Mixing
Understanding u-substitution theorem for integration when “u” is non-linear
$begingroup$
Integrating
$int frac{dx}{sqrt{2x-1}-sqrt[4]{2x-1}}$
by using the substitution $2x-1=u^4$ (non-linear u) yields
$int frac{2u^3}{u^2-u}du$.
How are these steps justified by the substitution theorem?
The substitution theorem states that if $u=varphi(x)$, $f:Itomathbb{R}$ continuous, $varphi(x)$ differentiable, then
$int f(varphi(x))varphi '(x)dx=int f(u)du$
Directly applying it to introductory problem would result in
$frac{1}{2}intfrac{d(u^4)}{u^2-u}$. I don't know how to proceed from here.
real-analysis calculus integration
edited Jan 22 at 22:54
Bernard
122k741116
asked Jan 22 at 22:50
Bruno RibarićBruno Ribarić
82
$endgroup$
add a comment |
$begingroup$
Integrating
$int frac{dx}{sqrt{2x-1}-sqrt[4]{2x-1}}$
by using the substitution $2x-1=u^4$ (non-linear u) yields
$int frac{2u^3}{u^2-u}du$.
How are these steps justified by the substitution theorem?
The substitution theorem states that if $u=varphi(x)$, $f:Itomathbb{R}$ continuous, $varphi(x)$ differentiable, then
$int f(varphi(x))varphi '(x)dx=int f(u)du$
Directly applying it to introductory problem would result in
$frac{1}{2}intfrac{d(u^4)}{u^2-u}$. I don't know how to proceed from here.
real-analysis calculus integration
edited Jan 22 at 22:54
Bernard
122k741116
asked Jan 22 at 22:50
Bruno RibarićBruno Ribarić
82
$endgroup$
$begingroup$
What is $d(u^4)$? Try to compute it
$endgroup$
– user289143
Jan 22 at 22:54
$begingroup$
Thank you for your input! I know of $d(u^4)$ only as the notation saying we're differentiating by the variable $u^4$. I don't know how to do any computations with it.
$endgroup$
– Bruno Ribarić
Jan 22 at 22:57
add a comment |
$begingroup$
Integrating
$int frac{dx}{sqrt{2x-1}-sqrt[4]{2x-1}}$
by using the substitution $2x-1=u^4$ (non-linear u) yields
$int frac{2u^3}{u^2-u}du$.
How are these steps justified by the substitution theorem?
The substitution theorem states that if $u=varphi(x)$, $f:Itomathbb{R}$ continuous, $varphi(x)$ differentiable, then
$int f(varphi(x))varphi '(x)dx=int f(u)du$
Directly applying it to introductory problem would result in
$frac{1}{2}intfrac{d(u^4)}{u^2-u}$. I don't know how to proceed from here.
real-analysis calculus integration
edited Jan 22 at 22:54
Bernard
122k741116
asked Jan 22 at 22:50
Bruno RibarićBruno Ribarić
82
$endgroup$
Integrating
$int frac{dx}{sqrt{2x-1}-sqrt[4]{2x-1}}$
by using the substitution $2x-1=u^4$ (non-linear u) yields
$int frac{2u^3}{u^2-u}du$.
How are these steps justified by the substitution theorem?
The substitution theorem states that if $u=varphi(x)$, $f:Itomathbb{R}$ continuous, $varphi(x)$ differentiable, then
$int f(varphi(x))varphi '(x)dx=int f(u)du$
Directly applying it to introductory problem would result in
$frac{1}{2}intfrac{d(u^4)}{u^2-u}$. I don't know how to proceed from here.
real-analysis calculus integration
real-analysis calculus integration
edited Jan 22 at 22:54
Bernard
122k741116
asked Jan 22 at 22:50
Bruno RibarićBruno Ribarić
82
edited Jan 22 at 22:54
Bernard
122k741116
asked Jan 22 at 22:50
Bruno RibarićBruno Ribarić
82
edited Jan 22 at 22:54
Bernard
122k741116
edited Jan 22 at 22:54
Bernard
122k741116
edited Jan 22 at 22:54
Bernard
122k741116
122k741116
asked Jan 22 at 22:50
Bruno RibarićBruno Ribarić
82
asked Jan 22 at 22:50
Bruno RibarićBruno Ribarić
82
asked Jan 22 at 22:50
Bruno RibarićBruno Ribarić
82
82
$begingroup$
What is $d(u^4)$? Try to compute it
$endgroup$
– user289143
Jan 22 at 22:54
$begingroup$
Thank you for your input! I know of $d(u^4)$ only as the notation saying we're differentiating by the variable $u^4$. I don't know how to do any computations with it.
$endgroup$
– Bruno Ribarić
Jan 22 at 22:57
add a comment |
$begingroup$
What is $d(u^4)$? Try to compute it
$endgroup$
– user289143
Jan 22 at 22:54
$begingroup$
Thank you for your input! I know of $d(u^4)$ only as the notation saying we're differentiating by the variable $u^4$. I don't know how to do any computations with it.
$endgroup$
– Bruno Ribarić
Jan 22 at 22:57
$begingroup$
What is $d(u^4)$? Try to compute it
$endgroup$
– user289143
Jan 22 at 22:54
$begingroup$
What is $d(u^4)$? Try to compute it
$endgroup$
– user289143
Jan 22 at 22:54
$begingroup$
Thank you for your input! I know of $d(u^4)$ only as the notation saying we're differentiating by the variable $u^4$. I don't know how to do any computations with it.
$endgroup$
– Bruno Ribarić
Jan 22 at 22:57
$begingroup$
Thank you for your input! I know of $d(u^4)$ only as the notation saying we're differentiating by the variable $u^4$. I don't know how to do any computations with it.
$endgroup$
– Bruno Ribarić
Jan 22 at 22:57
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Well, the rules of differential calculus assert that $;mathrm d(u^4)=4u^3,mathrm du$, and differentiating the relation $2x-1=u^4$ yields $;2mathrm d x=4u^3,mathrm d u$, so we have to replace in the integral $,mathrm dx$ with $2u^3,mathrm du$, and the integral becomes indeed
$$int frac{2u^3}{u^2-u},mathrm du=int frac{2u^2}{u-1},mathrm du.$$
Thus the integral has become the integral of a rational function. As the degree of the numerator is greater than the degree of the denominator, one has first to perform the euclidean division:
$$2u^2=(2u+2)(u-1)
+2implies frac{2u^2}{u-1}=2(u+1)+frac 2{u-1}.$$
Can you proceed?
answered Jan 22 at 23:07
BernardBernard
122k741116
$endgroup$
$begingroup$
Ah, I guess my question is then "why is $d(u^4)=4u^3du$ "which is a different topic. I know how to calculate the integral, but I was wondering about the fundamentals. I get that the derivative of $u^4$ is $4u^3$ but I am uncomfortable with treating differentiation "as a fraction" and dealing with $du$ as a single element, not as a part of something like $frac{du}{dx}$.
$endgroup$
– Bruno Ribarić
Jan 22 at 23:13
$begingroup$
By definition, the differential of a (single variable) function is the derivative of this function, times the differential of the variable, which is a symbol representing a small increment of the variable. Is it clear?
$endgroup$
– Bernard
Jan 22 at 23:17
$begingroup$
Yes thank you for your time.
$endgroup$
– Bruno Ribarić
Jan 22 at 23:28
add a comment |
$begingroup$
We have that $frac{d}{du}u^4=4u^3$. If we abuse notation somewhat, we get that $frac{d(u^4)}{du}=4u^3$, thus $d(u^4)=4u^3{du}$. Combining the $frac12$ in the front with the $4u^3$ give $2u^3$.
answered Jan 22 at 23:17
AcccumulationAcccumulation
7,1252619
$endgroup$
add a comment |
$begingroup$
$$int frac{1}{sqrt{2x-1}-sqrt[4]{2x-1}},dx$$
Letting $2x-1 = u^4$ means that $(2x-1)' = (u^4)' iff color{blue}{2,dx = 4u^3,du}$.
Hence you have that $$frac 12int frac {color{blue}{2dx}}{sqrt{2x-1}- sqrt[4]{2x-1}} = frac 12int frac {color{blue}{4u^3,du}}{u^2 - u} = 2 int frac{u^2}{u-1},du$$
$$= int left(2(u+1)+frac 2{u-1}right),du$$
$$= u^2 + 2u + 2ln|u-1| + C = u^2 + 2u+ ln(u-1)^2+C.$$
Now, remember that $u = sqrt[4]{2x-1}$.
answered Jan 22 at 23:05
jordan_glenjordan_glen
1
$endgroup$
$begingroup$
This does not show how the substitution theorem is applied to yield that result. I appreciate your input though.
$endgroup$
– Bruno Ribarić
Jan 22 at 23:08
$begingroup$
Yes, in fact, it does show that.
$endgroup$
– jordan_glen
Jan 22 at 23:14
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Well, the rules of differential calculus assert that $;mathrm d(u^4)=4u^3,mathrm du$, and differentiating the relation $2x-1=u^4$ yields $;2mathrm d x=4u^3,mathrm d u$, so we have to replace in the integral $,mathrm dx$ with $2u^3,mathrm du$, and the integral becomes indeed
$$int frac{2u^3}{u^2-u},mathrm du=int frac{2u^2}{u-1},mathrm du.$$
Thus the integral has become the integral of a rational function. As the degree of the numerator is greater than the degree of the denominator, one has first to perform the euclidean division:
$$2u^2=(2u+2)(u-1)
+2implies frac{2u^2}{u-1}=2(u+1)+frac 2{u-1}.$$
Can you proceed?
answered Jan 22 at 23:07
BernardBernard
122k741116
$endgroup$
$begingroup$
Ah, I guess my question is then "why is $d(u^4)=4u^3du$ "which is a different topic. I know how to calculate the integral, but I was wondering about the fundamentals. I get that the derivative of $u^4$ is $4u^3$ but I am uncomfortable with treating differentiation "as a fraction" and dealing with $du$ as a single element, not as a part of something like $frac{du}{dx}$.
$endgroup$
– Bruno Ribarić
Jan 22 at 23:13
$begingroup$
By definition, the differential of a (single variable) function is the derivative of this function, times the differential of the variable, which is a symbol representing a small increment of the variable. Is it clear?
$endgroup$
– Bernard
Jan 22 at 23:17
$begingroup$
Yes thank you for your time.
$endgroup$
– Bruno Ribarić
Jan 22 at 23:28
add a comment |
$begingroup$
Well, the rules of differential calculus assert that $;mathrm d(u^4)=4u^3,mathrm du$, and differentiating the relation $2x-1=u^4$ yields $;2mathrm d x=4u^3,mathrm d u$, so we have to replace in the integral $,mathrm dx$ with $2u^3,mathrm du$, and the integral becomes indeed
$$int frac{2u^3}{u^2-u},mathrm du=int frac{2u^2}{u-1},mathrm du.$$
Thus the integral has become the integral of a rational function. As the degree of the numerator is greater than the degree of the denominator, one has first to perform the euclidean division:
$$2u^2=(2u+2)(u-1)
+2implies frac{2u^2}{u-1}=2(u+1)+frac 2{u-1}.$$
Can you proceed?
answered Jan 22 at 23:07
BernardBernard
122k741116
$endgroup$
$begingroup$
Ah, I guess my question is then "why is $d(u^4)=4u^3du$ "which is a different topic. I know how to calculate the integral, but I was wondering about the fundamentals. I get that the derivative of $u^4$ is $4u^3$ but I am uncomfortable with treating differentiation "as a fraction" and dealing with $du$ as a single element, not as a part of something like $frac{du}{dx}$.
$endgroup$
– Bruno Ribarić
Jan 22 at 23:13
$begingroup$
By definition, the differential of a (single variable) function is the derivative of this function, times the differential of the variable, which is a symbol representing a small increment of the variable. Is it clear?
$endgroup$
– Bernard
Jan 22 at 23:17
$begingroup$
Yes thank you for your time.
$endgroup$
– Bruno Ribarić
Jan 22 at 23:28
add a comment |
$begingroup$
Well, the rules of differential calculus assert that $;mathrm d(u^4)=4u^3,mathrm du$, and differentiating the relation $2x-1=u^4$ yields $;2mathrm d x=4u^3,mathrm d u$, so we have to replace in the integral $,mathrm dx$ with $2u^3,mathrm du$, and the integral becomes indeed
$$int frac{2u^3}{u^2-u},mathrm du=int frac{2u^2}{u-1},mathrm du.$$
Thus the integral has become the integral of a rational function. As the degree of the numerator is greater than the degree of the denominator, one has first to perform the euclidean division:
$$2u^2=(2u+2)(u-1)
+2implies frac{2u^2}{u-1}=2(u+1)+frac 2{u-1}.$$
Can you proceed?
answered Jan 22 at 23:07
BernardBernard
122k741116
$endgroup$
Well, the rules of differential calculus assert that $;mathrm d(u^4)=4u^3,mathrm du$, and differentiating the relation $2x-1=u^4$ yields $;2mathrm d x=4u^3,mathrm d u$, so we have to replace in the integral $,mathrm dx$ with $2u^3,mathrm du$, and the integral becomes indeed
$$int frac{2u^3}{u^2-u},mathrm du=int frac{2u^2}{u-1},mathrm du.$$
Thus the integral has become the integral of a rational function. As the degree of the numerator is greater than the degree of the denominator, one has first to perform the euclidean division:
$$2u^2=(2u+2)(u-1)
+2implies frac{2u^2}{u-1}=2(u+1)+frac 2{u-1}.$$
Can you proceed?
answered Jan 22 at 23:07
BernardBernard
122k741116
answered Jan 22 at 23:07
BernardBernard
122k741116
answered Jan 22 at 23:07
BernardBernard
122k741116
answered Jan 22 at 23:07
BernardBernard
122k741116
122k741116
$begingroup$
Ah, I guess my question is then "why is $d(u^4)=4u^3du$ "which is a different topic. I know how to calculate the integral, but I was wondering about the fundamentals. I get that the derivative of $u^4$ is $4u^3$ but I am uncomfortable with treating differentiation "as a fraction" and dealing with $du$ as a single element, not as a part of something like $frac{du}{dx}$.
$endgroup$
– Bruno Ribarić
Jan 22 at 23:13
$begingroup$
By definition, the differential of a (single variable) function is the derivative of this function, times the differential of the variable, which is a symbol representing a small increment of the variable. Is it clear?
$endgroup$
– Bernard
Jan 22 at 23:17
$begingroup$
Yes thank you for your time.
$endgroup$
– Bruno Ribarić
Jan 22 at 23:28
add a comment |
$begingroup$
Ah, I guess my question is then "why is $d(u^4)=4u^3du$ "which is a different topic. I know how to calculate the integral, but I was wondering about the fundamentals. I get that the derivative of $u^4$ is $4u^3$ but I am uncomfortable with treating differentiation "as a fraction" and dealing with $du$ as a single element, not as a part of something like $frac{du}{dx}$.
$endgroup$
– Bruno Ribarić
Jan 22 at 23:13
$begingroup$
By definition, the differential of a (single variable) function is the derivative of this function, times the differential of the variable, which is a symbol representing a small increment of the variable. Is it clear?
$endgroup$
– Bernard
Jan 22 at 23:17
$begingroup$
Yes thank you for your time.
$endgroup$
– Bruno Ribarić
Jan 22 at 23:28
$begingroup$
Ah, I guess my question is then "why is $d(u^4)=4u^3du$ "which is a different topic. I know how to calculate the integral, but I was wondering about the fundamentals. I get that the derivative of $u^4$ is $4u^3$ but I am uncomfortable with treating differentiation "as a fraction" and dealing with $du$ as a single element, not as a part of something like $frac{du}{dx}$.
$endgroup$
– Bruno Ribarić
Jan 22 at 23:13
$begingroup$
Ah, I guess my question is then "why is $d(u^4)=4u^3du$ "which is a different topic. I know how to calculate the integral, but I was wondering about the fundamentals. I get that the derivative of $u^4$ is $4u^3$ but I am uncomfortable with treating differentiation "as a fraction" and dealing with $du$ as a single element, not as a part of something like $frac{du}{dx}$.
$endgroup$
– Bruno Ribarić
Jan 22 at 23:13
$begingroup$
By definition, the differential of a (single variable) function is the derivative of this function, times the differential of the variable, which is a symbol representing a small increment of the variable. Is it clear?
$endgroup$
– Bernard
Jan 22 at 23:17
$begingroup$
By definition, the differential of a (single variable) function is the derivative of this function, times the differential of the variable, which is a symbol representing a small increment of the variable. Is it clear?
$endgroup$
– Bernard
Jan 22 at 23:17
$begingroup$
Yes thank you for your time.
$endgroup$
– Bruno Ribarić
Jan 22 at 23:28
$begingroup$
Yes thank you for your time.
$endgroup$
– Bruno Ribarić
Jan 22 at 23:28
add a comment |
$begingroup$
We have that $frac{d}{du}u^4=4u^3$. If we abuse notation somewhat, we get that $frac{d(u^4)}{du}=4u^3$, thus $d(u^4)=4u^3{du}$. Combining the $frac12$ in the front with the $4u^3$ give $2u^3$.
answered Jan 22 at 23:17
AcccumulationAcccumulation
7,1252619
$endgroup$
add a comment |
$begingroup$
We have that $frac{d}{du}u^4=4u^3$. If we abuse notation somewhat, we get that $frac{d(u^4)}{du}=4u^3$, thus $d(u^4)=4u^3{du}$. Combining the $frac12$ in the front with the $4u^3$ give $2u^3$.
answered Jan 22 at 23:17
AcccumulationAcccumulation
7,1252619
$endgroup$
add a comment |
$begingroup$
We have that $frac{d}{du}u^4=4u^3$. If we abuse notation somewhat, we get that $frac{d(u^4)}{du}=4u^3$, thus $d(u^4)=4u^3{du}$. Combining the $frac12$ in the front with the $4u^3$ give $2u^3$.
answered Jan 22 at 23:17
AcccumulationAcccumulation
7,1252619
$endgroup$
We have that $frac{d}{du}u^4=4u^3$. If we abuse notation somewhat, we get that $frac{d(u^4)}{du}=4u^3$, thus $d(u^4)=4u^3{du}$. Combining the $frac12$ in the front with the $4u^3$ give $2u^3$.
answered Jan 22 at 23:17
AcccumulationAcccumulation
7,1252619
answered Jan 22 at 23:17
AcccumulationAcccumulation
7,1252619
answered Jan 22 at 23:17
AcccumulationAcccumulation
7,1252619
answered Jan 22 at 23:17
AcccumulationAcccumulation
7,1252619
7,1252619
add a comment |
add a comment |
$begingroup$
$$int frac{1}{sqrt{2x-1}-sqrt[4]{2x-1}},dx$$
Letting $2x-1 = u^4$ means that $(2x-1)' = (u^4)' iff color{blue}{2,dx = 4u^3,du}$.
Hence you have that $$frac 12int frac {color{blue}{2dx}}{sqrt{2x-1}- sqrt[4]{2x-1}} = frac 12int frac {color{blue}{4u^3,du}}{u^2 - u} = 2 int frac{u^2}{u-1},du$$
$$= int left(2(u+1)+frac 2{u-1}right),du$$
$$= u^2 + 2u + 2ln|u-1| + C = u^2 + 2u+ ln(u-1)^2+C.$$
Now, remember that $u = sqrt[4]{2x-1}$.
answered Jan 22 at 23:05
jordan_glenjordan_glen
1
$endgroup$
$begingroup$
This does not show how the substitution theorem is applied to yield that result. I appreciate your input though.
$endgroup$
– Bruno Ribarić
Jan 22 at 23:08
$begingroup$
Yes, in fact, it does show that.
$endgroup$
– jordan_glen
Jan 22 at 23:14
add a comment |
$begingroup$
$$int frac{1}{sqrt{2x-1}-sqrt[4]{2x-1}},dx$$
Letting $2x-1 = u^4$ means that $(2x-1)' = (u^4)' iff color{blue}{2,dx = 4u^3,du}$.
Hence you have that $$frac 12int frac {color{blue}{2dx}}{sqrt{2x-1}- sqrt[4]{2x-1}} = frac 12int frac {color{blue}{4u^3,du}}{u^2 - u} = 2 int frac{u^2}{u-1},du$$
$$= int left(2(u+1)+frac 2{u-1}right),du$$
$$= u^2 + 2u + 2ln|u-1| + C = u^2 + 2u+ ln(u-1)^2+C.$$
Now, remember that $u = sqrt[4]{2x-1}$.
answered Jan 22 at 23:05
jordan_glenjordan_glen
1
$endgroup$
$begingroup$
This does not show how the substitution theorem is applied to yield that result. I appreciate your input though.
$endgroup$
– Bruno Ribarić
Jan 22 at 23:08
$begingroup$
Yes, in fact, it does show that.
$endgroup$
– jordan_glen
Jan 22 at 23:14
add a comment |
$begingroup$
$$int frac{1}{sqrt{2x-1}-sqrt[4]{2x-1}},dx$$
Letting $2x-1 = u^4$ means that $(2x-1)' = (u^4)' iff color{blue}{2,dx = 4u^3,du}$.
Hence you have that $$frac 12int frac {color{blue}{2dx}}{sqrt{2x-1}- sqrt[4]{2x-1}} = frac 12int frac {color{blue}{4u^3,du}}{u^2 - u} = 2 int frac{u^2}{u-1},du$$
$$= int left(2(u+1)+frac 2{u-1}right),du$$
$$= u^2 + 2u + 2ln|u-1| + C = u^2 + 2u+ ln(u-1)^2+C.$$
Now, remember that $u = sqrt[4]{2x-1}$.
answered Jan 22 at 23:05
jordan_glenjordan_glen
1
$endgroup$
$$int frac{1}{sqrt{2x-1}-sqrt[4]{2x-1}},dx$$
Letting $2x-1 = u^4$ means that $(2x-1)' = (u^4)' iff color{blue}{2,dx = 4u^3,du}$.
Hence you have that $$frac 12int frac {color{blue}{2dx}}{sqrt{2x-1}- sqrt[4]{2x-1}} = frac 12int frac {color{blue}{4u^3,du}}{u^2 - u} = 2 int frac{u^2}{u-1},du$$
$$= int left(2(u+1)+frac 2{u-1}right),du$$
$$= u^2 + 2u + 2ln|u-1| + C = u^2 + 2u+ ln(u-1)^2+C.$$
Now, remember that $u = sqrt[4]{2x-1}$.
answered Jan 22 at 23:05
jordan_glenjordan_glen
1
edited Jan 22 at 23:22
answered Jan 22 at 23:05
jordan_glenjordan_glen
1
answered Jan 22 at 23:05
jordan_glenjordan_glen
1
answered Jan 22 at 23:05
jordan_glenjordan_glen
1
1
$begingroup$
This does not show how the substitution theorem is applied to yield that result. I appreciate your input though.
$endgroup$
– Bruno Ribarić
Jan 22 at 23:08
$begingroup$
Yes, in fact, it does show that.
$endgroup$
– jordan_glen
Jan 22 at 23:14
add a comment |
$begingroup$
This does not show how the substitution theorem is applied to yield that result. I appreciate your input though.
$endgroup$
– Bruno Ribarić
Jan 22 at 23:08
$begingroup$
Yes, in fact, it does show that.
$endgroup$
– jordan_glen
Jan 22 at 23:14
$begingroup$
This does not show how the substitution theorem is applied to yield that result. I appreciate your input though.
$endgroup$
– Bruno Ribarić
Jan 22 at 23:08
$begingroup$
This does not show how the substitution theorem is applied to yield that result. I appreciate your input though.
$endgroup$
– Bruno Ribarić
Jan 22 at 23:08
$begingroup$
Yes, in fact, it does show that.
$endgroup$
– jordan_glen
Jan 22 at 23:14
$begingroup$
Yes, in fact, it does show that.
$endgroup$
– jordan_glen
Jan 22 at 23:14
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$begingroup$
What is $d(u^4)$? Try to compute it
$endgroup$
– user289143
Jan 22 at 22:54
$begingroup$
Thank you for your input! I know of $d(u^4)$ only as the notation saying we're differentiating by the variable $u^4$. I don't know how to do any computations with it.
$endgroup$
– Bruno Ribarić
Jan 22 at 22:57