Calculate the area between the curve y = x³ e^(-x²)

Calculate the area between the curve $y = x^3 e^{-x^2}$ and its asymptote.

We will use odd symmetry and a substitution reducing the tail integral to a Gamma-function form.

Step 1: Find the asymptote

As $x\to\pm\infty$: $y=x^3e^{-x^2}\to0$ (exponential decay dominates polynomial growth). The asymptote is $y=0$.

Step 2: Analyze the function

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

Symmetry: $y(-x) = (-x)^3e^{-x^2} = -x^3e^{-x^2} = -y(x)$. The function is odd, so the curve is symmetric about the origin.

Step 3: Set up the area integral

\[A = \int_{-\infty}^{\infty} |y| \, dx = 2\int_0^{\infty} x^3 e^{-x^2} \, dx\]

Step 4: Evaluate the integral

Let $u=x^2$, $du=2x\,dx$, so $x\,dx=\frac12du$ and $x^3\,dx = x^2\cdot x\,dx = \frac{u}{2}du$. When $x=0$: $u=0$; when $x=\infty$: $u=\infty$.

\[\int_0^{\infty} x^3 e^{-x^2} \, dx = \frac{1}{2}\int_0^{\infty} u e^{-u} \, du\]

Step 5: Evaluate $\int_0^\infty ue^{-u}\,du$

By parts (or recognizing $\Gamma(2)=1!=1$):

\[\int_0^{\infty} u e^{-u} \, du = [-u e^{-u}]_0^{\infty} + \int_0^{\infty} e^{-u} \, du = 0 + 1 = 1\]

Step 6: Calculate the total area

\[\int_0^{\infty} x^3 e^{-x^2} \, dx = \frac{1}{2}\cdot1 = \frac12, \qquad A = 2\cdot\frac12 = 1\]

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

Summary

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

Key insight: Despite mixing polynomial and exponential terms, $y=x^3e^{-x^2}$ encloses exactly $1$ square unit with its asymptote; the odd symmetry ensures equal contributions from either side of the origin.