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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) 2e2+2e3+ln(2)2 e^{2}+2 e-3+\ln (2)
(B) 23e2-3 e
(C) 64ln(2)e226-4 \ln (2)-\frac{e^{2}}{2}
(D) 5+2ln(2)5+2 \ln (2)
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Answer from Sia
Posted 2 months ago
Solution by Steps
step 2
The inner integral is 0ex2x+2ydy\int_{0}^{e^{x}} -2x + 2y \, dy
step 3
Integrating with respect to yy, we get 2xy+y2-2xy + y^2 evaluated from 00 to exe^{x}
step 4
Plugging in the limits of integration, we obtain 2xex+(ex)2(0)-2xe^{x} + (e^{x})^2 - (0)
step 5
Simplifying, we have 2xex+e2x-2xe^{x} + e^{2x}
step 6
Now, we integrate the result of the inner integral with respect to xx from 11 to ln(2)\ln(2)
step 7
The outer integral is 1ln(2)(2xex+e2x)dx\int_{1}^{\ln(2)} (-2xe^{x} + e^{2x}) \, dx
step 8
Integrating term by term, we find the antiderivatives: 2xex-2xe^{x} integrates to (2xex2ex)-(2xe^{x} - 2e^{x}) and e2xe^{2x} integrates to 12e2x\frac{1}{2}e^{2x}
step 9
Evaluating these antiderivatives from 11 to ln(2)\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
Key Concept
Double Integration
The key concept is to perform double integration by first integrating the inner integral with respect to yy, and then integrating the result with respect to xx. Each step involves finding the antiderivative, evaluating at the limits of integration, and simplifying.

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