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Definite integration = exerise solution with substitution
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I am trying to integrate the following:
$$ int_0^infty y^2 fracytheta e^frac-y^22theta dy $$
Then
$$ int_0^inftyfracy^3theta e^frac-y^22theta dy $$
I tried the u-sibsitution also also integration bz parts but cannot get to the right solution.
Our professor provided us with solution in terms of x:
$$ int_0^infty x^2 fracxtheta e^frac-x^22theta = -int_0^infty x^2 de ^fracx^22theta = int_0^infty e^frac-x^22theta dx^2$$
Taking substitution: $ u=fracx^22theta$
$$ 2theta int_0^inftye^-u = 2theta $$
QUESTIONS:
Where does the de come from in the second steo? What does it mean?
I would very much appriciate a detailed breakdown of the steps/operations. I have only recently started working with integrations.
calculus integration definite-integrals
asked Jul 29 at 10:13
user1607
608
add a comment |Â
up vote
0
down vote
favorite
I am trying to integrate the following:
$$ int_0^infty y^2 fracytheta e^frac-y^22theta dy $$
Then
$$ int_0^inftyfracy^3theta e^frac-y^22theta dy $$
I tried the u-sibsitution also also integration bz parts but cannot get to the right solution.
Our professor provided us with solution in terms of x:
$$ int_0^infty x^2 fracxtheta e^frac-x^22theta = -int_0^infty x^2 de ^fracx^22theta = int_0^infty e^frac-x^22theta dx^2$$
Taking substitution: $ u=fracx^22theta$
$$ 2theta int_0^inftye^-u = 2theta $$
QUESTIONS:
Where does the de come from in the second steo? What does it mean?
I would very much appriciate a detailed breakdown of the steps/operations. I have only recently started working with integrations.
calculus integration definite-integrals
asked Jul 29 at 10:13
user1607
608
1
Your professor meant $$fracddx (e^x^2/2 theta) = fracxtheta e^x^2/2 theta implies d(e^x^2/2 theta) = fracxtheta e^x^2/2 theta dx$$ This notation it is used quite frequently in mathematics, though I personally don't use it. Also, I'm not sure why the integral in $y$ has an $s$ in the numerator of the power of the exponential.
– Mattos
Jul 29 at 10:55
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to integrate the following:
$$ int_0^infty y^2 fracytheta e^frac-y^22theta dy $$
Then
$$ int_0^inftyfracy^3theta e^frac-y^22theta dy $$
I tried the u-sibsitution also also integration bz parts but cannot get to the right solution.
Our professor provided us with solution in terms of x:
$$ int_0^infty x^2 fracxtheta e^frac-x^22theta = -int_0^infty x^2 de ^fracx^22theta = int_0^infty e^frac-x^22theta dx^2$$
Taking substitution: $ u=fracx^22theta$
$$ 2theta int_0^inftye^-u = 2theta $$
QUESTIONS:
Where does the de come from in the second steo? What does it mean?
I would very much appriciate a detailed breakdown of the steps/operations. I have only recently started working with integrations.
calculus integration definite-integrals
asked Jul 29 at 10:13
user1607
608
I am trying to integrate the following:
$$ int_0^infty y^2 fracytheta e^frac-y^22theta dy $$
Then
$$ int_0^inftyfracy^3theta e^frac-y^22theta dy $$
I tried the u-sibsitution also also integration bz parts but cannot get to the right solution.
Our professor provided us with solution in terms of x:
$$ int_0^infty x^2 fracxtheta e^frac-x^22theta = -int_0^infty x^2 de ^fracx^22theta = int_0^infty e^frac-x^22theta dx^2$$
Taking substitution: $ u=fracx^22theta$
$$ 2theta int_0^inftye^-u = 2theta $$
QUESTIONS:
Where does the de come from in the second steo? What does it mean?
I would very much appriciate a detailed breakdown of the steps/operations. I have only recently started working with integrations.
calculus integration definite-integrals
asked Jul 29 at 10:13
user1607
608
edited Jul 29 at 14:03
asked Jul 29 at 10:13
user1607
608
asked Jul 29 at 10:13
user1607
608
asked Jul 29 at 10:13
user1607
608
608
1
Your professor meant $$fracddx (e^x^2/2 theta) = fracxtheta e^x^2/2 theta implies d(e^x^2/2 theta) = fracxtheta e^x^2/2 theta dx$$ This notation it is used quite frequently in mathematics, though I personally don't use it. Also, I'm not sure why the integral in $y$ has an $s$ in the numerator of the power of the exponential.
– Mattos
Jul 29 at 10:55
add a comment |Â
1
Your professor meant $$fracddx (e^x^2/2 theta) = fracxtheta e^x^2/2 theta implies d(e^x^2/2 theta) = fracxtheta e^x^2/2 theta dx$$ This notation it is used quite frequently in mathematics, though I personally don't use it. Also, I'm not sure why the integral in $y$ has an $s$ in the numerator of the power of the exponential.
– Mattos
Jul 29 at 10:55
1
1
Your professor meant $$fracddx (e^x^2/2 theta) = fracxtheta e^x^2/2 theta implies d(e^x^2/2 theta) = fracxtheta e^x^2/2 theta dx$$ This notation it is used quite frequently in mathematics, though I personally don't use it. Also, I'm not sure why the integral in $y$ has an $s$ in the numerator of the power of the exponential.
– Mattos
Jul 29 at 10:55
Your professor meant $$fracddx (e^x^2/2 theta) = fracxtheta e^x^2/2 theta implies d(e^x^2/2 theta) = fracxtheta e^x^2/2 theta dx$$ This notation it is used quite frequently in mathematics, though I personally don't use it. Also, I'm not sure why the integral in $y$ has an $s$ in the numerator of the power of the exponential.
– Mattos
Jul 29 at 10:55
add a comment |Â
1 Answer
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beginalign
int_0^infty y^2 fracytheta e^frac-y^2stheta dy
&= int_0^infty y^2 left(fracytheta e^frac-y^2sthetaright) dy \
&= int_0^infty y^2 left(fracs-2cdotfrac-2ystheta e^frac-y^2sthetaright) dy \
&= -fracs2int_0^infty y^2 cdot fracddyleft( e^frac-y^2sthetaright) dy \
&= -fracs2int_1^0 y^2 dleft( e^frac-y^2sthetaright) \
&= -fracs2int_1^0 y^2 du, quad u=e^frac-y^2sthetaRightarrow y^2=-sthetaln u \
&= -fracs2int_1^0 -sthetaln u , du \
&= fracs^2thetaint_1^0 ln u , du \
&= fracs^2thetaleft[left[uln uright]_1^0-int_1^0 ucdotfrac1u , duright] \
&= fracs^2thetaleft[(0-0)+left[uright]_0^1right] \
&= fracs^2theta
endalign
answered Jul 29 at 11:45
Karn Watcharasupat
3,7992426
I appologize i just noticed i made a mistake when typing the exercises. There is no $s$, it was supposed to be a 2, So then: $$ int_0^infty y^2 (fracythetae^-fracy^22theta)dy$$. Thank you for the elaborate answer.
– user1607
Jul 29 at 14:07
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
beginalign
int_0^infty y^2 fracytheta e^frac-y^2stheta dy
&= int_0^infty y^2 left(fracytheta e^frac-y^2sthetaright) dy \
&= int_0^infty y^2 left(fracs-2cdotfrac-2ystheta e^frac-y^2sthetaright) dy \
&= -fracs2int_0^infty y^2 cdot fracddyleft( e^frac-y^2sthetaright) dy \
&= -fracs2int_1^0 y^2 dleft( e^frac-y^2sthetaright) \
&= -fracs2int_1^0 y^2 du, quad u=e^frac-y^2sthetaRightarrow y^2=-sthetaln u \
&= -fracs2int_1^0 -sthetaln u , du \
&= fracs^2thetaint_1^0 ln u , du \
&= fracs^2thetaleft[left[uln uright]_1^0-int_1^0 ucdotfrac1u , duright] \
&= fracs^2thetaleft[(0-0)+left[uright]_0^1right] \
&= fracs^2theta
endalign
answered Jul 29 at 11:45
Karn Watcharasupat
3,7992426
I appologize i just noticed i made a mistake when typing the exercises. There is no $s$, it was supposed to be a 2, So then: $$ int_0^infty y^2 (fracythetae^-fracy^22theta)dy$$. Thank you for the elaborate answer.
– user1607
Jul 29 at 14:07
add a comment |Â
up vote
1
down vote
accepted
beginalign
int_0^infty y^2 fracytheta e^frac-y^2stheta dy
&= int_0^infty y^2 left(fracytheta e^frac-y^2sthetaright) dy \
&= int_0^infty y^2 left(fracs-2cdotfrac-2ystheta e^frac-y^2sthetaright) dy \
&= -fracs2int_0^infty y^2 cdot fracddyleft( e^frac-y^2sthetaright) dy \
&= -fracs2int_1^0 y^2 dleft( e^frac-y^2sthetaright) \
&= -fracs2int_1^0 y^2 du, quad u=e^frac-y^2sthetaRightarrow y^2=-sthetaln u \
&= -fracs2int_1^0 -sthetaln u , du \
&= fracs^2thetaint_1^0 ln u , du \
&= fracs^2thetaleft[left[uln uright]_1^0-int_1^0 ucdotfrac1u , duright] \
&= fracs^2thetaleft[(0-0)+left[uright]_0^1right] \
&= fracs^2theta
endalign
answered Jul 29 at 11:45
Karn Watcharasupat
3,7992426
I appologize i just noticed i made a mistake when typing the exercises. There is no $s$, it was supposed to be a 2, So then: $$ int_0^infty y^2 (fracythetae^-fracy^22theta)dy$$. Thank you for the elaborate answer.
– user1607
Jul 29 at 14:07
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
beginalign
int_0^infty y^2 fracytheta e^frac-y^2stheta dy
&= int_0^infty y^2 left(fracytheta e^frac-y^2sthetaright) dy \
&= int_0^infty y^2 left(fracs-2cdotfrac-2ystheta e^frac-y^2sthetaright) dy \
&= -fracs2int_0^infty y^2 cdot fracddyleft( e^frac-y^2sthetaright) dy \
&= -fracs2int_1^0 y^2 dleft( e^frac-y^2sthetaright) \
&= -fracs2int_1^0 y^2 du, quad u=e^frac-y^2sthetaRightarrow y^2=-sthetaln u \
&= -fracs2int_1^0 -sthetaln u , du \
&= fracs^2thetaint_1^0 ln u , du \
&= fracs^2thetaleft[left[uln uright]_1^0-int_1^0 ucdotfrac1u , duright] \
&= fracs^2thetaleft[(0-0)+left[uright]_0^1right] \
&= fracs^2theta
endalign
answered Jul 29 at 11:45
Karn Watcharasupat
3,7992426
beginalign
int_0^infty y^2 fracytheta e^frac-y^2stheta dy
&= int_0^infty y^2 left(fracytheta e^frac-y^2sthetaright) dy \
&= int_0^infty y^2 left(fracs-2cdotfrac-2ystheta e^frac-y^2sthetaright) dy \
&= -fracs2int_0^infty y^2 cdot fracddyleft( e^frac-y^2sthetaright) dy \
&= -fracs2int_1^0 y^2 dleft( e^frac-y^2sthetaright) \
&= -fracs2int_1^0 y^2 du, quad u=e^frac-y^2sthetaRightarrow y^2=-sthetaln u \
&= -fracs2int_1^0 -sthetaln u , du \
&= fracs^2thetaint_1^0 ln u , du \
&= fracs^2thetaleft[left[uln uright]_1^0-int_1^0 ucdotfrac1u , duright] \
&= fracs^2thetaleft[(0-0)+left[uright]_0^1right] \
&= fracs^2theta
endalign
answered Jul 29 at 11:45
Karn Watcharasupat
3,7992426
answered Jul 29 at 11:45
Karn Watcharasupat
3,7992426
answered Jul 29 at 11:45
Karn Watcharasupat
3,7992426
answered Jul 29 at 11:45
Karn Watcharasupat
3,7992426
3,7992426
I appologize i just noticed i made a mistake when typing the exercises. There is no $s$, it was supposed to be a 2, So then: $$ int_0^infty y^2 (fracythetae^-fracy^22theta)dy$$. Thank you for the elaborate answer.
– user1607
Jul 29 at 14:07
add a comment |Â
I appologize i just noticed i made a mistake when typing the exercises. There is no $s$, it was supposed to be a 2, So then: $$ int_0^infty y^2 (fracythetae^-fracy^22theta)dy$$. Thank you for the elaborate answer.
– user1607
Jul 29 at 14:07
I appologize i just noticed i made a mistake when typing the exercises. There is no $s$, it was supposed to be a 2, So then: $$ int_0^infty y^2 (fracythetae^-fracy^22theta)dy$$. Thank you for the elaborate answer.
– user1607
Jul 29 at 14:07
I appologize i just noticed i made a mistake when typing the exercises. There is no $s$, it was supposed to be a 2, So then: $$ int_0^infty y^2 (fracythetae^-fracy^22theta)dy$$. Thank you for the elaborate answer.
– user1607
Jul 29 at 14:07
add a comment |Â
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1
Your professor meant $$fracddx (e^x^2/2 theta) = fracxtheta e^x^2/2 theta implies d(e^x^2/2 theta) = fracxtheta e^x^2/2 theta dx$$ This notation it is used quite frequently in mathematics, though I personally don't use it. Also, I'm not sure why the integral in $y$ has an $s$ in the numerator of the power of the exponential.
– Mattos
Jul 29 at 10:55