Question

Math

Posted 2 months ago

```
Evaluate the double integral.
$\int_{1}^{\ln (2)} \int_{0}^{e^{x}}-2 x+2 y d y d x=$
Choose 1 answer:
(A) $2 e^{2}+2 e-3+\ln (2)$
(B) $2-3 e$
(C) $6-4 \ln (2)-\frac{e^{2}}{2}$
(D) $5+2 \ln (2)$
```

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Answer from Sia

Posted 2 months ago

Solution by Steps

step 2

The inner integral is $\int_{0}^{e^{x}} -2x + 2y \, dy$

step 3

Integrating with respect to $y$, we get $-2xy + y^2$ evaluated from $0$ to $e^{x}$

step 4

Plugging in the limits of integration, we obtain $-2xe^{x} + (e^{x})^2 - (0)$

step 5

Simplifying, we have $-2xe^{x} + e^{2x}$

step 6

Now, we integrate the result of the inner integral with respect to $x$ from $1$ to $\ln(2)$

step 7

The outer integral is $\int_{1}^{\ln(2)} (-2xe^{x} + e^{2x}) \, dx$

step 8

Integrating term by term, we find the antiderivatives: $-2xe^{x}$ integrates to $-(2xe^{x} - 2e^{x})$ and $e^{2x}$ integrates to $\frac{1}{2}e^{2x}$

step 9

Evaluating these antiderivatives from $1$ to $\ln(2)$, we substitute the limits into the expressions

step 10

After substituting the limits and simplifying, we find the exact value of the double integral

C

Key Concept

Double Integration

Explanation

The key concept is to perform double integration by first integrating the inner integral with respect to $y$, and then integrating the result with respect to $x$. Each step involves finding the antiderivative, evaluating at the limits of integration, and simplifying.

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