Integrate $int_{-1}^{1}x^2sin^{2019}(x)e^{-x^4}dx$
$begingroup$
I'm following a course on analysis and I am supposed to compute the following integral:
$int_{-1}^{1}x^2sin^{2019}(x)e^{-x^4}dx$.
I've been trying to use the integration by part but I've been stuck for a while. Any help would be highly appreciated.
Thanks
real-analysis calculus integration
edited Dec 31 '18 at 14:21
Bernard
119k740113
asked Dec 31 '18 at 14:15
jffijffi
828
$endgroup$
add a comment |
$begingroup$
I'm following a course on analysis and I am supposed to compute the following integral:
$int_{-1}^{1}x^2sin^{2019}(x)e^{-x^4}dx$.
I've been trying to use the integration by part but I've been stuck for a while. Any help would be highly appreciated.
Thanks
real-analysis calculus integration
edited Dec 31 '18 at 14:21
Bernard
119k740113
asked Dec 31 '18 at 14:15
jffijffi
828
$endgroup$
$begingroup$
Use math.stackexchange.com/questions/439851/… and the given function is odd
$endgroup$
– lab bhattacharjee
Dec 31 '18 at 14:18
add a comment |
$begingroup$
I'm following a course on analysis and I am supposed to compute the following integral:
$int_{-1}^{1}x^2sin^{2019}(x)e^{-x^4}dx$.
I've been trying to use the integration by part but I've been stuck for a while. Any help would be highly appreciated.
Thanks
real-analysis calculus integration
edited Dec 31 '18 at 14:21
Bernard
119k740113
asked Dec 31 '18 at 14:15
jffijffi
828
$endgroup$
I'm following a course on analysis and I am supposed to compute the following integral:
$int_{-1}^{1}x^2sin^{2019}(x)e^{-x^4}dx$.
I've been trying to use the integration by part but I've been stuck for a while. Any help would be highly appreciated.
Thanks
real-analysis calculus integration
real-analysis calculus integration
edited Dec 31 '18 at 14:21
Bernard
119k740113
asked Dec 31 '18 at 14:15
jffijffi
828
edited Dec 31 '18 at 14:21
Bernard
119k740113
asked Dec 31 '18 at 14:15
jffijffi
828
edited Dec 31 '18 at 14:21
Bernard
119k740113
edited Dec 31 '18 at 14:21
Bernard
119k740113
edited Dec 31 '18 at 14:21
Bernard
119k740113
119k740113
asked Dec 31 '18 at 14:15
jffijffi
828
asked Dec 31 '18 at 14:15
jffijffi
828
asked Dec 31 '18 at 14:15
jffijffi
828
828
$begingroup$
Use math.stackexchange.com/questions/439851/… and the given function is odd
$endgroup$
– lab bhattacharjee
Dec 31 '18 at 14:18
add a comment |
$begingroup$
Use math.stackexchange.com/questions/439851/… and the given function is odd
$endgroup$
– lab bhattacharjee
Dec 31 '18 at 14:18
$begingroup$
Use math.stackexchange.com/questions/439851/… and the given function is odd
$endgroup$
– lab bhattacharjee
Dec 31 '18 at 14:18
$begingroup$
Use math.stackexchange.com/questions/439851/… and the given function is odd
$endgroup$
– lab bhattacharjee
Dec 31 '18 at 14:18
add a comment |
2 Answers
2
active
oldest
votes
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Just substitute $x=-t$ and you will get that: $$I=int_{-1}^{1}x^2sin^{2019}(x)e^{-x^4}dx=-int_{-1}^{1}t^2sin^{2019}(t)e^{-t^4}dt=-I$$
$$I=-IRightarrow I=0$$
answered Dec 31 '18 at 14:20
ZackyZacky
5,4661856
$endgroup$
2
$begingroup$
Oh... right! I can't believe I was actually trying to compute the primitive. Thanks
$endgroup$
– jffi
Dec 31 '18 at 14:26
1
$begingroup$
I believe it's a good thing that you already tried to compute the primitive, that is how in the future you will be able to see faster when you can find a primitive for an integral, by trying and failing in the past and earning experience. So no worries at all, keep it up!
$endgroup$
– Zacky
Dec 31 '18 at 14:32
add a comment |
$begingroup$
The integrand is an odd function.
For any odd (integrable) function $f$, we have
$$int_{-a}^a f = 0$$
Therefore, the integral is equal to $0$.
answered Dec 31 '18 at 14:20
Math_QEDMath_QED
7,39531450
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Just substitute $x=-t$ and you will get that: $$I=int_{-1}^{1}x^2sin^{2019}(x)e^{-x^4}dx=-int_{-1}^{1}t^2sin^{2019}(t)e^{-t^4}dt=-I$$
$$I=-IRightarrow I=0$$
answered Dec 31 '18 at 14:20
ZackyZacky
5,4661856
$endgroup$
2
$begingroup$
Oh... right! I can't believe I was actually trying to compute the primitive. Thanks
$endgroup$
– jffi
Dec 31 '18 at 14:26
1
$begingroup$
I believe it's a good thing that you already tried to compute the primitive, that is how in the future you will be able to see faster when you can find a primitive for an integral, by trying and failing in the past and earning experience. So no worries at all, keep it up!
$endgroup$
– Zacky
Dec 31 '18 at 14:32
add a comment |
$begingroup$
Just substitute $x=-t$ and you will get that: $$I=int_{-1}^{1}x^2sin^{2019}(x)e^{-x^4}dx=-int_{-1}^{1}t^2sin^{2019}(t)e^{-t^4}dt=-I$$
$$I=-IRightarrow I=0$$
answered Dec 31 '18 at 14:20
ZackyZacky
5,4661856
$endgroup$
2
$begingroup$
Oh... right! I can't believe I was actually trying to compute the primitive. Thanks
$endgroup$
– jffi
Dec 31 '18 at 14:26
1
$begingroup$
I believe it's a good thing that you already tried to compute the primitive, that is how in the future you will be able to see faster when you can find a primitive for an integral, by trying and failing in the past and earning experience. So no worries at all, keep it up!
$endgroup$
– Zacky
Dec 31 '18 at 14:32
add a comment |
$begingroup$
Just substitute $x=-t$ and you will get that: $$I=int_{-1}^{1}x^2sin^{2019}(x)e^{-x^4}dx=-int_{-1}^{1}t^2sin^{2019}(t)e^{-t^4}dt=-I$$
$$I=-IRightarrow I=0$$
answered Dec 31 '18 at 14:20
ZackyZacky
5,4661856
$endgroup$
Just substitute $x=-t$ and you will get that: $$I=int_{-1}^{1}x^2sin^{2019}(x)e^{-x^4}dx=-int_{-1}^{1}t^2sin^{2019}(t)e^{-t^4}dt=-I$$
$$I=-IRightarrow I=0$$
answered Dec 31 '18 at 14:20
ZackyZacky
5,4661856
answered Dec 31 '18 at 14:20
ZackyZacky
5,4661856
answered Dec 31 '18 at 14:20
ZackyZacky
5,4661856
answered Dec 31 '18 at 14:20
ZackyZacky
5,4661856
5,4661856
2
$begingroup$
Oh... right! I can't believe I was actually trying to compute the primitive. Thanks
$endgroup$
– jffi
Dec 31 '18 at 14:26
1
$begingroup$
I believe it's a good thing that you already tried to compute the primitive, that is how in the future you will be able to see faster when you can find a primitive for an integral, by trying and failing in the past and earning experience. So no worries at all, keep it up!
$endgroup$
– Zacky
Dec 31 '18 at 14:32
add a comment |
2
$begingroup$
Oh... right! I can't believe I was actually trying to compute the primitive. Thanks
$endgroup$
– jffi
Dec 31 '18 at 14:26
1
$begingroup$
I believe it's a good thing that you already tried to compute the primitive, that is how in the future you will be able to see faster when you can find a primitive for an integral, by trying and failing in the past and earning experience. So no worries at all, keep it up!
$endgroup$
– Zacky
Dec 31 '18 at 14:32
2
2
$begingroup$
Oh... right! I can't believe I was actually trying to compute the primitive. Thanks
$endgroup$
– jffi
Dec 31 '18 at 14:26
$begingroup$
Oh... right! I can't believe I was actually trying to compute the primitive. Thanks
$endgroup$
– jffi
Dec 31 '18 at 14:26
1
1
$begingroup$
I believe it's a good thing that you already tried to compute the primitive, that is how in the future you will be able to see faster when you can find a primitive for an integral, by trying and failing in the past and earning experience. So no worries at all, keep it up!
$endgroup$
– Zacky
Dec 31 '18 at 14:32
$begingroup$
I believe it's a good thing that you already tried to compute the primitive, that is how in the future you will be able to see faster when you can find a primitive for an integral, by trying and failing in the past and earning experience. So no worries at all, keep it up!
$endgroup$
– Zacky
Dec 31 '18 at 14:32
add a comment |
$begingroup$
The integrand is an odd function.
For any odd (integrable) function $f$, we have
$$int_{-a}^a f = 0$$
Therefore, the integral is equal to $0$.
answered Dec 31 '18 at 14:20
Math_QEDMath_QED
7,39531450
$endgroup$
add a comment |
$begingroup$
The integrand is an odd function.
For any odd (integrable) function $f$, we have
$$int_{-a}^a f = 0$$
Therefore, the integral is equal to $0$.
answered Dec 31 '18 at 14:20
Math_QEDMath_QED
7,39531450
$endgroup$
add a comment |
$begingroup$
The integrand is an odd function.
For any odd (integrable) function $f$, we have
$$int_{-a}^a f = 0$$
Therefore, the integral is equal to $0$.
answered Dec 31 '18 at 14:20
Math_QEDMath_QED
7,39531450
$endgroup$
The integrand is an odd function.
For any odd (integrable) function $f$, we have
$$int_{-a}^a f = 0$$
Therefore, the integral is equal to $0$.
answered Dec 31 '18 at 14:20
Math_QEDMath_QED
7,39531450
answered Dec 31 '18 at 14:20
Math_QEDMath_QED
7,39531450
answered Dec 31 '18 at 14:20
Math_QEDMath_QED
7,39531450
answered Dec 31 '18 at 14:20
Math_QEDMath_QED
7,39531450
7,39531450
add a comment |
add a comment |
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$begingroup$
Use math.stackexchange.com/questions/439851/… and the given function is odd
$endgroup$
– lab bhattacharjee
Dec 31 '18 at 14:18