Substitution leads me to wrong result for the integral $int_0^xlambda e^-lambda xdx$
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I have a question that should have been trivial to solve, but for some reason I keep messing up. I need to calculate the following integral:
$$int_0^xlambda e^-lambda xdx$$
I initially tried to solve this with the substitution:
$$u=-lambda x$$
This leads to the following:
$$
beginaligned
x&=-fraculambda\
dx&=-fracdulambda
endaligned$$
My first (wrong) attempt:
$$
beginaligned
int_0^xlambda e^-lambda xdx &= lambdaint_0^frac-ulambda e^ufrac-dulambda\
& = -int_0^frac-ulambda e^udu = -[e^u]_0^frac-ulambda\
&= -[e^frac-ulambda-e^0] = 1 - e^frac-ulambda = 1 - e^x
endaligned
$$
Online tools have shown that the answer is supposed to be $1 - e^-lambda x$ however.
My second slightly modified attempt:
I reuse the steps from my first attempt all the way until I get:
$$-[e^u]_0^frac-ulambda$$
and now I replace the substitution first before continuing:
$$
beginaligned
-[e^u]_0^frac-ulambda &= -[e^-lambda x]_0^x\
& = -[e^-lambda x - e^0]\
& = 1 - e^-lambda x
endaligned
$$
What sorcery is this? I can't seem to recall a rule about substituting the variables back to the original before filling in the limits. Otherwise, what is the point of changing the limits along with the substitution? I must be missing something obvious here, but not obvious enough for me to notice.
Without substitution works without a problem:
$$
beginaligned
int_0^xlambda e^-lambda xdx &= -[e^-lambda x]_0^xdx\
& = -[e^-lambda x - 1]\
&= 1 - e^-lambda x
endaligned
$$
Why was my first attempt wrong?
integration definite-integrals substitution
edited Aug 11 at 16:50
AccidentalFourierTransform
1,325627
asked Aug 11 at 11:05
Babyburger
1206
 |Â
show 2 more comments
up vote
3
down vote
favorite
I have a question that should have been trivial to solve, but for some reason I keep messing up. I need to calculate the following integral:
$$int_0^xlambda e^-lambda xdx$$
I initially tried to solve this with the substitution:
$$u=-lambda x$$
This leads to the following:
$$
beginaligned
x&=-fraculambda\
dx&=-fracdulambda
endaligned$$
My first (wrong) attempt:
$$
beginaligned
int_0^xlambda e^-lambda xdx &= lambdaint_0^frac-ulambda e^ufrac-dulambda\
& = -int_0^frac-ulambda e^udu = -[e^u]_0^frac-ulambda\
&= -[e^frac-ulambda-e^0] = 1 - e^frac-ulambda = 1 - e^x
endaligned
$$
Online tools have shown that the answer is supposed to be $1 - e^-lambda x$ however.
My second slightly modified attempt:
I reuse the steps from my first attempt all the way until I get:
$$-[e^u]_0^frac-ulambda$$
and now I replace the substitution first before continuing:
$$
beginaligned
-[e^u]_0^frac-ulambda &= -[e^-lambda x]_0^x\
& = -[e^-lambda x - e^0]\
& = 1 - e^-lambda x
endaligned
$$
What sorcery is this? I can't seem to recall a rule about substituting the variables back to the original before filling in the limits. Otherwise, what is the point of changing the limits along with the substitution? I must be missing something obvious here, but not obvious enough for me to notice.
Without substitution works without a problem:
$$
beginaligned
int_0^xlambda e^-lambda xdx &= -[e^-lambda x]_0^xdx\
& = -[e^-lambda x - 1]\
&= 1 - e^-lambda x
endaligned
$$
Why was my first attempt wrong?
integration definite-integrals substitution
edited Aug 11 at 16:50
AccidentalFourierTransform
1,325627
asked Aug 11 at 11:05
Babyburger
1206
1
In cases like this, where you only use the substituion for the part of evaluating the indefinite integral, I would suggest to not change the borders of integration. Just leave them as they are and plug in the values into the resubstituted anti-derivative in the end. To change the limits is only a crucial point when you are working with the integral itself without computing the value.
– mrtaurho
Aug 11 at 11:11
It is not defined. You can not have $dx$ inside if you have $x$ in one of the integral limits.
– mathreadler
Aug 11 at 11:13
@mrtaurho. But $int_a^b f(x) , dx = x = ky = int_a^b f(ky) , k , dy$ is certainly wrong.
– md2perpe
Aug 11 at 11:14
1
+1 with @mathreadler : $(e^-lambda x)'=-lambda e^-lambda x$, so $int_0^X lambda e^-lambda xrm dx = left[-e^-lambda xright]_0^X = 1-e^-lambda X$.
– Nicolas FRANCOIS
Aug 11 at 11:15
1
It is not a good idea to use the symbol for the integration parameter in the integration boundaries.
– random
Aug 11 at 11:16
 |Â
show 2 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I have a question that should have been trivial to solve, but for some reason I keep messing up. I need to calculate the following integral:
$$int_0^xlambda e^-lambda xdx$$
I initially tried to solve this with the substitution:
$$u=-lambda x$$
This leads to the following:
$$
beginaligned
x&=-fraculambda\
dx&=-fracdulambda
endaligned$$
My first (wrong) attempt:
$$
beginaligned
int_0^xlambda e^-lambda xdx &= lambdaint_0^frac-ulambda e^ufrac-dulambda\
& = -int_0^frac-ulambda e^udu = -[e^u]_0^frac-ulambda\
&= -[e^frac-ulambda-e^0] = 1 - e^frac-ulambda = 1 - e^x
endaligned
$$
Online tools have shown that the answer is supposed to be $1 - e^-lambda x$ however.
My second slightly modified attempt:
I reuse the steps from my first attempt all the way until I get:
$$-[e^u]_0^frac-ulambda$$
and now I replace the substitution first before continuing:
$$
beginaligned
-[e^u]_0^frac-ulambda &= -[e^-lambda x]_0^x\
& = -[e^-lambda x - e^0]\
& = 1 - e^-lambda x
endaligned
$$
What sorcery is this? I can't seem to recall a rule about substituting the variables back to the original before filling in the limits. Otherwise, what is the point of changing the limits along with the substitution? I must be missing something obvious here, but not obvious enough for me to notice.
Without substitution works without a problem:
$$
beginaligned
int_0^xlambda e^-lambda xdx &= -[e^-lambda x]_0^xdx\
& = -[e^-lambda x - 1]\
&= 1 - e^-lambda x
endaligned
$$
Why was my first attempt wrong?
integration definite-integrals substitution
edited Aug 11 at 16:50
AccidentalFourierTransform
1,325627
asked Aug 11 at 11:05
Babyburger
1206
I have a question that should have been trivial to solve, but for some reason I keep messing up. I need to calculate the following integral:
$$int_0^xlambda e^-lambda xdx$$
I initially tried to solve this with the substitution:
$$u=-lambda x$$
This leads to the following:
$$
beginaligned
x&=-fraculambda\
dx&=-fracdulambda
endaligned$$
My first (wrong) attempt:
$$
beginaligned
int_0^xlambda e^-lambda xdx &= lambdaint_0^frac-ulambda e^ufrac-dulambda\
& = -int_0^frac-ulambda e^udu = -[e^u]_0^frac-ulambda\
&= -[e^frac-ulambda-e^0] = 1 - e^frac-ulambda = 1 - e^x
endaligned
$$
Online tools have shown that the answer is supposed to be $1 - e^-lambda x$ however.
My second slightly modified attempt:
I reuse the steps from my first attempt all the way until I get:
$$-[e^u]_0^frac-ulambda$$
and now I replace the substitution first before continuing:
$$
beginaligned
-[e^u]_0^frac-ulambda &= -[e^-lambda x]_0^x\
& = -[e^-lambda x - e^0]\
& = 1 - e^-lambda x
endaligned
$$
What sorcery is this? I can't seem to recall a rule about substituting the variables back to the original before filling in the limits. Otherwise, what is the point of changing the limits along with the substitution? I must be missing something obvious here, but not obvious enough for me to notice.
Without substitution works without a problem:
$$
beginaligned
int_0^xlambda e^-lambda xdx &= -[e^-lambda x]_0^xdx\
& = -[e^-lambda x - 1]\
&= 1 - e^-lambda x
endaligned
$$
Why was my first attempt wrong?
integration definite-integrals substitution
integration definite-integrals substitution
edited Aug 11 at 16:50
AccidentalFourierTransform
1,325627
asked Aug 11 at 11:05
Babyburger
1206
edited Aug 11 at 16:50
AccidentalFourierTransform
1,325627
asked Aug 11 at 11:05
Babyburger
1206
edited Aug 11 at 16:50
AccidentalFourierTransform
1,325627
edited Aug 11 at 16:50
AccidentalFourierTransform
1,325627
edited Aug 11 at 16:50
AccidentalFourierTransform
1,325627
1,325627
asked Aug 11 at 11:05
Babyburger
1206
asked Aug 11 at 11:05
Babyburger
1206
asked Aug 11 at 11:05
Babyburger
1206
1206
1
In cases like this, where you only use the substituion for the part of evaluating the indefinite integral, I would suggest to not change the borders of integration. Just leave them as they are and plug in the values into the resubstituted anti-derivative in the end. To change the limits is only a crucial point when you are working with the integral itself without computing the value.
– mrtaurho
Aug 11 at 11:11
It is not defined. You can not have $dx$ inside if you have $x$ in one of the integral limits.
– mathreadler
Aug 11 at 11:13
@mrtaurho. But $int_a^b f(x) , dx = x = ky = int_a^b f(ky) , k , dy$ is certainly wrong.
– md2perpe
Aug 11 at 11:14
1
+1 with @mathreadler : $(e^-lambda x)'=-lambda e^-lambda x$, so $int_0^X lambda e^-lambda xrm dx = left[-e^-lambda xright]_0^X = 1-e^-lambda X$.
– Nicolas FRANCOIS
Aug 11 at 11:15
1
It is not a good idea to use the symbol for the integration parameter in the integration boundaries.
– random
Aug 11 at 11:16
 |Â
show 2 more comments
1
In cases like this, where you only use the substituion for the part of evaluating the indefinite integral, I would suggest to not change the borders of integration. Just leave them as they are and plug in the values into the resubstituted anti-derivative in the end. To change the limits is only a crucial point when you are working with the integral itself without computing the value.
– mrtaurho
Aug 11 at 11:11
It is not defined. You can not have $dx$ inside if you have $x$ in one of the integral limits.
– mathreadler
Aug 11 at 11:13
@mrtaurho. But $int_a^b f(x) , dx = x = ky = int_a^b f(ky) , k , dy$ is certainly wrong.
– md2perpe
Aug 11 at 11:14
1
+1 with @mathreadler : $(e^-lambda x)'=-lambda e^-lambda x$, so $int_0^X lambda e^-lambda xrm dx = left[-e^-lambda xright]_0^X = 1-e^-lambda X$.
– Nicolas FRANCOIS
Aug 11 at 11:15
1
It is not a good idea to use the symbol for the integration parameter in the integration boundaries.
– random
Aug 11 at 11:16
1
1
In cases like this, where you only use the substituion for the part of evaluating the indefinite integral, I would suggest to not change the borders of integration. Just leave them as they are and plug in the values into the resubstituted anti-derivative in the end. To change the limits is only a crucial point when you are working with the integral itself without computing the value.
– mrtaurho
Aug 11 at 11:11
In cases like this, where you only use the substituion for the part of evaluating the indefinite integral, I would suggest to not change the borders of integration. Just leave them as they are and plug in the values into the resubstituted anti-derivative in the end. To change the limits is only a crucial point when you are working with the integral itself without computing the value.
– mrtaurho
Aug 11 at 11:11
It is not defined. You can not have $dx$ inside if you have $x$ in one of the integral limits.
– mathreadler
Aug 11 at 11:13
It is not defined. You can not have $dx$ inside if you have $x$ in one of the integral limits.
– mathreadler
Aug 11 at 11:13
@mrtaurho. But $int_a^b f(x) , dx = x = ky = int_a^b f(ky) , k , dy$ is certainly wrong.
– md2perpe
Aug 11 at 11:14
@mrtaurho. But $int_a^b f(x) , dx = x = ky = int_a^b f(ky) , k , dy$ is certainly wrong.
– md2perpe
Aug 11 at 11:14
1
1
+1 with @mathreadler : $(e^-lambda x)'=-lambda e^-lambda x$, so $int_0^X lambda e^-lambda xrm dx = left[-e^-lambda xright]_0^X = 1-e^-lambda X$.
– Nicolas FRANCOIS
Aug 11 at 11:15
+1 with @mathreadler : $(e^-lambda x)'=-lambda e^-lambda x$, so $int_0^X lambda e^-lambda xrm dx = left[-e^-lambda xright]_0^X = 1-e^-lambda X$.
– Nicolas FRANCOIS
Aug 11 at 11:15
1
1
It is not a good idea to use the symbol for the integration parameter in the integration boundaries.
– random
Aug 11 at 11:16
It is not a good idea to use the symbol for the integration parameter in the integration boundaries.
– random
Aug 11 at 11:16
 |Â
show 2 more comments
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
In your wrong attempt you wrote:
$$int_0^xlambda e^-lambda x mathrm dx = lambdaint_0^colorredfrac-ulambda e^ufrac-mathrm dulambda$$
when in fact you should have written:
$$int_0^xlambda e^-lambda x mathrm dx = lambdaint_0^colorredu e^ufrac-mathrm dulambda$$
When changing the bounds, ask yourself "when $x$ = lower bound, what is $u$? when $x$ = upper bound, what is $u$?"
The answer is "when $x=0$, $u=0$; when $x=x$, $u=u$".
answered Aug 11 at 11:15
Kenny Lau
19k2157
But it is still wrong (and confusing, see the OP) to write $int_0^x f(x)rm dx$...
– Nicolas FRANCOIS
Aug 11 at 11:17
Well $displaystyle int_0^x f(x) mathrm dx$ isn't necessarily wrong.
– Kenny Lau
Aug 11 at 11:27
For the sanity of students, it SHOULD be :-)
– Nicolas FRANCOIS
Aug 11 at 11:29
@KennyLau yes it is wrong. It makes no sense whatsoever.
– mathreadler
Aug 11 at 14:52
Ok, for some reason that took a while to sink in and now I feel stupid because how obvious it is (as I expected it would be). I shouldn't have written the upper bound in terms of x=..., but keep it in the form of u=.... Thanks a lot!
– Babyburger
Aug 11 at 16:49
add a comment |Â
up vote
2
down vote
One mistake at the start with the substitution $u=-lambda x$ you should get $$dx = -fracdulambda $$
but for some reason you get
$$dx = -fracdux$$
If you keep going from that spot (and change integral limit to for example $X$ as proposed by Nicolas Francois) you should be able to solve it.
answered Aug 11 at 11:19
mathreadler
13.9k72058
1
Thanks for pointing that out! It was supposed to be lambda, but I made a typing mistake with my OP. I edited it.
– Babyburger
Aug 11 at 16:29
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
In your wrong attempt you wrote:
$$int_0^xlambda e^-lambda x mathrm dx = lambdaint_0^colorredfrac-ulambda e^ufrac-mathrm dulambda$$
when in fact you should have written:
$$int_0^xlambda e^-lambda x mathrm dx = lambdaint_0^colorredu e^ufrac-mathrm dulambda$$
When changing the bounds, ask yourself "when $x$ = lower bound, what is $u$? when $x$ = upper bound, what is $u$?"
The answer is "when $x=0$, $u=0$; when $x=x$, $u=u$".
answered Aug 11 at 11:15
Kenny Lau
19k2157
But it is still wrong (and confusing, see the OP) to write $int_0^x f(x)rm dx$...
– Nicolas FRANCOIS
Aug 11 at 11:17
Well $displaystyle int_0^x f(x) mathrm dx$ isn't necessarily wrong.
– Kenny Lau
Aug 11 at 11:27
For the sanity of students, it SHOULD be :-)
– Nicolas FRANCOIS
Aug 11 at 11:29
@KennyLau yes it is wrong. It makes no sense whatsoever.
– mathreadler
Aug 11 at 14:52
Ok, for some reason that took a while to sink in and now I feel stupid because how obvious it is (as I expected it would be). I shouldn't have written the upper bound in terms of x=..., but keep it in the form of u=.... Thanks a lot!
– Babyburger
Aug 11 at 16:49
add a comment |Â
up vote
4
down vote
accepted
In your wrong attempt you wrote:
$$int_0^xlambda e^-lambda x mathrm dx = lambdaint_0^colorredfrac-ulambda e^ufrac-mathrm dulambda$$
when in fact you should have written:
$$int_0^xlambda e^-lambda x mathrm dx = lambdaint_0^colorredu e^ufrac-mathrm dulambda$$
When changing the bounds, ask yourself "when $x$ = lower bound, what is $u$? when $x$ = upper bound, what is $u$?"
The answer is "when $x=0$, $u=0$; when $x=x$, $u=u$".
answered Aug 11 at 11:15
Kenny Lau
19k2157
But it is still wrong (and confusing, see the OP) to write $int_0^x f(x)rm dx$...
– Nicolas FRANCOIS
Aug 11 at 11:17
Well $displaystyle int_0^x f(x) mathrm dx$ isn't necessarily wrong.
– Kenny Lau
Aug 11 at 11:27
For the sanity of students, it SHOULD be :-)
– Nicolas FRANCOIS
Aug 11 at 11:29
@KennyLau yes it is wrong. It makes no sense whatsoever.
– mathreadler
Aug 11 at 14:52
Ok, for some reason that took a while to sink in and now I feel stupid because how obvious it is (as I expected it would be). I shouldn't have written the upper bound in terms of x=..., but keep it in the form of u=.... Thanks a lot!
– Babyburger
Aug 11 at 16:49
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
In your wrong attempt you wrote:
$$int_0^xlambda e^-lambda x mathrm dx = lambdaint_0^colorredfrac-ulambda e^ufrac-mathrm dulambda$$
when in fact you should have written:
$$int_0^xlambda e^-lambda x mathrm dx = lambdaint_0^colorredu e^ufrac-mathrm dulambda$$
When changing the bounds, ask yourself "when $x$ = lower bound, what is $u$? when $x$ = upper bound, what is $u$?"
The answer is "when $x=0$, $u=0$; when $x=x$, $u=u$".
answered Aug 11 at 11:15
Kenny Lau
19k2157
In your wrong attempt you wrote:
$$int_0^xlambda e^-lambda x mathrm dx = lambdaint_0^colorredfrac-ulambda e^ufrac-mathrm dulambda$$
when in fact you should have written:
$$int_0^xlambda e^-lambda x mathrm dx = lambdaint_0^colorredu e^ufrac-mathrm dulambda$$
When changing the bounds, ask yourself "when $x$ = lower bound, what is $u$? when $x$ = upper bound, what is $u$?"
The answer is "when $x=0$, $u=0$; when $x=x$, $u=u$".
answered Aug 11 at 11:15
Kenny Lau
19k2157
answered Aug 11 at 11:15
Kenny Lau
19k2157
answered Aug 11 at 11:15
Kenny Lau
19k2157
answered Aug 11 at 11:15
Kenny Lau
19k2157
19k2157
But it is still wrong (and confusing, see the OP) to write $int_0^x f(x)rm dx$...
– Nicolas FRANCOIS
Aug 11 at 11:17
Well $displaystyle int_0^x f(x) mathrm dx$ isn't necessarily wrong.
– Kenny Lau
Aug 11 at 11:27
For the sanity of students, it SHOULD be :-)
– Nicolas FRANCOIS
Aug 11 at 11:29
@KennyLau yes it is wrong. It makes no sense whatsoever.
– mathreadler
Aug 11 at 14:52
Ok, for some reason that took a while to sink in and now I feel stupid because how obvious it is (as I expected it would be). I shouldn't have written the upper bound in terms of x=..., but keep it in the form of u=.... Thanks a lot!
– Babyburger
Aug 11 at 16:49
add a comment |Â
But it is still wrong (and confusing, see the OP) to write $int_0^x f(x)rm dx$...
– Nicolas FRANCOIS
Aug 11 at 11:17
Well $displaystyle int_0^x f(x) mathrm dx$ isn't necessarily wrong.
– Kenny Lau
Aug 11 at 11:27
For the sanity of students, it SHOULD be :-)
– Nicolas FRANCOIS
Aug 11 at 11:29
@KennyLau yes it is wrong. It makes no sense whatsoever.
– mathreadler
Aug 11 at 14:52
Ok, for some reason that took a while to sink in and now I feel stupid because how obvious it is (as I expected it would be). I shouldn't have written the upper bound in terms of x=..., but keep it in the form of u=.... Thanks a lot!
– Babyburger
Aug 11 at 16:49
But it is still wrong (and confusing, see the OP) to write $int_0^x f(x)rm dx$...
– Nicolas FRANCOIS
Aug 11 at 11:17
But it is still wrong (and confusing, see the OP) to write $int_0^x f(x)rm dx$...
– Nicolas FRANCOIS
Aug 11 at 11:17
Well $displaystyle int_0^x f(x) mathrm dx$ isn't necessarily wrong.
– Kenny Lau
Aug 11 at 11:27
Well $displaystyle int_0^x f(x) mathrm dx$ isn't necessarily wrong.
– Kenny Lau
Aug 11 at 11:27
For the sanity of students, it SHOULD be :-)
– Nicolas FRANCOIS
Aug 11 at 11:29
For the sanity of students, it SHOULD be :-)
– Nicolas FRANCOIS
Aug 11 at 11:29
@KennyLau yes it is wrong. It makes no sense whatsoever.
– mathreadler
Aug 11 at 14:52
@KennyLau yes it is wrong. It makes no sense whatsoever.
– mathreadler
Aug 11 at 14:52
Ok, for some reason that took a while to sink in and now I feel stupid because how obvious it is (as I expected it would be). I shouldn't have written the upper bound in terms of x=..., but keep it in the form of u=.... Thanks a lot!
– Babyburger
Aug 11 at 16:49
Ok, for some reason that took a while to sink in and now I feel stupid because how obvious it is (as I expected it would be). I shouldn't have written the upper bound in terms of x=..., but keep it in the form of u=.... Thanks a lot!
– Babyburger
Aug 11 at 16:49
add a comment |Â
up vote
2
down vote
One mistake at the start with the substitution $u=-lambda x$ you should get $$dx = -fracdulambda $$
but for some reason you get
$$dx = -fracdux$$
If you keep going from that spot (and change integral limit to for example $X$ as proposed by Nicolas Francois) you should be able to solve it.
answered Aug 11 at 11:19
mathreadler
13.9k72058
1
Thanks for pointing that out! It was supposed to be lambda, but I made a typing mistake with my OP. I edited it.
– Babyburger
Aug 11 at 16:29
add a comment |Â
up vote
2
down vote
One mistake at the start with the substitution $u=-lambda x$ you should get $$dx = -fracdulambda $$
but for some reason you get
$$dx = -fracdux$$
If you keep going from that spot (and change integral limit to for example $X$ as proposed by Nicolas Francois) you should be able to solve it.
answered Aug 11 at 11:19
mathreadler
13.9k72058
1
Thanks for pointing that out! It was supposed to be lambda, but I made a typing mistake with my OP. I edited it.
– Babyburger
Aug 11 at 16:29
add a comment |Â
up vote
2
down vote
up vote
2
down vote
One mistake at the start with the substitution $u=-lambda x$ you should get $$dx = -fracdulambda $$
but for some reason you get
$$dx = -fracdux$$
If you keep going from that spot (and change integral limit to for example $X$ as proposed by Nicolas Francois) you should be able to solve it.
answered Aug 11 at 11:19
mathreadler
13.9k72058
One mistake at the start with the substitution $u=-lambda x$ you should get $$dx = -fracdulambda $$
but for some reason you get
$$dx = -fracdux$$
If you keep going from that spot (and change integral limit to for example $X$ as proposed by Nicolas Francois) you should be able to solve it.
answered Aug 11 at 11:19
mathreadler
13.9k72058
answered Aug 11 at 11:19
mathreadler
13.9k72058
answered Aug 11 at 11:19
mathreadler
13.9k72058
answered Aug 11 at 11:19
mathreadler
13.9k72058
13.9k72058
1
Thanks for pointing that out! It was supposed to be lambda, but I made a typing mistake with my OP. I edited it.
– Babyburger
Aug 11 at 16:29
add a comment |Â
1
Thanks for pointing that out! It was supposed to be lambda, but I made a typing mistake with my OP. I edited it.
– Babyburger
Aug 11 at 16:29
1
1
Thanks for pointing that out! It was supposed to be lambda, but I made a typing mistake with my OP. I edited it.
– Babyburger
Aug 11 at 16:29
Thanks for pointing that out! It was supposed to be lambda, but I made a typing mistake with my OP. I edited it.
– Babyburger
Aug 11 at 16:29
add a comment |Â
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1
In cases like this, where you only use the substituion for the part of evaluating the indefinite integral, I would suggest to not change the borders of integration. Just leave them as they are and plug in the values into the resubstituted anti-derivative in the end. To change the limits is only a crucial point when you are working with the integral itself without computing the value.
– mrtaurho
Aug 11 at 11:11
It is not defined. You can not have $dx$ inside if you have $x$ in one of the integral limits.
– mathreadler
Aug 11 at 11:13
@mrtaurho. But $int_a^b f(x) , dx = x = ky = int_a^b f(ky) , k , dy$ is certainly wrong.
– md2perpe
Aug 11 at 11:14
1
+1 with @mathreadler : $(e^-lambda x)'=-lambda e^-lambda x$, so $int_0^X lambda e^-lambda xrm dx = left[-e^-lambda xright]_0^X = 1-e^-lambda X$.
– Nicolas FRANCOIS
Aug 11 at 11:15
1
It is not a good idea to use the symbol for the integration parameter in the integration boundaries.
– random
Aug 11 at 11:16