To solve the integral

$I = \int_0^1 \frac{y}{e^{2y}} \, dy$

using integration by parts, we follow the standard integration by parts formula:

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

Step 1: Choose $u$ and $dv$

Let:

• $u = y$ (since it simplifies nicely when differentiated),
• $dv = \frac{1}{e^{2y}} \, dy$ (since it integrates neatly).

Now we compute the derivatives and integrals:

• $du = dy$,
• To find $v$, integrate $\frac{1}{e^{2y}}$:

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

Step 2: Apply the integration by parts formula

Now, applying the formula:

$I = uv \bigg|_0^1 - \int_0^1 v \, du$

Substitute $u = y$, $v = -\frac{1}{2} e^{-2y}$, and $du = dy$:

$I = \left[ y \left( -\frac{1}{2} e^{-2y} \right) \right]_0^1 - \int_0^1 \left( -\frac{1}{2} e^{-2y} \right) \, dy$

Step 3: Evaluate the boundary terms

First, evaluate $y \left( -\frac{1}{2} e^{-2y} \right)$ at the limits 0 and 1:

= -\frac{1}{2} e^{-2}$$ ### Step 4: Evaluate the remaining integral Now, for the remaining integral: $$-\int_0^1 -\frac{1}{2} e^{-2y} \, dy = \frac{1}{2} \int_0^1 e^{-2y} \, dy$$ To integrate $$e^{-2y}$$, we have: $$\int e^{-2y} \, dy = \frac{1}{-2} e^{-2y}$$ Thus: $$\frac{1}{2} \int_0^1 e^{-2y} \, dy = \frac{1}{2} \left[ \frac{1}{-2} e^{-2y} \right]_0^1 = \frac{1}{2} \left( \frac{1}{-2} e^{-2} - \frac{1}{-2} e^{0} \right)$$ $$= \frac{1}{2} \left( -\frac{1}{2} e^{-2} + \frac{1}{2} \right) = \frac{1}{4} \left( 1 - e^{-2} \right)$$ ### Step 5: Combine the results Now, combining both terms: $$I = -\frac{1}{2} e^{-2} + \frac{1}{4} \left( 1 - e^{-2} \right)$$ Simplify: $$I = -\frac{1}{2} e^{-2} + \frac{1}{4} - \frac{1}{4} e^{-2}$$ $$I = \frac{1}{4} - \frac{3}{4} e^{-2}$$ Thus, the solution to the integral is: $$I = \frac{1}{4} - \frac{3}{4} e^{-2}$$ Would you like more details on any step, or have any questions about integration by parts? Here are some related questions: 1. How can we apply integration by parts when one of the terms is a product of trigonometric functions? 2. Can we solve integrals involving exponential functions without integration by parts? 3. How do we decide which term to choose as $$u$$ in integration by parts? 4. What happens if we choose $$v = y$$ and $$dv = e^{-2y}$$? 5. How can you verify this result numerically using a calculator? **Tip**: A good strategy for choosing $$u$$ in integration by parts is to pick the term that simplifies when you differentiate it, like polynomials or logarithmic terms.