Christmas Math Problem Set 2017 - Turtle-dove Curves

We can start by solving the equation $x^3 \ \sqrt{1 \ - \ x} \ = \ 0$ , and see that the graph of the turtle-dove's body is positive in the interval $[0, \ 1]$ , since these are the solutions to this equation, and therefore are the roots. We can then realize that this is a textbook example of the beta function. We know that:

\text{B}(y, \ z) \ = \ \displaystyle\int_{0}^{1} \ x^{y \ - \ 1} \ (1 \ - \ x)^{z \ - \ 1} \ \text{dx} \tag*{}

To find the area of this section of the graph, we can simply evaluate the integral. Using the relationship:

\text{B}(y, \ z) \ = \ \dfrac{\Gamma (y) \ \Gamma (z)}{\Gamma (y \ + \ z)} \tag*{}

We can solve this integral. We can say that for the integral:

\displaystyle\int_{0}^{1} \ x^3 \ \sqrt{1 \ - \ x} \ \text{dx} \ \Rightarrow \ y \ = \ 4, \ z \ = \ \frac{3}{2} \tag*{}

Then this can be written as $\text{B}(4, \ \frac{3}{2})$ . This means that:

\text{B}(4, \ \frac{3}{2}) \ = \ \dfrac{\Gamma (4) \ \Gamma (\frac{3}{2})}{\Gamma (\frac{11}{2})} \tag*{}

We know that $\Gamma (x) \ = \ (x \ - \ 1)!$ and $\Gamma (x \ + \ 1) \ = \ x \Gamma (x)$ . This means that we can start to simplify this expression. First of all, $\Gamma (4) \ = \ (4 \ - \ 1)! \ = \ 6$ . We can now use the reflection formula to find $\Gamma (\frac{3}{2})$ . The reflection formula is defined as:

\Gamma (x) \ \Gamma (1 \ - \ x) \ = \ \dfrac{\pi}{\sin (\pi x)} \tag*{}

We can begin to simplify $\Gamma (\frac{3}{2}$ by seeing that it is equal to $\frac{1}{2} \ \Gamma (\frac{1}{2})$ . We can then see that through the reflection formula:

\Gamma (\frac{1}{2}) \ \Gamma (\frac{1}{2}) \ = \ (\Gamma (\frac{1}{2}))^2 \ = \ \dfrac{\pi}{\sin (\frac{\pi}{2})} \ = \ \pi \tag*{}

This means that $\Gamma (\frac{1}{2}) \ = \ \sqrt{\pi}$ . Finally, we can use the formula, $\Gamma (x \ + \ 1) \ = \ x \Gamma (x)$ , to find the value of the denominator. We can continue to iterate using this formula to get: $\Gamma (\frac{11}{2}) \ = \ \frac{9}{2} \ \cdot \ \frac{9}{2} \ \cdot \ \frac{9}{2} \ \cdot \ \frac{9}{2} \ \cdot \ \frac{9}{2} \ \Gamma (\frac{1}{2})$ . We know that $\Gamma (\frac{1}{2}) \ = \ \sqrt{\pi}$ . We can evalute the product of all of these terms to obtain $\frac{945 \sqrt{\pi}}{32}$ . We can plug this back into the original fraction to get $\frac{96}{945}$ . Simplifying this, we get $\frac{32}{315}$ . Adding the numerator and the denominator, we get $347$ .

The problem asks for $A = \int_0^1 x^3\sqrt{1-x}\:dx.$ To simplify the integral, we try the substitution $u = \sqrt{1 - x}$ . This gives $du = -dx/2\sqrt{1-x}$ so that $dx = -2u\:du$ , and the limits of integration are flipped: $A = 2 \int_0^1 (1-u^2)^3 u^2 \:du.$ Expand this: $A = 2\int_0^1 (u^2 - 3u^4 + 3u^6 - u^8)\:du = 2\left[\frac13u^3 - \frac 35 u^5 + \frac 37 u^7 - \frac 19 u^9\right]_0^1 \\ = 2\left(\frac 13 - \frac 35 + \frac 37 - \frac 19\right) = 2\times\frac{105 - 189 + 135 - 35}{315} = \frac{32}{315}.$ We therefore post our answer as $32 + 315 = \boxed{347}$ .