Evaluate the integral: integral x e^(-x²) dx
Evaluate the integral: $$ \int x e^{-x^2} \, dx. $$
Explanation of Method
The factor of $x$ is proportional to the derivative of the exponent $-x^2$, so the substitution $u=x^2$ converts the exponential integrand into $e^{-u}$, a standard antiderivative.
- Substitute $u=x^2$ to match the exponent.
- Convert $x\,dx$ into a constant multiple of $du$.
- Integrate the resulting exponential in $u$.
- Substitute back and add the constant of integration.
Step 1 · Substitution
We solve this integral using the method of substitution. Let:
$$ u = x^2. $$
Then, differentiate both sides with respect to $x$:
$$ \frac{du}{dx} = 2x \quad \Rightarrow \quad du = 2x\,dx. $$
Thus, $x\,dx = \dfrac{du}{2}$.
Step 2 · Substitute and integrate
Now, substitute into the integral:
$$ \int x e^{-x^2}\,dx = \int e^{-u}\cdot \frac{du}{2} = \frac{1}{2}\int e^{-u}\,du. $$
The integral of $e^{-u}$ is straightforward:
$$ \int e^{-u}\,du = -e^{-u}. $$
Thus, the integral becomes:
$$ \frac{1}{2}\left(-e^{-u}\right) = -\frac{1}{2}e^{-u}. $$
Step 3 · Substitute back
Finally, substitute back $u = x^2$ to return to the variable $x$:
$$ -\frac{1}{2}e^{-x^2} + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $u = x^2$ |
| $x\,dx$ in terms of $du$ | $\tfrac12 du$ |
| Antiderivative | $-\tfrac12 e^{-u}$ |
Additional Notes
- The integrand is defined for all real $x$, since exponential functions are defined everywhere.
- This is the classic "Gaussian-type" substitution — any integrand of the form $x\cdot f(x^2)$ becomes $\tfrac12 f(u)$ after substituting $u=x^2$.
- The result can be checked by differentiating $-\tfrac12 e^{-x^2}$ with the chain rule and confirming it returns $xe^{-x^2}$.
Result
The solution to the integral is: