Mixing
Find integration with limit from 1 to 2
$begingroup$
Find the integration 
my try :
Unable to solve further
integration
asked Jan 27 at 2:12
catttcattt
242
$endgroup$
add a comment |
$begingroup$
Find the integration
my try :
Unable to solve further
integration
asked Jan 27 at 2:12
catttcattt
242
$endgroup$
$begingroup$
It may help to note that $$int_0^1frac{te^{2t}}{e(1-t)+te^{2t}}dt=int_0^1frac{te^{2t}+e(1-t)}{e(1-t)+te^{2t}}dt+eint_0^1frac{t-1}{e(1-t)+te^{2t}}dt\=1+eint_0^1frac{t-1}{e(1-t)+te^{2t}}dt$$
$endgroup$
– clathratus
Jan 27 at 6:54
add a comment |
$begingroup$
Find the integration
my try :
Unable to solve further
integration
asked Jan 27 at 2:12
catttcattt
242
$endgroup$
Find the integration
my try :
Unable to solve further
integration
integration
asked Jan 27 at 2:12
catttcattt
242
asked Jan 27 at 2:12
catttcattt
242
asked Jan 27 at 2:12
catttcattt
242
asked Jan 27 at 2:12
catttcattt
242
asked Jan 27 at 2:12
catttcattt
242
242
$begingroup$
It may help to note that $$int_0^1frac{te^{2t}}{e(1-t)+te^{2t}}dt=int_0^1frac{te^{2t}+e(1-t)}{e(1-t)+te^{2t}}dt+eint_0^1frac{t-1}{e(1-t)+te^{2t}}dt\=1+eint_0^1frac{t-1}{e(1-t)+te^{2t}}dt$$
$endgroup$
– clathratus
Jan 27 at 6:54
add a comment |
$begingroup$
It may help to note that $$int_0^1frac{te^{2t}}{e(1-t)+te^{2t}}dt=int_0^1frac{te^{2t}+e(1-t)}{e(1-t)+te^{2t}}dt+eint_0^1frac{t-1}{e(1-t)+te^{2t}}dt\=1+eint_0^1frac{t-1}{e(1-t)+te^{2t}}dt$$
$endgroup$
– clathratus
Jan 27 at 6:54
$begingroup$
It may help to note that $$int_0^1frac{te^{2t}}{e(1-t)+te^{2t}}dt=int_0^1frac{te^{2t}+e(1-t)}{e(1-t)+te^{2t}}dt+eint_0^1frac{t-1}{e(1-t)+te^{2t}}dt\=1+eint_0^1frac{t-1}{e(1-t)+te^{2t}}dt$$
$endgroup$
– clathratus
Jan 27 at 6:54
$begingroup$
It may help to note that $$int_0^1frac{te^{2t}}{e(1-t)+te^{2t}}dt=int_0^1frac{te^{2t}+e(1-t)}{e(1-t)+te^{2t}}dt+eint_0^1frac{t-1}{e(1-t)+te^{2t}}dt\=1+eint_0^1frac{t-1}{e(1-t)+te^{2t}}dt$$
$endgroup$
– clathratus
Jan 27 at 6:54
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Being almost blind, I have problems trying to read your notes.
What it seems to me is that there is a symmetry around $x=frac 32$. So, let $x=y+frac 32$ to make the integrand to be
$$f(y)=frac{e^{2 y} (2 y+1)}{1-2 y+e^{2 y} (2 y+1)}$$ which makes
$$f(y)+f(-y)=1$$
So, you do not need integration at all. The result is the area of a right triangle the vertices of coordinates being $(-frac 12,0)$, $(frac 12,0)$, $(frac 12,1)$.
Now, a simple result.
Now, just behind you and me, there is no antiderivative to the function. I suppose that this problem is just a trap !
answered Jan 27 at 6:45
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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$begingroup$
Being almost blind, I have problems trying to read your notes.
What it seems to me is that there is a symmetry around $x=frac 32$. So, let $x=y+frac 32$ to make the integrand to be
$$f(y)=frac{e^{2 y} (2 y+1)}{1-2 y+e^{2 y} (2 y+1)}$$ which makes
$$f(y)+f(-y)=1$$
So, you do not need integration at all. The result is the area of a right triangle the vertices of coordinates being $(-frac 12,0)$, $(frac 12,0)$, $(frac 12,1)$.
Now, a simple result.
Now, just behind you and me, there is no antiderivative to the function. I suppose that this problem is just a trap !
answered Jan 27 at 6:45
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
add a comment |
$begingroup$
Being almost blind, I have problems trying to read your notes.
What it seems to me is that there is a symmetry around $x=frac 32$. So, let $x=y+frac 32$ to make the integrand to be
$$f(y)=frac{e^{2 y} (2 y+1)}{1-2 y+e^{2 y} (2 y+1)}$$ which makes
$$f(y)+f(-y)=1$$
So, you do not need integration at all. The result is the area of a right triangle the vertices of coordinates being $(-frac 12,0)$, $(frac 12,0)$, $(frac 12,1)$.
Now, a simple result.
Now, just behind you and me, there is no antiderivative to the function. I suppose that this problem is just a trap !
answered Jan 27 at 6:45
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
add a comment |
$begingroup$
Being almost blind, I have problems trying to read your notes.
What it seems to me is that there is a symmetry around $x=frac 32$. So, let $x=y+frac 32$ to make the integrand to be
$$f(y)=frac{e^{2 y} (2 y+1)}{1-2 y+e^{2 y} (2 y+1)}$$ which makes
$$f(y)+f(-y)=1$$
So, you do not need integration at all. The result is the area of a right triangle the vertices of coordinates being $(-frac 12,0)$, $(frac 12,0)$, $(frac 12,1)$.
Now, a simple result.
Now, just behind you and me, there is no antiderivative to the function. I suppose that this problem is just a trap !
answered Jan 27 at 6:45
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
Being almost blind, I have problems trying to read your notes.
What it seems to me is that there is a symmetry around $x=frac 32$. So, let $x=y+frac 32$ to make the integrand to be
$$f(y)=frac{e^{2 y} (2 y+1)}{1-2 y+e^{2 y} (2 y+1)}$$ which makes
$$f(y)+f(-y)=1$$
So, you do not need integration at all. The result is the area of a right triangle the vertices of coordinates being $(-frac 12,0)$, $(frac 12,0)$, $(frac 12,1)$.
Now, a simple result.
Now, just behind you and me, there is no antiderivative to the function. I suppose that this problem is just a trap !
answered Jan 27 at 6:45
Claude LeiboviciClaude Leibovici
124k1158135
answered Jan 27 at 6:45
Claude LeiboviciClaude Leibovici
124k1158135
answered Jan 27 at 6:45
Claude LeiboviciClaude Leibovici
124k1158135
answered Jan 27 at 6:45
Claude LeiboviciClaude Leibovici
124k1158135
124k1158135
add a comment |
add a comment |
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$begingroup$
It may help to note that $$int_0^1frac{te^{2t}}{e(1-t)+te^{2t}}dt=int_0^1frac{te^{2t}+e(1-t)}{e(1-t)+te^{2t}}dt+eint_0^1frac{t-1}{e(1-t)+te^{2t}}dt\=1+eint_0^1frac{t-1}{e(1-t)+te^{2t}}dt$$
$endgroup$
– clathratus
Jan 27 at 6:54