# Under between the x- and y-axis of the graph of $y=x^n$

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.

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 $$3 \displaystyle\int_{1}^{a}x^ndx = n \displaystyle\int_{1}^{a}x.x^{n-1}dx$$ $$n=3$$

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 $$a > 1$$.

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

Now solve from this equations for $$a$$ .