Evaluate the integral: J = integral x³e^(-x²) dx
Evaluate the integral: $$ J = \int x^3e^{-x^2}\,dx. $$
Explanation of Method
Substituting $u=x^2$ converts $x^3\,dx$ into $\tfrac12u\,du$ — a plain polynomial in $u$, not a fractional power — which reduces the whole integral to an elementary polynomial-times-exponential form resolved by a single integration by parts.
- Substitute $u=x^2$ and express $x^3\,dx$ in terms of $u$.
- Apply integration by parts to $\int ue^{-u}\,du$.
- Substitute back and simplify.
Step 1 · Substitution
Let $u=x^2$, so $du=2x\,dx$. Then $x^3\,dx = x^2\cdot x\,dx = u\cdot\dfrac{du}{2}$, giving:
$$ J = \int x^3e^{-x^2}\,dx = \frac12\int ue^{-u}\,du. $$
Step 2 · Integration by parts
Let $v=u$, $dw=e^{-u}du$, so $dv=du$, $w=-e^{-u}$:
$$ \int ue^{-u}\,du = -ue^{-u} + \int e^{-u}\,du = -ue^{-u} - e^{-u}. $$
Step 3 · Combine
$$ \frac12\int ue^{-u}\,du = -\frac12\left(u+1\right)e^{-u} + C. $$
Step 4 · Substitute back
Substituting back $u=x^2$:
$$ J = -\frac12\left(x^2+1\right)e^{-x^2} + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $u=x^2$ |
| $x^3\,dx$ in terms of $u$ | $\tfrac12u\,du$ |
| Antiderivative | $-\tfrac12(u+1)e^{-u}$ |
Additional Notes
- The integrand is defined for all real $x$.
- A common pitfall is miscomputing $x^3\,dx$ in terms of $u=x^2$: since $x=\sqrt u$, it's tempting to write $x^3=u^{3/2}$ and $dx=du/(2\sqrt u)$ separately, which (if combined incorrectly) can produce a fractional power like $u^{1/2}$ instead of the correct plain $u$. Grouping $x^3\,dx=x^2\cdot(x\,dx)=u\cdot\tfrac12du$ directly avoids this and keeps the integral elementary — no incomplete Gamma function is needed here, since the odd power of $x$ makes the substitution exact.
- The result can be checked by differentiating $-\tfrac12(x^2+1)e^{-x^2}$: this gives $-xe^{-x^2}+x(x^2+1)e^{-x^2}=x^3e^{-x^2}$, matching the integrand.
Result
The solution to the integral is: