question 1\nevaluate\n\\(\\int_{0}^{\\ln 1}\\int_{\\ln 1}^{0}e^{x}\\sin(e^{y})e^{y}dxdy\\)\n\\(\\bigcirc…

Explanation:

Step1: Evaluate the inner - integral with respect to (x)

First, integrate (\int_{\ln 1}^{0}e^{x}\sin(e^{y})e^{y}dx). Since (\sin(e^{y})e^{y}) is treated as a constant with respect to (x), and (\int e^{x}dx = e^{x}+C). Then (\int_{\ln 1}^{0}e^{x}\sin(e^{y})e^{y}dx=\sin(e^{y})e^{y}\int_{\ln 1}^{0}e^{x}dx). We know that (\int_{\ln 1}^{0}e^{x}dx=e^{x}\big|_{\ln 1}^{0}=e^{0}-e^{\ln 1}=1 - 1=0).

Step2: Evaluate the outer - integral

Now we have (\int_{0}^{\ln 1}\left(\int_{\ln 1}^{0}e^{x}\sin(e^{y})e^{y}dx\right)dy=\int_{0}^{\ln 1}0dy = 0).

Answer:

D. 0