Infinite Slices, Finite Values

The area bounded by the curve $f(x)$ , the $y$ - axis, the vertical assymptote for $f(x)$ (which is $(x = 25)$ ) and the $x$ - axis is: $\int_{0}^{25} \frac{x}{\sqrt{25 -x}} dx =$ Integrating by parts $= \left[-2x \cdot (\sqrt{25 - x})\right]_{0}^{25} + 2 \cdot \int_{0}^{25} \sqrt{25 -x} dx = 2 \cdot \int_{0}^{25} \sqrt{25 -x} dx =$ $= \left[\frac{-4}{3} \cdot (25 - x)^{\frac{3}{2}}\right]_0^{25} = \frac{500}{3} = A \Rightarrow 3A = \boxed{500}$

Required Area is:

$\displaystyle \int_{0}^{25} \frac{x}{\sqrt{25-x}} dx$

$\displaystyle \int_{0}^{25} \frac{x-25+25}{\sqrt{25-x}} dx$

$\displaystyle \int_{0}^{25} -1 \times \sqrt{25-x} + \frac{25}{\sqrt{25-x}} dx$

$\displaystyle \Rightarrow -1\times \frac{-1\times2}{3}\times \big( (\sqrt{25-x})^3 \big)_{0}^{25} + 25\times(- 1)\times2\times \big( \sqrt{25-x} \big)_{0}^{25}$

$\displaystyle \Rightarrow \frac{-250}{3} + 250$

$\displaystyle \Rightarrow A = \frac{500}{3}$

$\displaystyle \Rightarrow 3A = \boxed{500}$

The vertical asymptote of f(x) is found by setting the denominator to 0 and solving for values of x; it follows then that the area is bounded between x=0 and x=25.

Thus, our integral takes the following form:

$\displaystyle 3\int_0^{25} \frac{x}{ \sqrt{25 - x}} dx$

Using the substitution $u = 25 - x$ , we can state that $x = u - 25$ and $dx = - du$ . Thus we can transform the integral like so;

$\displaystyle -3\int_{25}^{0} \frac{25-u}{ \sqrt{u}} du = \displaystyle -3\int_{25}^{0} 25u^{-\frac{1}{2}} - u^{\frac{2}{3}} du$

Solving this simple integral gives us the following;

$-3\left[ 50\sqrt{u} - \frac{2}{3}u^{\frac{3}{2}} \right]_{25}^0 \\ = -3(0 -( 250 - \frac{250}{3}) = 500$

Hence, the area is $\boxed{500}$ .

This is my first provided solution on Brilliant! Let me know what you think. :)