To evaluate the integral $\int_{0}^{1} x e^{-x^2 + 1} \, dx$, we can use substitution and properties of exponential functions. Let's work through it step by step:

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Step 1: Simplify the exponential

Rewrite $e^{-x^2 + 1}$ as $e^{-x^2} \cdot e^{1}$, which simplifies the expression:

$\int_{0}^{1} x e^{-x^2 + 1} \, dx = e \int_{0}^{1} x e^{-x^2} \, dx$

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Step 2: Substitution

Let $u = -x^2$, which implies: $du = -2x \, dx \quad \text{or} \quad x \, dx = -\frac{1}{2} du$

When $x = 0$, $u = 0$. When $x = 1$, $u = -1$. The limits of integration change accordingly.

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Step 3: Transform the integral

Substituting $x \, dx = -\frac{1}{2} du$ and $e^{-x^2} = e^{u}$, we get: $e \int_{0}^{1} x e^{-x^2} \, dx = e \int_{0}^{-1} e^{u} \left(-\frac{1}{2}\right) du$

Rearrange the integral: $e \int_{0}^{1} x e^{-x^2} \, dx = -\frac{e}{2} \int_{0}^{-1} e^{u} \, du$

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Step 4: Evaluate the integral

The integral of $e^{u}$ is simply $e^{u}$: $-\frac{e}{2} \int_{0}^{-1} e^{u} \, du = -\frac{e}{2} \left[ e^{u} \right]_{0}^{-1}$

Compute the limits: [ -\frac{e}{2} \left[ e^{-1} - e^{0} \right] = -\frac{e}{2} \left[\frac{1}{e} - 1 \right] ]

Simplify: $-\frac{e}{2} \left[\frac{1}{e} - 1 \right] = -\frac{e}{2} \cdot \left(-1 + \frac{1}{e}\right) = \frac{e}{2} \cdot \left(1 - \frac{1}{e}\right)$

$= \frac{e}{2} \cdot \frac{e-1}{e} = \frac{e-1}{2}$

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Final Answer:

$\int_{0}^{1} x e^{-x^2 + 1} \, dx = \frac{e-1}{2}$

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Let me know if you'd like a more detailed breakdown or help with related integrals!

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Related Questions:

1. What is the significance of substitution in solving integrals like this one?
2. Can integration by parts be used for similar problems? Why or why not?
3. How does the presence of exponential functions affect the difficulty of integration?
4. What happens if the bounds of the integral change?
5. How can this integral be evaluated numerically if symbolic methods fail?

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Tip:

Always check if a substitution simplifies the integral by making the variable transformations clear and manageable.