Find the area between line and curve Point $A$ lies on the curve $g(x) = x$ And lies on a value $x > 0$. From point $A$ a line is drawn parallel to the $y$-axis. The second drawn graph is $f(x)=x^3$.
Calculate the coordinates of point $A$ so that the area of ​​$B$ and $C$ will be equal.

x must be less than 1, according to this picture, right? But I don't know how to get the point A.

Should I use this formula?
$$A_{C-B}=\int^{x}_{0} \pi x^{2}-\int^{x}_{0} \pi x^{6}$$, where $x<1$.
$$\int^{x}_{0} \pi x^{2}dx=\pi \int^{x}_{0} x^{2}dx=\pi \left[ \frac{x^{3}}{3} \right]^{x}_{0}  =\pi \left[ \left( \frac{x^{3}}{3} -\frac{0^{3}}{3} \right)  \right]  =\frac{\pi x^{3}}{3}$$
$$\int^{x}_{0} \pi x^{6}dx=\pi \int^{x}_{0} x^{6}dx=\pi \left[ \frac{x^{7}}{7} \right]^{x}_{0}  =\pi \left[ \left( \frac{x^{7}}{7} -\frac{0^{7}}{7} \right)  \right]  =\frac{\pi x^{7}}{7}$$

Let abscissa of $A$ be $A_x$.
We need to calculate coordinates of $A$ such that area of $B$ = area of $C$
$$\begin{align}\implies& \int_0^{A_x} x^3 \ dx = \int_0^{A_x} x - x^3\ dx\\\implies &\frac{(A_x)^4}{4 } = \frac{(A_x)^2}{2} - \frac{(A_x)^4}{4}\\\implies & \frac{(A_x)^4}{2} = \frac{(A_x)^2}{2}\\\implies & (A_x)^2(A_x^2 - 1)  = 0\end{align}$$
Thus $A_x$ is $1$ indeed.
Therefore coordinates of $A$ are $(1, 1).$

I don't agree with the graph. But the area of regions $B$ and $C$ are equal when coordinates of $A$ are $(1, 1).$
