Determine the average value of f(x,y) = ex+y over the region R=[0,2]X[0,2].
Let f(x,y)=ex+yf(x,y)=e^{x+y}f(x,y)=ex+y and R=[0,2]×[0,2]R=[0,2]\times [0,2]R=[0,2]×[0,2] .
The average value of fff over RRR is favg=1AreaR∬Rf(x,y)dAf_\text{avg}=\frac{1}{\text{Area}_R}\iint _R f(x,y)dAfavg=AreaR1∬Rf(x,y)dA .
AreaR=22=4\text{Area}_R=2^2=4AreaR=22=4
∬Rf(x,y)dA=∫02∫02ex+ydxdy=∫02∫02ex⋅eydxdy=∫02exdx⋅∫02eydy=ex∣02 ⋅ ey∣02=(e2−1)2\iint _R f(x,y)dA=\int\limits_0^2\int\limits_0^2e^{x+y}dxdy=\int\limits_0^2 \int\limits_0^2 e^{x}\cdot e^ydx dy= \int\limits_0^2e^{x}dx \cdot \int\limits_0^2e^ydy=e^x\big|^2_0\ \cdot \ e^y\big|_0^2=(e^2-1)^2∬Rf(x,y)dA=0∫20∫2ex+ydxdy=0∫20∫2ex⋅eydxdy=0∫2exdx⋅0∫2eydy=ex∣∣02 ⋅ ey∣∣02=(e2−1)2
Answer: favg=(e2−1)24f_{\text{avg}}=\frac{(e^2-1)^2}{4}favg=4(e2−1)2 .
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