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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 $$Y=3\int _1 ^a x^n dx$$

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.

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Hint: $X = \displaystyle\int_{1}^{a}x^ndx = \displaystyle\int_{1}^{a^n}y^{1/n}dy = \displaystyle\frac{Y}{3}$

Substitute $y^{1/n} = x$ to arrive at $$3 \displaystyle\int_{1}^{a}x^ndx = n \displaystyle\int_{1}^{a}x.x^{n-1}dx$$ $$n=3$$

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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 $a > 1$.

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$$ \huge X=\frac {a^{n+1}-1}{n+1}=Y/3=\frac {a^{n+1}-1}{3+\frac {3}{n}}$$

Now solve from this equations for $a$ .

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