Evaluate the iterated integral by choosing the order of integration.
1

0
2

1
xe^(x + 4y) dy dx

To evaluate the iterated integral, we can choose the order of integration to simplify the computation. The given integral is:

\[ \int_{0}^{1} \int_{1}^{2} x e^{x + 4y} \, dy \, dx \]

Let's switch the order of integration. First, we need to determine the new bounds of integration. The original bounds are \( x \) from \( 0 \) to \( 1 \) and \( y \) from \( 1 \) to \( 2 \). We can express these bounds in the following way:

1. \( y \) goes from \( 1 \) to \( 2 \).
2. For a fixed \( y \), \( x \) goes from \( 0 \) to \( 1 \).

Now, the iterated integral becomes:

\[ \int_{1}^{2} \int_{0}^{1} x e^{x + 4y} \, dx \, dy \]

We evaluate the inner integral first:

\[ \int_{0}^{1} x e^{x + 4y} \, dx \]

Let's use integration by parts to solve this. Set \( u = x \) and \( dv = e^{x + 4y} dx \). Then, we have:

\[ du = dx \quad \text{and} \quad v = \int e^{x + 4y} dx = e^{x + 4y} \]

Using the integration by parts formula \(\int u \, dv = uv - \int v \, du\), we get:

\[ \int_{0}^{1} x e^{x + 4y} \, dx = \left[ x e^{x + 4y} \right]_{0}^{1} - \int_{0}^{1} e^{x + 4y} \, dx \]

Evaluate the boundary terms:

\[ \left[ x e^{x + 4y} \right]_{0}^{1} = 1 \cdot e^{1 + 4y} - 0 \cdot e^{0 + 4y} = e^{1 + 4y} \]

Now, we need to evaluate the remaining integral:

\[ \int_{0}^{1} e^{x + 4y} \, dx \]

Since \( e^{x + 4y} \) can be separated as \( e^{x} e^{4y} \), and \( e^{4y} \) is a constant with respect to \( x \), we get:

\[ \int_{0}^{1} e^{x + 4y} \, dx = e^{4y} \int_{0}^{1} e^{x} \, dx \]

Evaluate the integral \( \int_{0}^{1} e^{x} \, dx \):

\[ \int_{0}^{1} e^{x} \, dx = \left[ e^{x} \right]_{0}^{1} = e^{1} - e^{0} = e - 1 \]

So,

\[ \int_{0}^{1} e^{x + 4y} \, dx = e^{4y} (e - 1) \]

Putting it all together:

\[ \int_{0}^{1} x e^{x + 4y} \, dx = e^{1 + 4y} - e^{4y} (e - 1) \]

Simplify:

\[ \int_{0}^{1} x e^{x + 4y} \, dx = e^{1 + 4y} - (e - 1) e^{4y} = e^{4y} (e - 1) \]

Now, we evaluate the outer integral:

\[ \int_{1}^{2} e^{4y} (e - 1) \, dy \]

Factor out the constant:

\[ (e - 1) \int_{1}^{2} e^{4y} \, dy \]

Evaluate \( \int_{1}^{2} e^{4y} \, dy \):

\[ \int_{1}^{2} e^{4y} \, dy = \frac{1}{4} e^{4y} \Big|_{1}^{2} = \frac{1}{4} (e^{8} - e^{4}) \]

Thus,

\[ (e - 1) \cdot \frac{1}{4} (e^{8} - e^{4}) = \frac{e - 1}{4} (e^{8} - e^{4}) \]

Therefore, the value of the iterated integral is:

\[ \boxed{\frac{e - 1}{4} (e^{8} - e^{4})} \]

Would you like more details or have any questions? Here are some related questions to expand your understanding:

1. How does the order of integration affect the ease of solving an iterated integral?
2. What are some common techniques for changing the order of integration in double integrals?
3. Can you explain the integration by parts technique used in solving integrals?
4. How does the exponential function property \( e^{a + b} = e^{a} e^{b} \) simplify integration?
5. What are some applications of iterated integrals in physics and engineering?

**Tip:** When dealing with double integrals, always consider if changing the order of integration might simplify the problem.

Evaluate the iterated integral by choosing the order of integration: ∫ from 0 to 1 ∫ from 1 to 2 xe^(x + 4y) dy dx

Discover how to evaluate an iterated integral using integration by parts and the properties of exponential functions. This detailed solution demonstrates switching the order of integration and solving step-by-step for the iterated integral ∫ from 0 to 1 ∫ from 1 to 2 xe^(x + 4y) dy dx. Perfect for college-level calculus students.

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