Find the area bounded by the curve y = e^(-x²)(x + x³)

Find the area bounded by the curve $y = e^{-x^2}(x + x^3)$ and its asymptote.

We will use the curve's odd symmetry to reduce the problem to two standard Gaussian-type integrals.

Step 1: Find the asymptote

As $x\to\pm\infty$, $e^{-x^2}$ decays faster than any polynomial grows, so $y=e^{-x^2}(x+x^3)\to0$. The asymptote is $y=0$.

Step 2: Analyze the curve's sign

\[y = xe^{-x^2}(1+x^2)\]

For $x>0$: $y>0$. For $x<0$: $y<0$. At $x=0$: $y=0$.

Step 3: Recognize the odd symmetry

Since $y$ contains only odd powers of $x$ multiplied by the even function $e^{-x^2}$: $y(-x)=-y(x)$, so $y$ is an odd function. By symmetry, the lobe below the axis (for $x<0$) has exactly the same area as the lobe above the axis (for $x>0$):

\[A = 2\int_0^{\infty} y\,dx = 2\int_0^{\infty} e^{-x^2}(x+x^3)\,dx\]

Step 4: Evaluate $\int_0^\infty xe^{-x^2}\,dx$

Let $u=x^2$, $du=2x\,dx$:

\[\int_0^{\infty} xe^{-x^2}\,dx = \frac12\int_0^{\infty} e^{-u}\,du = \frac12\]

Step 5: Evaluate $\int_0^\infty x^3e^{-x^2}\,dx$

Writing $x^3e^{-x^2}=x^2\cdot xe^{-x^2}$ and substituting $u=x^2$:

\[\int_0^{\infty} x^3e^{-x^2}\,dx = \frac12\int_0^{\infty} ue^{-u}\,du = \frac12\Gamma(2) = \frac12\]

Step 6: Calculate the total area

\[A = 2\left(\frac12+\frac12\right) = 2\]

\[\boxed{A = 2 \text{ square units}}\]

Summary

\[\boxed{A = 2 \text{ square units}}\]

Key insight: The curve oscillates in sign around the $x$-axis; the exponential decay $e^{-x^2}$ tames the polynomial factor enough that the total enclosed area converges to the clean value $2$, illustrating the balance between polynomial growth and Gaussian decay.