Area with a Twist(Corrected)

Calculus Level 5

The area bounded by $y=f(x)$ , $x=\frac{1}{2}$ , $x=\frac{\sqrt{3}}{2},$ and the $x$ -axis is $A$ square units, where $f(x) = x+ \dfrac{2}{3}x^3 + \dfrac{2}{3}\cdot\dfrac{4}{5}x^5 + \dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}x^7+ \cdots$ and $|x| < 1.$

If $A$ can be written as $\dfrac{\pi^a}{b}$ , where $a$ and $b$ are positive integers, find $a + b$ .

The answer is 26.

1 solution

Kunal Gupta
May 15, 2015

Let's begin; Now, $f(x) = x+ \dfrac{2}{3}x^3 + \dfrac{2}{3}\cdot\dfrac{4}{5}x^5 + \dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}x^7+ \cdots \infty ; \\ f'(x) = 1+ x \left[2x + \dfrac{2}{3}.4x^3 + \dfrac{2}{3}\dfrac{4} {5}6x^5+\cdots\right] \\ =1+ x\dfrac{d}{dx}(xf(x)) = 1+ xf(x) +x^2 f'(x)\\ (1-x^2)f'(x)=1+xf(x);$ Also $f(0)=0$ Now,by solving the L.D.E we get

$f(x)=\dfrac{\sin^{-1} x}{\sqrt{1-x^2}}$

Now Area is trivial , it comes out to be $\dfrac{\pi^2}{24}$ So, $a+b = \boxed{26}$ $\large{ \mathbb{Q.E.D}}$

the answers can be infinite, you should make the terms as co prime integers , btw nice solution

Kyle Finch - 6 years, 1 month ago

Great solution! I couldn't think of it and had to use Beta Function. Damn!

Kartik Sharma - 5 years, 8 months ago

Area with a Twist(Corrected)

The answer is 26.

1 solution

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