y^2=x^3(1-x)^6 . How to find the area of this equation. Thank you in advance
Probably, you want to find the area between the curve and the x-axis.
This is $\displaystyle y=\pm\sqrt{x^3(1-x)^6}$
Above the x-axis it's $\displaystyle y=\sqrt{x^3(1-x)^6}$
This is zero when x=0 and when x=1. For negative x, this would be complex.
Beyond x=1, the curve travels to infinity.
Therefore, integrate $\displaystyle y=\sqrt{x^3(1-x)^6}$ from x=0 to x=1,
and double your result if you want to include the region below the x-axis.
You can start by simplifying the powers of x.