Mixing
Is this not a well-defined integral?
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
If we consider: $int_a^bxdx$, where $a=sinx$, $b=cosx$.
How can we evaluate this particular integral, if $a$ and $b$ are both functions of $x$, which is the variable with respect to which we are integrating?
calculus
asked 40 mins ago
Julia Kim
262
New contributor
Julia Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
3
down vote
favorite
If we consider: $int_a^bxdx$, where $a=sinx$, $b=cosx$.
How can we evaluate this particular integral, if $a$ and $b$ are both functions of $x$, which is the variable with respect to which we are integrating?
calculus
asked 40 mins ago
Julia Kim
262
New contributor
Julia Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
If we consider: $int_a^bxdx$, where $a=sinx$, $b=cosx$.
How can we evaluate this particular integral, if $a$ and $b$ are both functions of $x$, which is the variable with respect to which we are integrating?
calculus
asked 40 mins ago
Julia Kim
262
New contributor
Julia Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
If we consider: $int_a^bxdx$, where $a=sinx$, $b=cosx$.
How can we evaluate this particular integral, if $a$ and $b$ are both functions of $x$, which is the variable with respect to which we are integrating?
calculus
calculus
asked 40 mins ago
Julia Kim
262
New contributor
Julia Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 40 mins ago
Julia Kim
262
New contributor
Julia Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 40 mins ago
Julia Kim
262
New contributor
Julia Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 40 mins ago
Julia Kim
262
asked 40 mins ago
Julia Kim
262
262
New contributor
Julia Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Julia Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Julia Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
6
down vote
The variable inside the integral is a "dummy" in that it could be replaced by any other symbol. I think you could interpret $$int_sin x^cos x x , dx$$ as a sloppy way of writing, but having the same meaning as, $$int_sin x^cos x y , dy.$$
answered 38 mins ago
Umberto P.
36.5k12861
add a comment |Â
up vote
1
down vote
Here is another interpretation.
Formally if $bgeq a$ then $int_a^bxdx$ is a notation for $intmathbf1_(a,b](x)xdx$.
Applying that here leads to: $$int_sin x^cos xxdx=intmathbf1_(sin x,cos x](x)xdx$$
I do not dare to say that this is the correct way of interpreting, but it illustrates at least that your question is a good question.
Personally I go for the interpretation of Umberto.
answered 33 mins ago
drhab
92.1k542124
add a comment |Â
up vote
0
down vote
Alternatively to Umberto P. you could also view it as:
$int_A f(x) dx$ where $A$ is a set, for example an area or volume, that is further distorted by your function $f(x)$.
In your case you would simply plug in your constraints: $A = a in mathbbR $ and for $f(x)$ you have of course: $f(x) = x$
answered 30 mins ago
Imago
1,2211919
add a comment |Â
up vote
0
down vote
Doesn't matter. Just your final solution will also be a function of x.
Proceed the usual way: find the indefinite integral $$int xdx=fracx^22$$
put limits: $a=sin(x)$ and $b=cos(x)$
$$=fraccos^2x2-fracsin^2x2$$
$$=fraccos(2x)2$$
answered 35 mins ago
omega
1,3472918
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
The variable inside the integral is a "dummy" in that it could be replaced by any other symbol. I think you could interpret $$int_sin x^cos x x , dx$$ as a sloppy way of writing, but having the same meaning as, $$int_sin x^cos x y , dy.$$
answered 38 mins ago
Umberto P.
36.5k12861
add a comment |Â
up vote
6
down vote
The variable inside the integral is a "dummy" in that it could be replaced by any other symbol. I think you could interpret $$int_sin x^cos x x , dx$$ as a sloppy way of writing, but having the same meaning as, $$int_sin x^cos x y , dy.$$
answered 38 mins ago
Umberto P.
36.5k12861
add a comment |Â
up vote
6
down vote
up vote
6
down vote
The variable inside the integral is a "dummy" in that it could be replaced by any other symbol. I think you could interpret $$int_sin x^cos x x , dx$$ as a sloppy way of writing, but having the same meaning as, $$int_sin x^cos x y , dy.$$
answered 38 mins ago
Umberto P.
36.5k12861
The variable inside the integral is a "dummy" in that it could be replaced by any other symbol. I think you could interpret $$int_sin x^cos x x , dx$$ as a sloppy way of writing, but having the same meaning as, $$int_sin x^cos x y , dy.$$
answered 38 mins ago
Umberto P.
36.5k12861
answered 38 mins ago
Umberto P.
36.5k12861
answered 38 mins ago
Umberto P.
36.5k12861
answered 38 mins ago
Umberto P.
36.5k12861
36.5k12861
add a comment |Â
add a comment |Â
up vote
1
down vote
Here is another interpretation.
Formally if $bgeq a$ then $int_a^bxdx$ is a notation for $intmathbf1_(a,b](x)xdx$.
Applying that here leads to: $$int_sin x^cos xxdx=intmathbf1_(sin x,cos x](x)xdx$$
I do not dare to say that this is the correct way of interpreting, but it illustrates at least that your question is a good question.
Personally I go for the interpretation of Umberto.
answered 33 mins ago
drhab
92.1k542124
add a comment |Â
up vote
1
down vote
Here is another interpretation.
Formally if $bgeq a$ then $int_a^bxdx$ is a notation for $intmathbf1_(a,b](x)xdx$.
Applying that here leads to: $$int_sin x^cos xxdx=intmathbf1_(sin x,cos x](x)xdx$$
I do not dare to say that this is the correct way of interpreting, but it illustrates at least that your question is a good question.
Personally I go for the interpretation of Umberto.
answered 33 mins ago
drhab
92.1k542124
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Here is another interpretation.
Formally if $bgeq a$ then $int_a^bxdx$ is a notation for $intmathbf1_(a,b](x)xdx$.
Applying that here leads to: $$int_sin x^cos xxdx=intmathbf1_(sin x,cos x](x)xdx$$
I do not dare to say that this is the correct way of interpreting, but it illustrates at least that your question is a good question.
Personally I go for the interpretation of Umberto.
answered 33 mins ago
drhab
92.1k542124
Here is another interpretation.
Formally if $bgeq a$ then $int_a^bxdx$ is a notation for $intmathbf1_(a,b](x)xdx$.
Applying that here leads to: $$int_sin x^cos xxdx=intmathbf1_(sin x,cos x](x)xdx$$
I do not dare to say that this is the correct way of interpreting, but it illustrates at least that your question is a good question.
Personally I go for the interpretation of Umberto.
answered 33 mins ago
drhab
92.1k542124
edited 25 mins ago
answered 33 mins ago
drhab
92.1k542124
answered 33 mins ago
drhab
92.1k542124
answered 33 mins ago
drhab
92.1k542124
92.1k542124
add a comment |Â
add a comment |Â
up vote
0
down vote
Alternatively to Umberto P. you could also view it as:
$int_A f(x) dx$ where $A$ is a set, for example an area or volume, that is further distorted by your function $f(x)$.
In your case you would simply plug in your constraints: $A = a in mathbbR $ and for $f(x)$ you have of course: $f(x) = x$
answered 30 mins ago
Imago
1,2211919
add a comment |Â
up vote
0
down vote
Alternatively to Umberto P. you could also view it as:
$int_A f(x) dx$ where $A$ is a set, for example an area or volume, that is further distorted by your function $f(x)$.
In your case you would simply plug in your constraints: $A = a in mathbbR $ and for $f(x)$ you have of course: $f(x) = x$
answered 30 mins ago
Imago
1,2211919
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Alternatively to Umberto P. you could also view it as:
$int_A f(x) dx$ where $A$ is a set, for example an area or volume, that is further distorted by your function $f(x)$.
In your case you would simply plug in your constraints: $A = a in mathbbR $ and for $f(x)$ you have of course: $f(x) = x$
answered 30 mins ago
Imago
1,2211919
Alternatively to Umberto P. you could also view it as:
$int_A f(x) dx$ where $A$ is a set, for example an area or volume, that is further distorted by your function $f(x)$.
In your case you would simply plug in your constraints: $A = a in mathbbR $ and for $f(x)$ you have of course: $f(x) = x$
answered 30 mins ago
Imago
1,2211919
answered 30 mins ago
Imago
1,2211919
answered 30 mins ago
Imago
1,2211919
answered 30 mins ago
Imago
1,2211919
1,2211919
add a comment |Â
add a comment |Â
up vote
0
down vote
Doesn't matter. Just your final solution will also be a function of x.
Proceed the usual way: find the indefinite integral $$int xdx=fracx^22$$
put limits: $a=sin(x)$ and $b=cos(x)$
$$=fraccos^2x2-fracsin^2x2$$
$$=fraccos(2x)2$$
answered 35 mins ago
omega
1,3472918
add a comment |Â
up vote
0
down vote
Doesn't matter. Just your final solution will also be a function of x.
Proceed the usual way: find the indefinite integral $$int xdx=fracx^22$$
put limits: $a=sin(x)$ and $b=cos(x)$
$$=fraccos^2x2-fracsin^2x2$$
$$=fraccos(2x)2$$
answered 35 mins ago
omega
1,3472918
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Doesn't matter. Just your final solution will also be a function of x.
Proceed the usual way: find the indefinite integral $$int xdx=fracx^22$$
put limits: $a=sin(x)$ and $b=cos(x)$
$$=fraccos^2x2-fracsin^2x2$$
$$=fraccos(2x)2$$
answered 35 mins ago
omega
1,3472918
Doesn't matter. Just your final solution will also be a function of x.
Proceed the usual way: find the indefinite integral $$int xdx=fracx^22$$
put limits: $a=sin(x)$ and $b=cos(x)$
$$=fraccos^2x2-fracsin^2x2$$
$$=fraccos(2x)2$$
answered 35 mins ago
omega
1,3472918
edited 7 mins ago
answered 35 mins ago
omega
1,3472918
answered 35 mins ago
omega
1,3472918
answered 35 mins ago
omega
1,3472918
1,3472918
add a comment |Â
add a comment |Â
Julia Kim is a new contributor. Be nice, and check out our Code of Conduct.
Julia Kim is a new contributor. Be nice, and check out our Code of Conduct.
Julia Kim is a new contributor. Be nice, and check out our Code of Conduct.
Julia Kim is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2972491%2fis-this-not-a-well-defined-integral%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password