https://r2.erweima.ai/mathbot/image/1725898906912f933xf1g.jpg

The integral given in the image is:

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

To solve this, we will use integration by parts, which is based on the formula:

\[
\int u \, dv = uv - \int v \, du
\]

For this problem, we can choose:
- \( u = x \) (so that \( du = dx \)),
- \( dv = e^{2x} dx \) (so that \( v = \frac{e^{2x}}{2} \), because the derivative of \( e^{2x} \) is \( 2e^{2x} \)).

Now, applying the formula:

\[
\int x e^{2x} \, dx = \left( x \cdot \frac{e^{2x}}{2} \right) - \int \frac{e^{2x}}{2} \, dx
\]

The second integral is straightforward:

\[
\int \frac{e^{2x}}{2} \, dx = \frac{1}{4} e^{2x}
\]

So the integral becomes:

\[
x \cdot \frac{e^{2x}}{2} - \frac{1}{4} e^{2x} + C
\]

Now we substitute the limits of integration from 0 to 1:

\[
\left[ \left( x \cdot \frac{e^{2x}}{2} - \frac{1}{4} e^{2x} \right) \right]_0^1
\]

At \( x = 1 \):

\[
\left( 1 \cdot \frac{e^{2}}{2} - \frac{1}{4} e^{2} \right) = \frac{e^2}{2} - \frac{e^2}{4} = \frac{e^2}{4}
\]

At \( x = 0 \):

\[
\left( 0 \cdot \frac{e^{0}}{2} - \frac{1}{4} e^{0} \right) = 0 - \frac{1}{4} = -\frac{1}{4}
\]

Now, subtract the result at \( x = 0 \) from the result at \( x = 1 \):

\[
\frac{e^2}{4} - \left( -\frac{1}{4} \right) = \frac{e^2}{4} + \frac{1}{4}
\]

Thus, the final result is:

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

Would you like more details, or do you have any questions?

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Here are 5 related questions that might interest you:

1. How does integration by parts work in other cases?
2. What are the common mistakes made when choosing \( u \) and \( dv \)?
3. Can this problem be solved using another method, like substitution?
4. How does the result change if the limits of integration are different?
5. How would this integral look if the exponent were \( 3x \) instead of \( 2x \)?

**Tip:** Integration by parts is often useful when you see a product of functions, especially when one function simplifies upon differentiation, like \( x \) here.

Learn how to solve the definite integral ∫₀¹ x e²ˣ dx using integration by parts. This detailed step-by-step solution covers key concepts of calculus, including the integration by parts formula and exponential function integration. Suitable for college-level calculus students.

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