To evaluate the integral

$\int_{0}^{1} -\frac{2y}{e^{2y}} \, dy,$

we can first simplify the expression and then use integration by parts.

Step 1: Simplify the Integral

Rewrite the integral as:

$\int_{0}^{1} -2y e^{-2y} \, dy.$

Step 2: Integration by Parts

To apply integration by parts, we set:

• Let $u = y$ so that $du = dy$.
• Let $dv = -2e^{-2y} \, dy$ so that $v = \int -2e^{-2y} \, dy$.

Calculating $v$:

$v = \int -2e^{-2y} \, dy = e^{-2y}.$

Now, using the integration by parts formula:

$\int u \, dv = uv - \int v \, du,$

we have:

$u = y, \quad dv = -2e^{-2y} \, dy, \quad du = dy, \quad v = e^{-2y}.$

Substituting into the formula gives:

$\int_{0}^{1} -2y e^{-2y} \, dy = \left[y e^{-2y}\right]_0^1 - \int_{0}^{1} e^{-2y} \, dy.$

Step 3: Evaluate Each Term

1. Evaluate $\left[y e^{-2y}\right]_0^1$:

   $\left[ y e^{-2y} \right]_0^1 = (1 \cdot e^{-2}) - (0 \cdot e^0) = e^{-2}.$

2. Evaluate $\int_{0}^{1} e^{-2y} \, dy$:

   $\int e^{-2y} \, dy = -\frac{1}{2} e^{-2y}.$

   Thus,

   $\left[ -\frac{1}{2} e^{-2y} \right]_0^1 = -\frac{1}{2} \left( e^{-2} - e^0 \right) = -\frac{1}{2} \left( e^{-2} - 1 \right).$

Step 4: Combine Results

$\int_{0}^{1} -2y e^{-2y} \, dy = e^{-2} - \left( -\frac{1}{2} (e^{-2} - 1) \right).$

Simplifying:

$= e^{-2} + \frac{1}{2} - \frac{1}{2} e^{-2}.$

$= \frac{1}{2} e^{-2} + \frac{1}{2}.$

$= \frac{1}{2} (e^{-2} + 1).$

Final Answer

$\int_{0}^{1} -\frac{2y}{e^{2y}} \, dy = \frac{1}{2} (e^{-2} + 1).$

Would you like more details, or do you have any questions? Here are some related questions to consider:

1. How does the choice of $u$ and $dv$ affect the integration by parts process?
2. What are alternative methods for integrating similar types of integrals?
3. How would you solve this integral if the limits of integration were different?
4. What are some common mistakes to avoid when using integration by parts?
5. How do you interpret the result of this integral in a physical or real-world context?

Tip: When using integration by parts, carefully choosing $u$ and $dv$ can simplify the problem significantly. Functions that simplify when differentiated (like polynomials) are often good choices for $u$.