finding the area using integration?

**takermaster35** · February 23rd 2010, 04:14 AM

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

**NonCommAlg** · February 23rd 2010, 04:59 AM

Originally Posted by takermaster35

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

$2 \int_0^1 x \sqrt{x} (1-x)^3 \ dx.$

**Archie Meade** · February 23rd 2010, 05:11 AM

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.

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