Mixing
Under between the x- and y-axis of the graph of $y=x^n$
$begingroup$
Consider the graph of $y=x^n$ for $n>1$ and $x>1$. The area bound between the curve and the x-axis between $1$ and $a$ is one third the area between the curve and the y-axis between the values of $1$ and $a^n$.
If we let the area between the x-axis be $X$ and the area between the y-axis be $Y$.
$$X=int _1 ^a x^n dx qquad Y=int _1 ^{a^n} y^{1/n} dy$$
and as $Y=3X$ then
$$Y=3int _1 ^a x^n dx$$
So I've integrated both X and Y, but how do I go from here to get that the value of $n$ should be 3 as there seems to be too many variables in the way.
integration area
asked Jan 27 at 11:07
H.LinkhornH.Linkhorn
473213
$endgroup$
add a comment |
$begingroup$
Consider the graph of $y=x^n$ for $n>1$ and $x>1$. The area bound between the curve and the x-axis between $1$ and $a$ is one third the area between the curve and the y-axis between the values of $1$ and $a^n$.
If we let the area between the x-axis be $X$ and the area between the y-axis be $Y$.
$$X=int _1 ^a x^n dx qquad Y=int _1 ^{a^n} y^{1/n} dy$$
and as $Y=3X$ then
$$Y=3int _1 ^a x^n dx$$
So I've integrated both X and Y, but how do I go from here to get that the value of $n$ should be 3 as there seems to be too many variables in the way.
integration area
asked Jan 27 at 11:07
H.LinkhornH.Linkhorn
473213
$endgroup$
add a comment |
$begingroup$
Consider the graph of $y=x^n$ for $n>1$ and $x>1$. The area bound between the curve and the x-axis between $1$ and $a$ is one third the area between the curve and the y-axis between the values of $1$ and $a^n$.
If we let the area between the x-axis be $X$ and the area between the y-axis be $Y$.
$$X=int _1 ^a x^n dx qquad Y=int _1 ^{a^n} y^{1/n} dy$$
and as $Y=3X$ then
$$Y=3int _1 ^a x^n dx$$
So I've integrated both X and Y, but how do I go from here to get that the value of $n$ should be 3 as there seems to be too many variables in the way.
integration area
asked Jan 27 at 11:07
H.LinkhornH.Linkhorn
473213
$endgroup$
Consider the graph of $y=x^n$ for $n>1$ and $x>1$. The area bound between the curve and the x-axis between $1$ and $a$ is one third the area between the curve and the y-axis between the values of $1$ and $a^n$.
If we let the area between the x-axis be $X$ and the area between the y-axis be $Y$.
$$X=int _1 ^a x^n dx qquad Y=int _1 ^{a^n} y^{1/n} dy$$
and as $Y=3X$ then
$$Y=3int _1 ^a x^n dx$$
So I've integrated both X and Y, but how do I go from here to get that the value of $n$ should be 3 as there seems to be too many variables in the way.
integration area
integration area
asked Jan 27 at 11:07
H.LinkhornH.Linkhorn
473213
asked Jan 27 at 11:07
H.LinkhornH.Linkhorn
473213
asked Jan 27 at 11:07
H.LinkhornH.Linkhorn
473213
asked Jan 27 at 11:07
H.LinkhornH.Linkhorn
473213
asked Jan 27 at 11:07
H.LinkhornH.Linkhorn
473213
473213
add a comment |
add a comment |
3 Answers
3
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oldest
votes
$begingroup$
Hint: $X = displaystyleint_{1}^{a}x^ndx = displaystyleint_{1}^{a^n}y^{1/n}dy = displaystylefrac{Y}{3}$
Substitute $y^{1/n} = x$ to arrive at $$3 displaystyleint_{1}^{a}x^ndx = n displaystyleint_{1}^{a}x.x^{n-1}dx$$
$$n=3$$
answered Jan 27 at 11:29
Exp ikxExp ikx
4489
$endgroup$
add a comment |
$begingroup$
I'll answer to the question as I understand it.
Using the notation of the OP
Find $n$ such that $Y = 3X$.
Then, we have
$$
int_{1}^{a^{n}} y^{1/n} dy = int_{1}^{a} nz^{n} dz = n X
$$
where I set
$$z = y^{1/n} implies y = z^{n} implies dy = nz^{n-1} dz.
$$
Therefore, $nX = Y = 3X iff n = 3$ as
$$
X = int_{1}^{a} x^{n} dx = frac{a^{n+1}-1}{n} neq 0
$$
being $a > 1$.
answered Jan 27 at 11:20
FedericoFederico
918313
$endgroup$
add a comment |
$begingroup$
$$ huge X=frac {a^{n+1}-1}{n+1}=Y/3=frac {a^{n+1}-1}{3+frac {3}{n}}$$
Now solve from this equations for $a$ .
answered Jan 27 at 11:22
Bijayan RayBijayan Ray
136112
$endgroup$
add a comment |
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: $X = displaystyleint_{1}^{a}x^ndx = displaystyleint_{1}^{a^n}y^{1/n}dy = displaystylefrac{Y}{3}$
Substitute $y^{1/n} = x$ to arrive at $$3 displaystyleint_{1}^{a}x^ndx = n displaystyleint_{1}^{a}x.x^{n-1}dx$$
$$n=3$$
answered Jan 27 at 11:29
Exp ikxExp ikx
4489
$endgroup$
add a comment |
$begingroup$
Hint: $X = displaystyleint_{1}^{a}x^ndx = displaystyleint_{1}^{a^n}y^{1/n}dy = displaystylefrac{Y}{3}$
Substitute $y^{1/n} = x$ to arrive at $$3 displaystyleint_{1}^{a}x^ndx = n displaystyleint_{1}^{a}x.x^{n-1}dx$$
$$n=3$$
answered Jan 27 at 11:29
Exp ikxExp ikx
4489
$endgroup$
add a comment |
$begingroup$
Hint: $X = displaystyleint_{1}^{a}x^ndx = displaystyleint_{1}^{a^n}y^{1/n}dy = displaystylefrac{Y}{3}$
Substitute $y^{1/n} = x$ to arrive at $$3 displaystyleint_{1}^{a}x^ndx = n displaystyleint_{1}^{a}x.x^{n-1}dx$$
$$n=3$$
answered Jan 27 at 11:29
Exp ikxExp ikx
4489
$endgroup$
Hint: $X = displaystyleint_{1}^{a}x^ndx = displaystyleint_{1}^{a^n}y^{1/n}dy = displaystylefrac{Y}{3}$
Substitute $y^{1/n} = x$ to arrive at $$3 displaystyleint_{1}^{a}x^ndx = n displaystyleint_{1}^{a}x.x^{n-1}dx$$
$$n=3$$
answered Jan 27 at 11:29
Exp ikxExp ikx
4489
answered Jan 27 at 11:29
Exp ikxExp ikx
4489
answered Jan 27 at 11:29
Exp ikxExp ikx
4489
answered Jan 27 at 11:29
Exp ikxExp ikx
4489
4489
add a comment |
add a comment |
$begingroup$
I'll answer to the question as I understand it.
Using the notation of the OP
Find $n$ such that $Y = 3X$.
Then, we have
$$
int_{1}^{a^{n}} y^{1/n} dy = int_{1}^{a} nz^{n} dz = n X
$$
where I set
$$z = y^{1/n} implies y = z^{n} implies dy = nz^{n-1} dz.
$$
Therefore, $nX = Y = 3X iff n = 3$ as
$$
X = int_{1}^{a} x^{n} dx = frac{a^{n+1}-1}{n} neq 0
$$
being $a > 1$.
answered Jan 27 at 11:20
FedericoFederico
918313
$endgroup$
add a comment |
$begingroup$
I'll answer to the question as I understand it.
Using the notation of the OP
Find $n$ such that $Y = 3X$.
Then, we have
$$
int_{1}^{a^{n}} y^{1/n} dy = int_{1}^{a} nz^{n} dz = n X
$$
where I set
$$z = y^{1/n} implies y = z^{n} implies dy = nz^{n-1} dz.
$$
Therefore, $nX = Y = 3X iff n = 3$ as
$$
X = int_{1}^{a} x^{n} dx = frac{a^{n+1}-1}{n} neq 0
$$
being $a > 1$.
answered Jan 27 at 11:20
FedericoFederico
918313
$endgroup$
add a comment |
$begingroup$
I'll answer to the question as I understand it.
Using the notation of the OP
Find $n$ such that $Y = 3X$.
Then, we have
$$
int_{1}^{a^{n}} y^{1/n} dy = int_{1}^{a} nz^{n} dz = n X
$$
where I set
$$z = y^{1/n} implies y = z^{n} implies dy = nz^{n-1} dz.
$$
Therefore, $nX = Y = 3X iff n = 3$ as
$$
X = int_{1}^{a} x^{n} dx = frac{a^{n+1}-1}{n} neq 0
$$
being $a > 1$.
answered Jan 27 at 11:20
FedericoFederico
918313
$endgroup$
I'll answer to the question as I understand it.
Using the notation of the OP
Find $n$ such that $Y = 3X$.
Then, we have
$$
int_{1}^{a^{n}} y^{1/n} dy = int_{1}^{a} nz^{n} dz = n X
$$
where I set
$$z = y^{1/n} implies y = z^{n} implies dy = nz^{n-1} dz.
$$
Therefore, $nX = Y = 3X iff n = 3$ as
$$
X = int_{1}^{a} x^{n} dx = frac{a^{n+1}-1}{n} neq 0
$$
being $a > 1$.
answered Jan 27 at 11:20
FedericoFederico
918313
answered Jan 27 at 11:20
FedericoFederico
918313
answered Jan 27 at 11:20
FedericoFederico
918313
answered Jan 27 at 11:20
FedericoFederico
918313
918313
add a comment |
add a comment |
$begingroup$
$$ huge X=frac {a^{n+1}-1}{n+1}=Y/3=frac {a^{n+1}-1}{3+frac {3}{n}}$$
Now solve from this equations for $a$ .
answered Jan 27 at 11:22
Bijayan RayBijayan Ray
136112
$endgroup$
add a comment |
$begingroup$
$$ huge X=frac {a^{n+1}-1}{n+1}=Y/3=frac {a^{n+1}-1}{3+frac {3}{n}}$$
Now solve from this equations for $a$ .
answered Jan 27 at 11:22
Bijayan RayBijayan Ray
136112
$endgroup$
add a comment |
$begingroup$
$$ huge X=frac {a^{n+1}-1}{n+1}=Y/3=frac {a^{n+1}-1}{3+frac {3}{n}}$$
Now solve from this equations for $a$ .
answered Jan 27 at 11:22
Bijayan RayBijayan Ray
136112
$endgroup$
$$ huge X=frac {a^{n+1}-1}{n+1}=Y/3=frac {a^{n+1}-1}{3+frac {3}{n}}$$
Now solve from this equations for $a$ .
answered Jan 27 at 11:22
Bijayan RayBijayan Ray
136112
edited Jan 28 at 15:41
answered Jan 27 at 11:22
Bijayan RayBijayan Ray
136112
answered Jan 27 at 11:22
Bijayan RayBijayan Ray
136112
answered Jan 27 at 11:22
Bijayan RayBijayan Ray
136112
136112
add a comment |
add a comment |
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