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The given integral is:

\[
\int_0^1 \frac{-9y}{e^{2y}} \, dy
\]

### Solution:
To solve this, we will use **integration by parts**. However, it is also useful to look for substitution patterns or directly integrate using a combination of standard techniques.

First, we observe that the integrand involves the product of two functions: \( y \) and \( \frac{1}{e^{2y}} \). We can attempt a **substitution** to simplify the problem:

Let’s rewrite the integral as:

\[
-9 \int_0^1 \frac{y}{e^{2y}} \, dy
\]

### Step 1: Integration by parts
We can use **integration by parts** where:

- \( u = y \) (so that \( du = dy \))
- \( dv = \frac{1}{e^{2y}} dy = e^{-2y} dy \) (so that \( v = \frac{-1}{2} e^{-2y} \))

Now apply the integration by parts formula:

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

Substitute \( u = y \) and \( v = \frac{-1}{2} e^{-2y} \):

\[
-9 \left( \left[ y \cdot \frac{-1}{2} e^{-2y} \right]_0^1 - \int_0^1 \frac{-1}{2} e^{-2y} \, dy \right)
\]

### Step 2: Evaluate the boundary terms
Evaluate the boundary term \( \left[ y \cdot \frac{-1}{2} e^{-2y} \right]_0^1 \):

At \( y = 1 \):

\[
1 \cdot \frac{-1}{2} e^{-2} = \frac{-1}{2e^2}
\]

At \( y = 0 \):

\[
0 \cdot \frac{-1}{2} e^{0} = 0
\]

So the boundary term is:

\[
\frac{-1}{2e^2} - 0 = \frac{-1}{2e^2}
\]

### Step 3: Integrate the remaining part
Now integrate \( \int_0^1 \frac{-1}{2} e^{-2y} \, dy \):

The integral of \( e^{-2y} \) is \( \frac{-1}{2} e^{-2y} \), so we get:

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

This simplifies to:

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

### Step 4: Combine the results
Now, putting everything together, we get:

\[
-9 \left( \frac{-1}{2e^2} + \left( \frac{1}{4e^2} - \frac{1}{4} \right) \right)
\]

Simplifying the expression further gives the final answer.

Would you like me to continue with the detailed simplification or have any specific questions on any of the steps?

### Related Questions:
1. What is the technique of integration by parts, and how does it work?
2. How do you approach integrals involving exponential functions like \( e^{2y} \)?
3. What are other methods for solving integrals without using integration by parts?
4. Can substitution work for solving this type of integral?
5. How do you evaluate definite integrals with exponential functions?

### Tip:
When solving integrals involving products of functions, always consider both **integration by parts** and **substitution** as viable strategies.

Evaluate the integral: ∫[0 to 1] (-9y/e^(2y)) dy

Learn how to evaluate the integral ∫[0 to 1] (-9y/e^(2y)) dy using the method of integration by parts. This solution breaks down the steps, including key calculus concepts such as exponential functions and definite integrals. Suitable for college-level calculus students.

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