# Math Help - finding the area using integration?

1. ## finding the area using integration?

y^2=x^3(1-x)^6 . How to find the area of this equation. Thank you in advance

2. Originally Posted by takermaster35
y^2=x^3(1-x)^6 . How to find the area of this equation. Thank you in advance
$2 \int_0^1 x \sqrt{x} (1-x)^3 \ dx.$

3. Originally Posted by takermaster35
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 $y=\pm\sqrt{x^3(1-x)^6}$

Above the x-axis it's $y=\sqrt{x^3(1-x)^6}$

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 $y=\sqrt{x^3(1-x)^6}$ 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.