Find the area bounded by the arc of the curve y = (1

Find the area bounded by the arc of the curve $y = (1 - x^2)\ln(1 - x)$ and the positive part of the $x$-axis.

We will substitute $u=1-x$ to turn the integral into two standard "power times logarithm" integrals.

Step 1: Determine the domain and sign

The function $y=(1-x^2)\ln(1-x)$ requires $1-x>0$, i.e. $x<1$; we consider the "positive part of the $x$-axis" as $x\in[0,1)$. For $x\in(0,1)$: $1-x^2>0$ while $\ln(1-x)<0$, so $y<0$ throughout $(0,1)$. At $x=0$: $y=0$.

Step 2: Behavior as $x\to1^-$

Writing $1-x^2=(1-x)(1+x)$ and letting $u=1-x\to0^+$:

\[(1-x^2)\ln(1-x) = u(2-u)\ln u \sim 2u\ln u \to 0\]

(since $u\ln u\to0$ as $u\to0^+$), so the curve returns to the axis smoothly at $x=1$ — there is no singularity to worry about.

Step 3: Set up the area integral

Since the curve is below the axis on $(0,1)$:

\[A = \int_0^1 |y|\,dx = -\int_0^1 (1-x^2)\ln(1-x)\,dx\]

Step 4: Substitute $u = 1 - x$

With $u=1-x$ (so $x=1-u$, $dx=-du$, and the limits $x:0\to1$ become $u:1\to0$):

\[1-x^2 = (1-x)(1+x) = u(2-u) = 2u - u^2\]

\[\int_0^1 (1-x^2)\ln(1-x)\,dx = \int_0^1 (2u - u^2)\ln u \, du = 2\int_0^1 u\ln u\,du - \int_0^1 u^2\ln u\,du\]

Step 5: Use the standard result $\int_0^1 u^n\ln u\,du = -\frac{1}{(n+1)^2}$

This follows from integrating by parts with $v=\ln u$, $dw=u^n\,du$:

\[\int_0^1 u^n\ln u\,du = \left[\frac{u^{n+1}}{n+1}\ln u\right]_0^1 - \int_0^1 \frac{u^n}{n+1}\,du = 0 - \frac{1}{(n+1)^2} = -\frac{1}{(n+1)^2}\]

(the boundary term vanishes at both ends: $u^{n+1}\ln u\to0$ as $u\to0^+$, and $\ln1=0$ at $u=1$). Applying this with $n=1$ and $n=2$:

\[\int_0^1 u\ln u\,du = -\frac{1}{4}, \qquad \int_0^1 u^2\ln u\,du = -\frac{1}{9}\]

Step 6: Combine

\[\int_0^1 (1-x^2)\ln(1-x)\,dx = 2\left(-\frac14\right) - \left(-\frac19\right) = -\frac12+\frac19 = -\frac{7}{18}\]

Therefore:

\[A = -\left(-\frac{7}{18}\right) = \frac{7}{18}\]

\[\boxed{A = \frac{7}{18} \text{ square units}}\]

Step 7: Numerical value

\[A = \frac{7}{18} \approx 0.389 \text{ square units}\]

Verification

A direct numerical (Simpson's rule) estimate of $\int_0^1(1-x^2)\ln(1-x)\,dx$ gives approximately $-0.387$, matching $-\frac{7}{18}\approx-0.389$ closely. ✓

Summary

\[\boxed{A = \frac{7}{18} \approx 0.389 \text{ square units}}\]

Key insight: Substituting $u=1-x$ turns the integrand into a polynomial times $\ln u$ on $[0,1]$, where the standard formula $\int_0^1 u^n\ln u\,du=-\frac{1}{(n+1)^2}$ applies directly — sidestepping the divergent-looking terms a naive integration by parts in the original variable can produce.