked Dec 9 at 17:12
Winter
492421
Please consider the following case:
$$I = int^1_{-1}x^2dx$$
$$u(x) = x^2 rightarrow du = 2x,dx$$
$$u(-1) = 1, u(1) = 1$$
So
$$I = int^1_1frac{u}{2sqrt u} du = 0$$
Obviously the problem here is to only consider the positive root of u. I don't know how to consider both roots. This example is trivial but I have another example where such substitution would be really helpful:
$$I = int^1_{-1}frac{x^2(1 - x^2)^frac{3}{2}}{3} - frac{x^2(1 - x^2)^frac{5}{2}}{5} - frac{x^4(1 - x^2)^frac{3}{2}}{3} dx$$
I don't want you to solve it for me using another method, I know how to use an integral solver online. My question is how to properly do the change of variable.
integration definite-integrals change-of-variable
integration definite-integrals change-of-variable
asked Dec 9 at 17:12
Winter
492421
asked Dec 9 at 17:12
Winter
492421
edited Dec 9 at 19:41
asked Dec 9 at 17:12
Winter
492421
asked Dec 9 at 17:12
Winter
492421
asked Dec 9 at 17:12
Winter
492421
492421
The derivative of $x^2$ isn't 2
– orange
Dec 9 at 19:06
@orange Thanks, corrected
– Winter
Dec 9 at 19:42
related: $u$-substitution always evaluates to $0$
– Dando18
Dec 10 at 2:45
add a comment |
The derivative of $x^2$ isn't 2
– orange
Dec 9 at 19:06
@orange Thanks, corrected
– Winter
Dec 9 at 19:42
related: $u$-substitution always evaluates to $0$
– Dando18
Dec 10 at 2:45
The derivative of $x^2$ isn't 2
– orange
Dec 9 at 19:06
The derivative of $x^2$ isn't 2
– orange
Dec 9 at 19:06
@orange Thanks, corrected
– Winter
Dec 9 at 19:42
@orange Thanks, corrected
– Winter
Dec 9 at 19:42
related: $u$-substitution always evaluates to $0$
– Dando18
Dec 10 at 2:45
related: $u$-substitution always evaluates to $0$
– Dando18
Dec 10 at 2:45
add a comment |
2 Answers
2
active
oldest
votes
In general you need to use a one-to-one change of variable; thus for instance if you were to compute $int_{-1}^1 x^2 dx$ you would split into $int_{-1}^0 x^2 dx + int_0^1 x^2 dx$. Then each of those integrals, after substitution and simplification, becomes $int_0^1 frac{1}{2} u^{1/2} du$.
(That said, there are some situations where by basically breaking the rules twice you cancel out your mistakes and get the right answer; see Why can the trig sub $x=cosh(theta)$ be used to solve the integral: $int frac{1}{(x^2-1)^{3/2}}dx$? for instance.)
answered Dec 9 at 17:20
Ian
67.3k25386
It is not necessary for changes of variable to be one-to-one, see my comment here: math.stackexchange.com/a/1101421/5316
– Matthew Towers
Dec 9 at 19:39
1
@MatthewTowers Yes, it is true that $int_a^b f'(x) dx$ is always $f(b) - f(a)$, regardless of injectivity, which means you can do various similar things also regardless of injectivity. What injectivity buys you is a lack of a singularity in $frac{dx}{du}$, which is a serious problem for, say, $u=x^2$ on $[a,b]$ with $a<0,b>0$.
– Ian
Dec 9 at 19:52
add a comment |
While it's certainly true that $du = 2x,dx$ over the whole region of integration, it's not true that $x = sqrt{u}$ over the region. This is only true for $x > 0$, so you can't apply it to the whole integral. Instead, you need to split it apart and use the correct transformation $x = -sqrt{u}$ for the region $x < 0$.
In general, if your change of variables $u(x)$ has a multivalued inverse, you'll need to split the integral into the regions where each possible inverse function $x(u)$ applies.
answered Dec 9 at 17:33
eyeballfrog
6,058629
add a comment |
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2 Answers
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2 Answers
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In general you need to use a one-to-one change of variable; thus for instance if you were to compute $int_{-1}^1 x^2 dx$ you would split into $int_{-1}^0 x^2 dx + int_0^1 x^2 dx$. Then each of those integrals, after substitution and simplification, becomes $int_0^1 frac{1}{2} u^{1/2} du$.
(That said, there are some situations where by basically breaking the rules twice you cancel out your mistakes and get the right answer; see Why can the trig sub $x=cosh(theta)$ be used to solve the integral: $int frac{1}{(x^2-1)^{3/2}}dx$? for instance.)
answered Dec 9 at 17:20
Ian
67.3k25386
It is not necessary for changes of variable to be one-to-one, see my comment here: math.stackexchange.com/a/1101421/5316
– Matthew Towers
Dec 9 at 19:39
1
@MatthewTowers Yes, it is true that $int_a^b f'(x) dx$ is always $f(b) - f(a)$, regardless of injectivity, which means you can do various similar things also regardless of injectivity. What injectivity buys you is a lack of a singularity in $frac{dx}{du}$, which is a serious problem for, say, $u=x^2$ on $[a,b]$ with $a<0,b>0$.
– Ian
Dec 9 at 19:52
add a comment |
In general you need to use a one-to-one change of variable; thus for instance if you were to compute $int_{-1}^1 x^2 dx$ you would split into $int_{-1}^0 x^2 dx + int_0^1 x^2 dx$. Then each of those integrals, after substitution and simplification, becomes $int_0^1 frac{1}{2} u^{1/2} du$.
(That said, there are some situations where by basically breaking the rules twice you cancel out your mistakes and get the right answer; see Why can the trig sub $x=cosh(theta)$ be used to solve the integral: $int frac{1}{(x^2-1)^{3/2}}dx$? for instance.)
answered Dec 9 at 17:20
Ian
67.3k25386
It is not necessary for changes of variable to be one-to-one, see my comment here: math.stackexchange.com/a/1101421/5316
– Matthew Towers
Dec 9 at 19:39
1
@MatthewTowers Yes, it is true that $int_a^b f'(x) dx$ is always $f(b) - f(a)$, regardless of injectivity, which means you can do various similar things also regardless of injectivity. What injectivity buys you is a lack of a singularity in $frac{dx}{du}$, which is a serious problem for, say, $u=x^2$ on $[a,b]$ with $a<0,b>0$.
– Ian
Dec 9 at 19:52
add a comment |
In general you need to use a one-to-one change of variable; thus for instance if you were to compute $int_{-1}^1 x^2 dx$ you would split into $int_{-1}^0 x^2 dx + int_0^1 x^2 dx$. Then each of those integrals, after substitution and simplification, becomes $int_0^1 frac{1}{2} u^{1/2} du$.
(That said, there are some situations where by basically breaking the rules twice you cancel out your mistakes and get the right answer; see Why can the trig sub $x=cosh(theta)$ be used to solve the integral: $int frac{1}{(x^2-1)^{3/2}}dx$? for instance.)
answered Dec 9 at 17:20
Ian
67.3k25386
In general you need to use a one-to-one change of variable; thus for instance if you were to compute $int_{-1}^1 x^2 dx$ you would split into $int_{-1}^0 x^2 dx + int_0^1 x^2 dx$. Then each of those integrals, after substitution and simplification, becomes $int_0^1 frac{1}{2} u^{1/2} du$.
(That said, there are some situations where by basically breaking the rules twice you cancel out your mistakes and get the right answer; see Why can the trig sub $x=cosh(theta)$ be used to solve the integral: $int frac{1}{(x^2-1)^{3/2}}dx$? for instance.)
answered Dec 9 at 17:20
Ian
67.3k25386
answered Dec 9 at 17:20
Ian
67.3k25386
answered Dec 9 at 17:20
Ian
67.3k25386
answered Dec 9 at 17:20
Ian
67.3k25386
67.3k25386
It is not necessary for changes of variable to be one-to-one, see my comment here: math.stackexchange.com/a/1101421/5316
– Matthew Towers
Dec 9 at 19:39
1
@MatthewTowers Yes, it is true that $int_a^b f'(x) dx$ is always $f(b) - f(a)$, regardless of injectivity, which means you can do various similar things also regardless of injectivity. What injectivity buys you is a lack of a singularity in $frac{dx}{du}$, which is a serious problem for, say, $u=x^2$ on $[a,b]$ with $a<0,b>0$.
– Ian
Dec 9 at 19:52
add a comment |
It is not necessary for changes of variable to be one-to-one, see my comment here: math.stackexchange.com/a/1101421/5316
– Matthew Towers
Dec 9 at 19:39
1
@MatthewTowers Yes, it is true that $int_a^b f'(x) dx$ is always $f(b) - f(a)$, regardless of injectivity, which means you can do various similar things also regardless of injectivity. What injectivity buys you is a lack of a singularity in $frac{dx}{du}$, which is a serious problem for, say, $u=x^2$ on $[a,b]$ with $a<0,b>0$.
– Ian
Dec 9 at 19:52
It is not necessary for changes of variable to be one-to-one, see my comment here: math.stackexchange.com/a/1101421/5316
– Matthew Towers
Dec 9 at 19:39
It is not necessary for changes of variable to be one-to-one, see my comment here: math.stackexchange.com/a/1101421/5316
– Matthew Towers
Dec 9 at 19:39
1
1
@MatthewTowers Yes, it is true that $int_a^b f'(x) dx$ is always $f(b) - f(a)$, regardless of injectivity, which means you can do various similar things also regardless of injectivity. What injectivity buys you is a lack of a singularity in $frac{dx}{du}$, which is a serious problem for, say, $u=x^2$ on $[a,b]$ with $a<0,b>0$.
– Ian
Dec 9 at 19:52
@MatthewTowers Yes, it is true that $int_a^b f'(x) dx$ is always $f(b) - f(a)$, regardless of injectivity, which means you can do various similar things also regardless of injectivity. What injectivity buys you is a lack of a singularity in $frac{dx}{du}$, which is a serious problem for, say, $u=x^2$ on $[a,b]$ with $a<0,b>0$.
– Ian
Dec 9 at 19:52
add a comment |
While it's certainly true that $du = 2x,dx$ over the whole region of integration, it's not true that $x = sqrt{u}$ over the region. This is only true for $x > 0$, so you can't apply it to the whole integral. Instead, you need to split it apart and use the correct transformation $x = -sqrt{u}$ for the region $x < 0$.
In general, if your change of variables $u(x)$ has a multivalued inverse, you'll need to split the integral into the regions where each possible inverse function $x(u)$ applies.
answered Dec 9 at 17:33
eyeballfrog
6,058629
add a comment |
While it's certainly true that $du = 2x,dx$ over the whole region of integration, it's not true that $x = sqrt{u}$ over the region. This is only true for $x > 0$, so you can't apply it to the whole integral. Instead, you need to split it apart and use the correct transformation $x = -sqrt{u}$ for the region $x < 0$.
In general, if your change of variables $u(x)$ has a multivalued inverse, you'll need to split the integral into the regions where each possible inverse function $x(u)$ applies.
answered Dec 9 at 17:33
eyeballfrog
6,058629
add a comment |
While it's certainly true that $du = 2x,dx$ over the whole region of integration, it's not true that $x = sqrt{u}$ over the region. This is only true for $x > 0$, so you can't apply it to the whole integral. Instead, you need to split it apart and use the correct transformation $x = -sqrt{u}$ for the region $x < 0$.
In general, if your change of variables $u(x)$ has a multivalued inverse, you'll need to split the integral into the regions where each possible inverse function $x(u)$ applies.
answered Dec 9 at 17:33
eyeballfrog
6,058629
While it's certainly true that $du = 2x,dx$ over the whole region of integration, it's not true that $x = sqrt{u}$ over the region. This is only true for $x > 0$, so you can't apply it to the whole integral. Instead, you need to split it apart and use the correct transformation $x = -sqrt{u}$ for the region $x < 0$.
In general, if your change of variables $u(x)$ has a multivalued inverse, you'll need to split the integral into the regions where each possible inverse function $x(u)$ applies.
answered Dec 9 at 17:33
eyeballfrog
6,058629
edited Dec 10 at 4:54
answered Dec 9 at 17:33
eyeballfrog
6,058629
answered Dec 9 at 17:33
eyeballfrog
6,058629
answered Dec 9 at 17:33
eyeballfrog
6,058629
6,058629
add a comment |
add a comment |
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The derivative of $x^2$ isn't 2
– orange
Dec 9 at 19:06
@orange Thanks, corrected
– Winter
Dec 9 at 19:42
related: $u$-substitution always evaluates to $0$
– Dando18
Dec 10 at 2:45