Let f ( x , y ) = e x + y f(x,y)=e^{x+y} f ( x , y ) = e x + y and R = [ 0 , 2 ] × [ 0 , 2 ] R=[0,2]\times [0,2] R = [ 0 , 2 ] × [ 0 , 2 ] .
The average value of f f f over R R R is f avg = 1 Area R ∬ R f ( x , y ) d A f_\text{avg}=\frac{1}{\text{Area}_R}\iint _R f(x,y)dA f avg = Area R 1 ∬ R f ( x , y ) d A .
Area R = 2 2 = 4 \text{Area}_R=2^2=4 Area R = 2 2 = 4
∬ R f ( x , y ) d A = ∫ 0 2 ∫ 0 2 e x + y d x d y = ∫ 0 2 ∫ 0 2 e x ⋅ e y d x d y = ∫ 0 2 e x d x ⋅ ∫ 0 2 e y d y = e x ∣ 0 2 ⋅ e y ∣ 0 2 = ( e 2 − 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 ∬ R f ( x , y ) d A = 0 ∫ 2 0 ∫ 2 e x + y d x d y = 0 ∫ 2 0 ∫ 2 e x ⋅ e y d x d y = 0 ∫ 2 e x d x ⋅ 0 ∫ 2 e y d y = e x ∣ ∣ 0 2 ⋅ e y ∣ ∣ 0 2 = ( e 2 − 1 ) 2
Answer: f avg = ( e 2 − 1 ) 2 4 f_{\text{avg}}=\frac{(e^2-1)^2}{4} f avg = 4 ( e 2 − 1 ) 2 .
Comments