Limits with definite integration
Clash Royale CLAN TAG#URR8PPP
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$Evaluate:$
$$lim_xrightarrow infty fracx int_0^x e^x^2 dx e^x^2$$
My attempt
First I dealt with numerator with integral :
As $$int_b^a f(x) dx= int_b^a f(a+b-x),dx$$
To get it as $$int_0^x 1cdot dx =x$$
So the limit turned to be
$$lim_nrightarrow infty frac x^2 e^x^2=0$$
Is it true?
limits definite-integrals
edited Sep 10 at 18:56
Bernard
112k636104
asked Sep 10 at 18:21
jayant98
709
 |Â
show 3 more comments
up vote
2
down vote
favorite
$Evaluate:$
$$lim_xrightarrow infty fracx int_0^x e^x^2 dx e^x^2$$
My attempt
First I dealt with numerator with integral :
As $$int_b^a f(x) dx= int_b^a f(a+b-x),dx$$
To get it as $$int_0^x 1cdot dx =x$$
So the limit turned to be
$$lim_nrightarrow infty frac x^2 e^x^2=0$$
Is it true?
limits definite-integrals
edited Sep 10 at 18:56
Bernard
112k636104
asked Sep 10 at 18:21
jayant98
709
Please get your formatting at least somewhat right before you post the question.
– Sobi
Sep 10 at 18:23
Sorry for that. But I am just learning about it recently. So, I may mistake but I am trying it to do it correct.
– jayant98
Sep 10 at 18:24
Note that the integral in the numerator does not makes sense since your interval of integration depends on the variable you are integrating with respect to. Do you perhaps want $$ int_0^x e^t^2, dt $$ in the numerator?
– Sobi
Sep 10 at 18:30
Okay. But in my text book it is given as I have typed. I can give you the picture of it if you want to see.
– jayant98
Sep 10 at 18:32
It is likely a typo on their side then. In any case, the computation of your integral is wrong. Do you know L'Hospital's rule?
– Sobi
Sep 10 at 18:35
 |Â
show 3 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
$Evaluate:$
$$lim_xrightarrow infty fracx int_0^x e^x^2 dx e^x^2$$
My attempt
First I dealt with numerator with integral :
As $$int_b^a f(x) dx= int_b^a f(a+b-x),dx$$
To get it as $$int_0^x 1cdot dx =x$$
So the limit turned to be
$$lim_nrightarrow infty frac x^2 e^x^2=0$$
Is it true?
limits definite-integrals
edited Sep 10 at 18:56
Bernard
112k636104
asked Sep 10 at 18:21
jayant98
709
$Evaluate:$
$$lim_xrightarrow infty fracx int_0^x e^x^2 dx e^x^2$$
My attempt
First I dealt with numerator with integral :
As $$int_b^a f(x) dx= int_b^a f(a+b-x),dx$$
To get it as $$int_0^x 1cdot dx =x$$
So the limit turned to be
$$lim_nrightarrow infty frac x^2 e^x^2=0$$
Is it true?
limits definite-integrals
limits definite-integrals
edited Sep 10 at 18:56
Bernard
112k636104
asked Sep 10 at 18:21
jayant98
709
edited Sep 10 at 18:56
Bernard
112k636104
asked Sep 10 at 18:21
jayant98
709
edited Sep 10 at 18:56
Bernard
112k636104
edited Sep 10 at 18:56
Bernard
112k636104
edited Sep 10 at 18:56
Bernard
112k636104
112k636104
asked Sep 10 at 18:21
jayant98
709
asked Sep 10 at 18:21
jayant98
709
asked Sep 10 at 18:21
jayant98
709
709
Please get your formatting at least somewhat right before you post the question.
– Sobi
Sep 10 at 18:23
Sorry for that. But I am just learning about it recently. So, I may mistake but I am trying it to do it correct.
– jayant98
Sep 10 at 18:24
Note that the integral in the numerator does not makes sense since your interval of integration depends on the variable you are integrating with respect to. Do you perhaps want $$ int_0^x e^t^2, dt $$ in the numerator?
– Sobi
Sep 10 at 18:30
Okay. But in my text book it is given as I have typed. I can give you the picture of it if you want to see.
– jayant98
Sep 10 at 18:32
It is likely a typo on their side then. In any case, the computation of your integral is wrong. Do you know L'Hospital's rule?
– Sobi
Sep 10 at 18:35
 |Â
show 3 more comments
Please get your formatting at least somewhat right before you post the question.
– Sobi
Sep 10 at 18:23
Sorry for that. But I am just learning about it recently. So, I may mistake but I am trying it to do it correct.
– jayant98
Sep 10 at 18:24
Note that the integral in the numerator does not makes sense since your interval of integration depends on the variable you are integrating with respect to. Do you perhaps want $$ int_0^x e^t^2, dt $$ in the numerator?
– Sobi
Sep 10 at 18:30
Okay. But in my text book it is given as I have typed. I can give you the picture of it if you want to see.
– jayant98
Sep 10 at 18:32
It is likely a typo on their side then. In any case, the computation of your integral is wrong. Do you know L'Hospital's rule?
– Sobi
Sep 10 at 18:35
Please get your formatting at least somewhat right before you post the question.
– Sobi
Sep 10 at 18:23
Please get your formatting at least somewhat right before you post the question.
– Sobi
Sep 10 at 18:23
Sorry for that. But I am just learning about it recently. So, I may mistake but I am trying it to do it correct.
– jayant98
Sep 10 at 18:24
Sorry for that. But I am just learning about it recently. So, I may mistake but I am trying it to do it correct.
– jayant98
Sep 10 at 18:24
Note that the integral in the numerator does not makes sense since your interval of integration depends on the variable you are integrating with respect to. Do you perhaps want $$ int_0^x e^t^2, dt $$ in the numerator?
– Sobi
Sep 10 at 18:30
Note that the integral in the numerator does not makes sense since your interval of integration depends on the variable you are integrating with respect to. Do you perhaps want $$ int_0^x e^t^2, dt $$ in the numerator?
– Sobi
Sep 10 at 18:30
Okay. But in my text book it is given as I have typed. I can give you the picture of it if you want to see.
– jayant98
Sep 10 at 18:32
Okay. But in my text book it is given as I have typed. I can give you the picture of it if you want to see.
– jayant98
Sep 10 at 18:32
It is likely a typo on their side then. In any case, the computation of your integral is wrong. Do you know L'Hospital's rule?
– Sobi
Sep 10 at 18:35
It is likely a typo on their side then. In any case, the computation of your integral is wrong. Do you know L'Hospital's rule?
– Sobi
Sep 10 at 18:35
 |Â
show 3 more comments
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Using L'Hospital's Rule yields
$$beginalign
lim_xtoinftyfracxint_0^x e^t^2,dte^x^2&=lim_xtoinftyfracint_0^x e^t^2,dt+xe^x^22xe^x^2\\
&=frac12+lim_xtoinftyfracint_0^x e^t^2,dt2xe^x^2\\
&=frac12+lim_xtoinftyfrace^x^22e^x^2+4x^2e^x^2\\
&=frac12
endalign$$
answered Sep 10 at 18:39
Mark Viola
127k1172167
Thanks for the solution. :)
– jayant98
Sep 10 at 18:46
You're welcome. My pleasure.
– Mark Viola
Sep 10 at 18:52
Why was this down voted??
– Mark Viola
Sep 11 at 14:19
add a comment |Â
up vote
2
down vote
By L'Hospital,
$$fracdisplaystyleint_0^xe^t^2dtdfrace^x^2xtofrace^x^2dfrac(2x^2-1)e^x^2x^2tofracx^22x^2-1tofrac12.$$
answered Sep 10 at 19:14
Yves Daoust
115k667210
Isolating the integral avoids the need for a double application of the rule.
– Yves Daoust
Sep 10 at 19:17
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Using L'Hospital's Rule yields
$$beginalign
lim_xtoinftyfracxint_0^x e^t^2,dte^x^2&=lim_xtoinftyfracint_0^x e^t^2,dt+xe^x^22xe^x^2\\
&=frac12+lim_xtoinftyfracint_0^x e^t^2,dt2xe^x^2\\
&=frac12+lim_xtoinftyfrace^x^22e^x^2+4x^2e^x^2\\
&=frac12
endalign$$
answered Sep 10 at 18:39
Mark Viola
127k1172167
Thanks for the solution. :)
– jayant98
Sep 10 at 18:46
You're welcome. My pleasure.
– Mark Viola
Sep 10 at 18:52
Why was this down voted??
– Mark Viola
Sep 11 at 14:19
add a comment |Â
up vote
2
down vote
accepted
Using L'Hospital's Rule yields
$$beginalign
lim_xtoinftyfracxint_0^x e^t^2,dte^x^2&=lim_xtoinftyfracint_0^x e^t^2,dt+xe^x^22xe^x^2\\
&=frac12+lim_xtoinftyfracint_0^x e^t^2,dt2xe^x^2\\
&=frac12+lim_xtoinftyfrace^x^22e^x^2+4x^2e^x^2\\
&=frac12
endalign$$
answered Sep 10 at 18:39
Mark Viola
127k1172167
Thanks for the solution. :)
– jayant98
Sep 10 at 18:46
You're welcome. My pleasure.
– Mark Viola
Sep 10 at 18:52
Why was this down voted??
– Mark Viola
Sep 11 at 14:19
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Using L'Hospital's Rule yields
$$beginalign
lim_xtoinftyfracxint_0^x e^t^2,dte^x^2&=lim_xtoinftyfracint_0^x e^t^2,dt+xe^x^22xe^x^2\\
&=frac12+lim_xtoinftyfracint_0^x e^t^2,dt2xe^x^2\\
&=frac12+lim_xtoinftyfrace^x^22e^x^2+4x^2e^x^2\\
&=frac12
endalign$$
answered Sep 10 at 18:39
Mark Viola
127k1172167
Using L'Hospital's Rule yields
$$beginalign
lim_xtoinftyfracxint_0^x e^t^2,dte^x^2&=lim_xtoinftyfracint_0^x e^t^2,dt+xe^x^22xe^x^2\\
&=frac12+lim_xtoinftyfracint_0^x e^t^2,dt2xe^x^2\\
&=frac12+lim_xtoinftyfrace^x^22e^x^2+4x^2e^x^2\\
&=frac12
endalign$$
answered Sep 10 at 18:39
Mark Viola
127k1172167
answered Sep 10 at 18:39
Mark Viola
127k1172167
answered Sep 10 at 18:39
Mark Viola
127k1172167
answered Sep 10 at 18:39
Mark Viola
127k1172167
127k1172167
Thanks for the solution. :)
– jayant98
Sep 10 at 18:46
You're welcome. My pleasure.
– Mark Viola
Sep 10 at 18:52
Why was this down voted??
– Mark Viola
Sep 11 at 14:19
add a comment |Â
Thanks for the solution. :)
– jayant98
Sep 10 at 18:46
You're welcome. My pleasure.
– Mark Viola
Sep 10 at 18:52
Why was this down voted??
– Mark Viola
Sep 11 at 14:19
Thanks for the solution. :)
– jayant98
Sep 10 at 18:46
Thanks for the solution. :)
– jayant98
Sep 10 at 18:46
You're welcome. My pleasure.
– Mark Viola
Sep 10 at 18:52
You're welcome. My pleasure.
– Mark Viola
Sep 10 at 18:52
Why was this down voted??
– Mark Viola
Sep 11 at 14:19
Why was this down voted??
– Mark Viola
Sep 11 at 14:19
add a comment |Â
up vote
2
down vote
By L'Hospital,
$$fracdisplaystyleint_0^xe^t^2dtdfrace^x^2xtofrace^x^2dfrac(2x^2-1)e^x^2x^2tofracx^22x^2-1tofrac12.$$
answered Sep 10 at 19:14
Yves Daoust
115k667210
Isolating the integral avoids the need for a double application of the rule.
– Yves Daoust
Sep 10 at 19:17
add a comment |Â
up vote
2
down vote
By L'Hospital,
$$fracdisplaystyleint_0^xe^t^2dtdfrace^x^2xtofrace^x^2dfrac(2x^2-1)e^x^2x^2tofracx^22x^2-1tofrac12.$$
answered Sep 10 at 19:14
Yves Daoust
115k667210
Isolating the integral avoids the need for a double application of the rule.
– Yves Daoust
Sep 10 at 19:17
add a comment |Â
up vote
2
down vote
up vote
2
down vote
By L'Hospital,
$$fracdisplaystyleint_0^xe^t^2dtdfrace^x^2xtofrace^x^2dfrac(2x^2-1)e^x^2x^2tofracx^22x^2-1tofrac12.$$
answered Sep 10 at 19:14
Yves Daoust
115k667210
By L'Hospital,
$$fracdisplaystyleint_0^xe^t^2dtdfrace^x^2xtofrace^x^2dfrac(2x^2-1)e^x^2x^2tofracx^22x^2-1tofrac12.$$
answered Sep 10 at 19:14
Yves Daoust
115k667210
answered Sep 10 at 19:14
Yves Daoust
115k667210
answered Sep 10 at 19:14
Yves Daoust
115k667210
answered Sep 10 at 19:14
Yves Daoust
115k667210
115k667210
Isolating the integral avoids the need for a double application of the rule.
– Yves Daoust
Sep 10 at 19:17
add a comment |Â
Isolating the integral avoids the need for a double application of the rule.
– Yves Daoust
Sep 10 at 19:17
Isolating the integral avoids the need for a double application of the rule.
– Yves Daoust
Sep 10 at 19:17
Isolating the integral avoids the need for a double application of the rule.
– Yves Daoust
Sep 10 at 19:17
add a comment |Â
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Please get your formatting at least somewhat right before you post the question.
– Sobi
Sep 10 at 18:23
Sorry for that. But I am just learning about it recently. So, I may mistake but I am trying it to do it correct.
– jayant98
Sep 10 at 18:24
Note that the integral in the numerator does not makes sense since your interval of integration depends on the variable you are integrating with respect to. Do you perhaps want $$ int_0^x e^t^2, dt $$ in the numerator?
– Sobi
Sep 10 at 18:30
Okay. But in my text book it is given as I have typed. I can give you the picture of it if you want to see.
– jayant98
Sep 10 at 18:32
It is likely a typo on their side then. In any case, the computation of your integral is wrong. Do you know L'Hospital's rule?
– Sobi
Sep 10 at 18:35