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) $5+2 \ln (2)$

Question

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) $5+2 \ln (2)$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ First, we integrate the inner integral with respect to $y$, keeping $x$ constant. ‖ ‖ 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.⚹◊What are the steps to evaluate a double integral with specified limits of integration for x and y variables?⍭ Generate me a similar question◊

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