Find the area bounded by the curve y = ln(1 - x - 2x²)

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

We will find the domain and the zeros of $y$ separately — they are different! — which splits the region into three lobes, each evaluated in closed form via the factorization $1-x-2x^2=(1-2x)(1+x)$.

Step 1: Determine the domain

We need $1-x-2x^2>0$, i.e. $2x^2+x-1<0$. Factoring:

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

This product is negative when the factors have opposite signs, i.e. for $-1 < x < \frac12$. So the curve is defined only on $\left(-1,\frac12\right)$, with $y\to-\infty$ at both endpoints (the argument of the log tends to $0^+$).

Step 2: Find where the curve meets the $x$-axis

Setting $y=0$ means the argument equals $1$, not $0$:

\[1-x-2x^2=1 \implies -x-2x^2=0 \implies x(1+2x)=0\]

So $y=0$ at $x=0$ and $x=-\frac12$. These two points split the domain $\left(-1,\frac12\right)$ into three lobes: $\left(-1,-\frac12\right)$, $\left(-\frac12,0\right)$, and $\left(0,\frac12\right)$.

Step 3: Determine the sign of $y$ on each lobe

At $x=-\frac34$ (left lobe): $1-x-2x^2 = 1+\frac34-\frac98=\frac58<1$, so $y=\ln\frac58<0$.

At $x=-\frac14$ (middle lobe): $1-x-2x^2=1+\frac14-\frac18=\frac98>1$, so $y=\ln\frac98>0$.

At $x=\frac14$ (right lobe): $1-x-2x^2=1-\frac14-\frac18=\frac58<1$, so $y=\ln\frac58<0$.

So the curve rises above the $x$-axis only on the middle lobe $\left(-\frac12,0\right)$, and dips below it (down to $-\infty$ at the domain's edges) on the two outer lobes.

Step 4: Set up the three sub-areas

Each bounded region between the curve and the $x$-axis is evaluated with the elementary antiderivative $\int\ln u\,du = u\ln u-u+C$, using the factored form $\ln(1-x-2x^2) = \ln(1-2x)+\ln(1+x)$:

\[\int\ln(1-2x)\,dx = -\tfrac12(1-2x)\ln(1-2x)+\tfrac12(1-2x)+C, \qquad \int\ln(1+x)\,dx = (1+x)\ln(1+x)-(1+x)+C\]

Step 5: Middle lobe (the true bounded region above the axis)

\[A_{\text{mid}} = \int_{-1/2}^{0} \ln(1-x-2x^2)\,dx = \left[-\tfrac12(1-2x)\ln(1-2x)+\tfrac12(1-2x) + (1+x)\ln(1+x)-(1+x)\right]_{-1/2}^{0}\]

At $x=0$: $-\tfrac12(1)(0)+\tfrac12(1) + (1)(0)-(1) = \tfrac12-1=-\tfrac12$.

At $x=-\tfrac12$: $-\tfrac12(2)\ln2+\tfrac12(2) + \tfrac12\ln\tfrac12-\tfrac12 = -\ln2+1-\tfrac12\ln2-\tfrac12 = \tfrac12-\tfrac32\ln2$.

\[A_{\text{mid}} = -\tfrac12-\left(\tfrac12-\tfrac32\ln2\right) = \tfrac32\ln2-1\]

\[\boxed{A_{\text{mid}} = \frac{3\ln2-2}{2} \approx 0.0397 \text{ square units}}\]

Step 6: The two outer lobes (by symmetry of construction, both equal)

Carrying out the same antiderivative on $\left(-1,-\frac12\right)$ and $\left(0,\frac12\right)$ (each integral is negative, so the enclosed area is its absolute value) gives, after simplification:

\[A_{\text{left}} = A_{\text{right}} = 1-\frac32\ln\frac32 \approx 0.3918 \text{ square units}\]

(the two outer lobes are equal in area, a consequence of the specific coefficients in the quadratic).

Step 7: Total area of all three regions

\[A_{\text{total}} = A_{\text{mid}} + 2A_{\text{left}} = \left(\frac32\ln2-1\right) + 2\left(1-\frac32\ln\frac32\right) = 1 - 3\ln3 + \frac92\ln2 \approx 0.8233 \text{ square units}\]

Summary

The curve $y=\ln(1-x-2x^2)$ bounds three separate lobes with the $x$-axis. The single genuinely self-contained region — bounded only by the curve and the axis on both sides, with no dependence on the domain's endpoints — is the middle lobe:

\[\boxed{A_{\text{mid}} = \frac{3\ln2-2}{2} \approx 0.0397 \text{ square units}}\]

with the two (equal) outer lobes each contributing $1-\frac32\ln\frac32\approx0.3918$, for a combined three-lobe total of $1-3\ln3+\frac92\ln2\approx0.8233$.

Key insight: The key simplification is factoring $1-x-2x^2=(1-2x)(1+x)$, which turns a seemingly special-function-requiring logarithmic integral into two elementary pieces. Note that the zeros of $y$ (at $x=0,-\frac12$) are different from the zeros of the argument (at $x=-1,\frac12$) — conflating the two is an easy mistake that leads to an incomplete or wrong answer.