Calculate the iterated integral: $\int_0^1 \int_1^4 \frac{2xe^x}{y} \, dy \, dx$

Neetesh Kumar | November 29, 2024

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This is the solution to Math 1D
Assignment: 15.1 Question Number 9
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Step-by-step solution:

Step 1: Evaluate the Inner Integral

The inner integral is:

$\int_1^4 \frac{2xe^x}{y} \, dy$

Since $2xe^x$ is independent of $y$, we can factor it out of the integral:

$2xe^x \int_1^4 \frac{1}{y} \, dy$

The integral of $\frac{1}{y}$ is $\ln|y|$, so:

$2xe^x \left[ \ln|y| \right]_1^4 = 2xe^x \left( \ln(4) - \ln(1) \right)$

We know that $\ln(1) = 0$, so this simplifies to:

$2xe^x \cdot \ln(4)$

Thus, the result of the inner integral is:

$2xe^x \cdot \ln(4)$

Step 2: Evaluate the Outer Integral

Now, we evaluate the outer integral:

$\int_0^1 2xe^x \cdot \ln(4) \, dx$

Since $\ln(4)$ is a constant, we can factor it out of the integral:

$\ln(4) \int_0^1 2xe^x \, dx$

Evaluate the Integral $\int_0^1 2xe^x \, dx$

We can solve this using integration by parts. Let:

• $u = 2x$, so $du = 2 \, dx$
• $dv = e^x \, dx$, so $v = e^x$

The integration by parts formula is:

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

Substituting, we get:

$\int 2xe^x \, dx = 2xe^x - \int 2e^x \, dx$

The integral of $2e^x$ is:

$\int 2e^x \, dx = 2e^x$

Thus, we have:

$\int 2xe^x \, dx = 2xe^x - 2e^x = 2e^x(x - 1)$

Now, evaluate from $x = 0$ to $x = 1$:

$\left[ 2e^x(x - 1) \right]_0^1 = 2e^1(1 - 1) - 2e^0(0 - 1) = 0 - 2(-1) = 2$

Step 3: Final Calculation

Now, substitute this result back into the outer integral:

$\ln(4) \times 2 = 2 \ln(4)$

Final Answer:

The value of the iterated integral is:

$\boxed{2 \ln(4)}$

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