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: $\int_{0}^{e^{x}}(-2x + 2y) dy$. ‖ ‖ step 2 ⋮ The integral of a constant $-2x$ with respect to $y$ is $-2xy$, and the integral of $2y$ with respect to $y$ is $y^2$. Evaluating from $0$ to $e^x$ gives $-2xe^x + e^{2x}$. ‖ ‖ step 3 ⋮ Now, we integrate the result of step 2 with respect to $x$ from $1$ to $\ln(2)$: $\int_{1}^{\ln(2)}(-2xe^x + e^{2x}) dx$. ‖ ‖ step 4 ⋮ The integral of $-2xe^x$ with respect to $x$ is $-2(e^x + xe^x)$, and the integral of $e^{2x}$ with respect to $x$ is $\frac{1}{2}e^{2x}$. Evaluating from $1$ to $\ln(2)$ gives the final result. ‖ ‖ step 5 ⋮ Substitute the limits into the antiderivatives and subtract the lower limit from the upper limit to find the value of the double integral. ‖ ***C*** ∻Key Concept∻ ⚹Double Integration⚹ ∻Explanation∻ ⚹The key concept in solving this problem is performing double integration, which involves integrating with respect to one variable first (holding the other constant), and then integrating the result with respect to the second variable.⚹◊What is the process for evaluating a double integral in this given problem⍭ Generate me a similar question◊

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