Calculate the area bounded by the curve y = x/(sqrt(1

Calculate the area bounded by the curve $y = \frac{x}{\sqrt{1 - x^2}}$, its asymptote, and the positive part of the $x$-axis.

We will evaluate this as an improper integral near the vertical asymptote at $x=1$.

Step 1: Determine the domain

We need $1-x^2>0 \implies -1 < x < 1$. For the positive part of the $x$-axis, $x\in(0,1)$.

Step 2: Find the asymptote

As $x\to1^-$: $y=\frac{x}{\sqrt{1-x^2}}\to+\infty$, a vertical asymptote at $x=1$. As $x\to0^+$: $y\to0$; the curve passes through the origin.

Step 3: Set up the improper integral

\[A = \int_0^1 \frac{x}{\sqrt{1 - x^2}} \, dx = \lim_{b \to 1^-} \int_0^b \frac{x}{\sqrt{1 - x^2}} \, dx\]

Step 4: Evaluate the integral

Let $u=1-x^2$, $du=-2x\,dx$, so $x\,dx=-\frac12du$. When $x=0$: $u=1$; when $x=b$: $u=1-b^2$.

\[\int_0^b \frac{x}{\sqrt{1 - x^2}} \, dx = -\frac{1}{2}\int_1^{1-b^2} u^{-1/2} \, du = -\left[u^{1/2}\right]_1^{1-b^2} = 1 - \sqrt{1-b^2}\]

Step 5: Take the limit

\[A = \lim_{b \to 1^-} (1 - \sqrt{1-b^2}) = 1 - 0 = 1\]

\[\boxed{A = 1 \text{ square unit}}\]

Step 6: Verification via the direct antiderivative

\[\int \frac{x}{\sqrt{1 - x^2}} \, dx = -\sqrt{1 - x^2} + C\]

\[A = \left[-\sqrt{1 - x^2}\right]_0^1 = 0 - (-1) = 1 \quad \checkmark\]

Summary

\[\boxed{A = 1 \text{ square unit}}\]

Key insight: Despite a vertical asymptote at $x=1$, the curve encloses exactly $1$ square unit with the $x$-axis; its antiderivative $-\sqrt{1-x^2}$ is the equation of a circular arc, linking this area to the geometry of a unit semicircle.