https://r2.erweima.ai/mathbot/image/1725900574781ve901s3u.jpg

The integral shown in the image is:

\[
\int_0^1 x e^{2x} \, dx
\]

This is a standard integration problem that can be solved using integration by parts.

### Step-by-step solution:

1. **Integration by Parts Formula**:
   \[
   \int u \, dv = uv - \int v \, du
   \]
   We need to choose parts of the integrand to assign as \( u \) and \( dv \). Let:
   - \( u = x \) (so that \( du = dx \))
   - \( dv = e^{2x} dx \) (so that \( v = \frac{1}{2}e^{2x} \), because the integral of \( e^{2x} \) is \( \frac{1}{2}e^{2x} \))

2. **Applying the formula**:
   \[
   \int_0^1 x e^{2x} dx = \left[ \frac{x}{2} e^{2x} \right]_0^1 - \int_0^1 \frac{1}{2} e^{2x} dx
   \]

3. **Evaluating the first term**:
   \[
   \left[ \frac{x}{2} e^{2x} \right]_0^1 = \frac{1}{2} e^2 - 0 = \frac{1}{2} e^2
   \]

4. **Evaluating the remaining integral**:
   \[
   \int_0^1 \frac{1}{2} e^{2x} dx = \frac{1}{2} \cdot \frac{1}{2} \left[ e^{2x} \right]_0^1 = \frac{1}{4} \left( e^2 - 1 \right)
   \]

5. **Final result**:
   \[
   \int_0^1 x e^{2x} dx = \frac{1}{2} e^2 - \frac{1}{4} (e^2 - 1) = \frac{1}{4} (e^2 + 1)
   \]

Thus, the value of the integral is:

\[
\boxed{\frac{1}{4}(e^2 + 1)}
\]

Would you like further details or have any questions?

Here are five related questions for further exploration:
1. How would the integration change if the limits were from 0 to infinity?
2. How can you verify the result of this integration using numerical methods?
3. What if the exponent were \( 3x \) instead of \( 2x \), how would the solution change?
4. How does integration by parts work for functions like \( x^2 e^{2x} \)?
5. What is the general formula for integrating \( x^n e^{kx} \) from 0 to 1?

**Tip**: When using integration by parts, always choose \( u \) to be a function that simplifies when differentiated and \( dv \) to be a function that is easy to integrate.

Evaluate the integral from 0 to 1 of x * e^(2x) dx.

Learn how to evaluate the integral ∫ x * e^(2x) dx from 0 to 1 using the integration by parts technique. This calculus problem explores key concepts in integration, including applying the formula and simplifying the result. Suitable for high school and college calculus students.

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