Evaluation of the double integral $int_{[0,1]×[0,1]} max{x, y} dxdy$Evaluate the integral $intint _{[0,1]^2}...
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Evaluation of the double integral $int_{[0,1]×[0,1]} max{x, y} dxdy$
Evaluate the integral $intint _{[0,1]^2} max left{x,yright} dx, dy$How can I solve the following integrals?a question about the evaluation of integralTroubles with double integraldouble integral on $[0,1]times [0,1]$If ${f_n}subset L_1([0,1])$, $f_nto f$ pointwise, and $sup_{n} int_{0}^{1} |f_n|max (0, log |f_n|)<infty$, then $f_nto f$ in $L_1$Compute $int_{-infty}^{infty}int_{-infty}^{infty}e^{-(x^2+(x-y)^2+y^2)}dxdy$Let $f:[0,1] rightarrow [0,1] $ then does $ f $ takes the value $int_{0}^{1} f^{2} (x)dx $Computing: $int_0^1int_sqrt{x}^1{frac{1}{sqrt{1+y^3}}dxdy}$Evaluate $int_{0}^{5} max ({{x}^{2},{6x-8}}),dx$Show that $int_{I} |f(x)|^p dx leq int_{I times I} |f(x)+g(y)|^{p}dxdy$ if $int_{I} g(y)dy = 0$.About the double integral and fubini's theorem
$begingroup$
Evaluate: $$int_{[0,1]×[0,1]} max{x, y} dxdy$$
I am totally stuck on it. How can I solve this?
real-analysis
asked Jan 16 '13 at 2:03
user58430user58430
3112
$endgroup$
add a comment |
$begingroup$
Evaluate: $$int_{[0,1]×[0,1]} max{x, y} dxdy$$
I am totally stuck on it. How can I solve this?
real-analysis
asked Jan 16 '13 at 2:03
user58430user58430
3112
$endgroup$
add a comment |
$begingroup$
Evaluate: $$int_{[0,1]×[0,1]} max{x, y} dxdy$$
I am totally stuck on it. How can I solve this?
real-analysis
asked Jan 16 '13 at 2:03
user58430user58430
3112
$endgroup$
Evaluate: $$int_{[0,1]×[0,1]} max{x, y} dxdy$$
I am totally stuck on it. How can I solve this?
real-analysis
real-analysis
asked Jan 16 '13 at 2:03
user58430user58430
3112
asked Jan 16 '13 at 2:03
user58430user58430
3112
edited Jan 16 '13 at 6:58
user17762
asked Jan 16 '13 at 2:03
user58430user58430
3112
asked Jan 16 '13 at 2:03
user58430user58430
3112
asked Jan 16 '13 at 2:03
user58430user58430
3112
3112
add a comment |
add a comment |
4 Answers
4
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$begingroup$
$$
begin{align*}
int_0^1 int_0^1 max(x, y) dx dy &= int_0^1 int_0^y y dx dy + int_0^1 int_y^1 x dx dy \
&= int_0^1 y^2 dy + int_0^1frac{1 - y^2}{2} dy \
&= frac{1}{2}int_0^1 (y^2 + 1) dy \
&= frac{1}{2}left( frac{1}{3} + 1right) = frac{2}{3}
end{align*}$$
answered Jan 16 '13 at 2:14
zrbeckerzrbecker
2,77311725
$endgroup$
add a comment |
$begingroup$
In general, the integral $$I_n = int_{[0,1]^n} max{x_1,x_2,ldots,x_n} dx_1 dx_2 ldots dx_n$$can be written as
$$I_n = n int_{x_1=0}^1 int_{0leq x_2,x_3,ldots,x_n leq x_1} x_1 dx_1 dx_2 cdots dx_n = n int_{x_1=0}^1 x_1^n dx_1 = dfrac{n}{n+1}$$
since $$max{x_1,x_2,ldots,x_n} = begin{cases} x_1; & x_k leq x_1\ x_2; & x_k leq x_2\ vdots& vdots\ x_n; & x_k leq x_nend{cases}$$
and this divides $[0,1]^n$ into $n$ regions.
$endgroup$
add a comment |
$begingroup$
Hint: break it into two parts, one where $x > y$ and the other $x le y$.
answered Jan 16 '13 at 2:06
Robert IsraelRobert Israel
326k23215469
$endgroup$
add a comment |
$begingroup$
In general, the integral
$$I_n=int_{[0,1]^n}max{x_1,cdots,x_n},mathrm{d} x_1cdotsmathrm{d}x_n$$
can be viewed as an integral of the largest order statistic, say $X^{(1)}$.
Let $x^{(1)}=max {x_1,cdots,x_n}$, the distribution of $X^{(1)}$ will be $$F_{X^{(1)}}(x)=F_{X_1}(x)timescdotstimes F_{X_n}(x)=x^n.$$
Hence
$$I_n=int_{[0,1]^n}max{x_1,cdots,x_n},mathrm{d} x_1cdotsmathrm{d}x_n=int_{[0,1]}x,mathrm{d}F_{X^{(1)}}(x)=int_0^1 nx^n,mathrm{d}x=frac{n}{n+1}.$$
answered 2 days ago
Xiangqian YangXiangqian Yang
1
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add a comment |
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4 Answers
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4 Answers
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$begingroup$
$$
begin{align*}
int_0^1 int_0^1 max(x, y) dx dy &= int_0^1 int_0^y y dx dy + int_0^1 int_y^1 x dx dy \
&= int_0^1 y^2 dy + int_0^1frac{1 - y^2}{2} dy \
&= frac{1}{2}int_0^1 (y^2 + 1) dy \
&= frac{1}{2}left( frac{1}{3} + 1right) = frac{2}{3}
end{align*}$$
answered Jan 16 '13 at 2:14
zrbeckerzrbecker
2,77311725
$endgroup$
add a comment |
$begingroup$
$$
begin{align*}
int_0^1 int_0^1 max(x, y) dx dy &= int_0^1 int_0^y y dx dy + int_0^1 int_y^1 x dx dy \
&= int_0^1 y^2 dy + int_0^1frac{1 - y^2}{2} dy \
&= frac{1}{2}int_0^1 (y^2 + 1) dy \
&= frac{1}{2}left( frac{1}{3} + 1right) = frac{2}{3}
end{align*}$$
answered Jan 16 '13 at 2:14
zrbeckerzrbecker
2,77311725
$endgroup$
add a comment |
$begingroup$
$$
begin{align*}
int_0^1 int_0^1 max(x, y) dx dy &= int_0^1 int_0^y y dx dy + int_0^1 int_y^1 x dx dy \
&= int_0^1 y^2 dy + int_0^1frac{1 - y^2}{2} dy \
&= frac{1}{2}int_0^1 (y^2 + 1) dy \
&= frac{1}{2}left( frac{1}{3} + 1right) = frac{2}{3}
end{align*}$$
answered Jan 16 '13 at 2:14
zrbeckerzrbecker
2,77311725
$endgroup$
$$
begin{align*}
int_0^1 int_0^1 max(x, y) dx dy &= int_0^1 int_0^y y dx dy + int_0^1 int_y^1 x dx dy \
&= int_0^1 y^2 dy + int_0^1frac{1 - y^2}{2} dy \
&= frac{1}{2}int_0^1 (y^2 + 1) dy \
&= frac{1}{2}left( frac{1}{3} + 1right) = frac{2}{3}
end{align*}$$
answered Jan 16 '13 at 2:14
zrbeckerzrbecker
2,77311725
answered Jan 16 '13 at 2:14
zrbeckerzrbecker
2,77311725
answered Jan 16 '13 at 2:14
zrbeckerzrbecker
2,77311725
answered Jan 16 '13 at 2:14
zrbeckerzrbecker
2,77311725
2,77311725
add a comment |
add a comment |
$begingroup$
In general, the integral $$I_n = int_{[0,1]^n} max{x_1,x_2,ldots,x_n} dx_1 dx_2 ldots dx_n$$can be written as
$$I_n = n int_{x_1=0}^1 int_{0leq x_2,x_3,ldots,x_n leq x_1} x_1 dx_1 dx_2 cdots dx_n = n int_{x_1=0}^1 x_1^n dx_1 = dfrac{n}{n+1}$$
since $$max{x_1,x_2,ldots,x_n} = begin{cases} x_1; & x_k leq x_1\ x_2; & x_k leq x_2\ vdots& vdots\ x_n; & x_k leq x_nend{cases}$$
and this divides $[0,1]^n$ into $n$ regions.
$endgroup$
add a comment |
$begingroup$
In general, the integral $$I_n = int_{[0,1]^n} max{x_1,x_2,ldots,x_n} dx_1 dx_2 ldots dx_n$$can be written as
$$I_n = n int_{x_1=0}^1 int_{0leq x_2,x_3,ldots,x_n leq x_1} x_1 dx_1 dx_2 cdots dx_n = n int_{x_1=0}^1 x_1^n dx_1 = dfrac{n}{n+1}$$
since $$max{x_1,x_2,ldots,x_n} = begin{cases} x_1; & x_k leq x_1\ x_2; & x_k leq x_2\ vdots& vdots\ x_n; & x_k leq x_nend{cases}$$
and this divides $[0,1]^n$ into $n$ regions.
$endgroup$
add a comment |
$begingroup$
In general, the integral $$I_n = int_{[0,1]^n} max{x_1,x_2,ldots,x_n} dx_1 dx_2 ldots dx_n$$can be written as
$$I_n = n int_{x_1=0}^1 int_{0leq x_2,x_3,ldots,x_n leq x_1} x_1 dx_1 dx_2 cdots dx_n = n int_{x_1=0}^1 x_1^n dx_1 = dfrac{n}{n+1}$$
since $$max{x_1,x_2,ldots,x_n} = begin{cases} x_1; & x_k leq x_1\ x_2; & x_k leq x_2\ vdots& vdots\ x_n; & x_k leq x_nend{cases}$$
and this divides $[0,1]^n$ into $n$ regions.
$endgroup$
In general, the integral $$I_n = int_{[0,1]^n} max{x_1,x_2,ldots,x_n} dx_1 dx_2 ldots dx_n$$can be written as
$$I_n = n int_{x_1=0}^1 int_{0leq x_2,x_3,ldots,x_n leq x_1} x_1 dx_1 dx_2 cdots dx_n = n int_{x_1=0}^1 x_1^n dx_1 = dfrac{n}{n+1}$$
since $$max{x_1,x_2,ldots,x_n} = begin{cases} x_1; & x_k leq x_1\ x_2; & x_k leq x_2\ vdots& vdots\ x_n; & x_k leq x_nend{cases}$$
and this divides $[0,1]^n$ into $n$ regions.
answered Jan 16 '13 at 2:17
user17762
add a comment |
add a comment |
$begingroup$
Hint: break it into two parts, one where $x > y$ and the other $x le y$.
answered Jan 16 '13 at 2:06
Robert IsraelRobert Israel
326k23215469
$endgroup$
add a comment |
$begingroup$
Hint: break it into two parts, one where $x > y$ and the other $x le y$.
answered Jan 16 '13 at 2:06
Robert IsraelRobert Israel
326k23215469
$endgroup$
add a comment |
$begingroup$
Hint: break it into two parts, one where $x > y$ and the other $x le y$.
answered Jan 16 '13 at 2:06
Robert IsraelRobert Israel
326k23215469
$endgroup$
Hint: break it into two parts, one where $x > y$ and the other $x le y$.
answered Jan 16 '13 at 2:06
Robert IsraelRobert Israel
326k23215469
answered Jan 16 '13 at 2:06
Robert IsraelRobert Israel
326k23215469
answered Jan 16 '13 at 2:06
Robert IsraelRobert Israel
326k23215469
answered Jan 16 '13 at 2:06
Robert IsraelRobert Israel
326k23215469
326k23215469
add a comment |
add a comment |
$begingroup$
In general, the integral
$$I_n=int_{[0,1]^n}max{x_1,cdots,x_n},mathrm{d} x_1cdotsmathrm{d}x_n$$
can be viewed as an integral of the largest order statistic, say $X^{(1)}$.
Let $x^{(1)}=max {x_1,cdots,x_n}$, the distribution of $X^{(1)}$ will be $$F_{X^{(1)}}(x)=F_{X_1}(x)timescdotstimes F_{X_n}(x)=x^n.$$
Hence
$$I_n=int_{[0,1]^n}max{x_1,cdots,x_n},mathrm{d} x_1cdotsmathrm{d}x_n=int_{[0,1]}x,mathrm{d}F_{X^{(1)}}(x)=int_0^1 nx^n,mathrm{d}x=frac{n}{n+1}.$$
answered 2 days ago
Xiangqian YangXiangqian Yang
1
$endgroup$
add a comment |
$begingroup$
In general, the integral
$$I_n=int_{[0,1]^n}max{x_1,cdots,x_n},mathrm{d} x_1cdotsmathrm{d}x_n$$
can be viewed as an integral of the largest order statistic, say $X^{(1)}$.
Let $x^{(1)}=max {x_1,cdots,x_n}$, the distribution of $X^{(1)}$ will be $$F_{X^{(1)}}(x)=F_{X_1}(x)timescdotstimes F_{X_n}(x)=x^n.$$
Hence
$$I_n=int_{[0,1]^n}max{x_1,cdots,x_n},mathrm{d} x_1cdotsmathrm{d}x_n=int_{[0,1]}x,mathrm{d}F_{X^{(1)}}(x)=int_0^1 nx^n,mathrm{d}x=frac{n}{n+1}.$$
answered 2 days ago
Xiangqian YangXiangqian Yang
1
$endgroup$
add a comment |
$begingroup$
In general, the integral
$$I_n=int_{[0,1]^n}max{x_1,cdots,x_n},mathrm{d} x_1cdotsmathrm{d}x_n$$
can be viewed as an integral of the largest order statistic, say $X^{(1)}$.
Let $x^{(1)}=max {x_1,cdots,x_n}$, the distribution of $X^{(1)}$ will be $$F_{X^{(1)}}(x)=F_{X_1}(x)timescdotstimes F_{X_n}(x)=x^n.$$
Hence
$$I_n=int_{[0,1]^n}max{x_1,cdots,x_n},mathrm{d} x_1cdotsmathrm{d}x_n=int_{[0,1]}x,mathrm{d}F_{X^{(1)}}(x)=int_0^1 nx^n,mathrm{d}x=frac{n}{n+1}.$$
answered 2 days ago
Xiangqian YangXiangqian Yang
1
$endgroup$
In general, the integral
$$I_n=int_{[0,1]^n}max{x_1,cdots,x_n},mathrm{d} x_1cdotsmathrm{d}x_n$$
can be viewed as an integral of the largest order statistic, say $X^{(1)}$.
Let $x^{(1)}=max {x_1,cdots,x_n}$, the distribution of $X^{(1)}$ will be $$F_{X^{(1)}}(x)=F_{X_1}(x)timescdotstimes F_{X_n}(x)=x^n.$$
Hence
$$I_n=int_{[0,1]^n}max{x_1,cdots,x_n},mathrm{d} x_1cdotsmathrm{d}x_n=int_{[0,1]}x,mathrm{d}F_{X^{(1)}}(x)=int_0^1 nx^n,mathrm{d}x=frac{n}{n+1}.$$
answered 2 days ago
Xiangqian YangXiangqian Yang
1
answered 2 days ago
Xiangqian YangXiangqian Yang
1
answered 2 days ago
Xiangqian YangXiangqian Yang
1
answered 2 days ago
Xiangqian YangXiangqian Yang
1
1
add a comment |
add a comment |
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