Integrate the function f over the given region.
-$f ( x , y ) = \sqrt { x } + \sqrt { y }$ over the rectangle $0 \leq x \leq 1,0 \leq y \leq 1$
A) $\frac { 1 } { 3 }$
B) $\frac { 4 } { 3 }$
C) $\frac { 2 } { 3 }$
D) $\frac { 8 } { 3 }$

Question

Quizplus · Accepted Answer

To integrate $f(x, y) = \sqrt{x} + \sqrt{y}$ over the rectangle $0 \leq x \leq 1, 0 \leq y \leq 1$, we separate the integral into two parts due to the addition in the function:$$\int_{0}^{1} \int_{0}^{1} (\sqrt{x} + \sqrt{y}) \, dx \, dy = \int_{0}^{1} \int_{0}^{1} \sqrt{x} \, dx \, dy + \int_{0}^{1} \int_{0}^{1} \sqrt{y} \, dx \, dy$$Solving each integral separately:$$\int_{0}^{1} \sqrt{x} \, dx = \frac{2}{3}x^{3/2} \Big|_{0}^{1} = \frac{2}{3}$$$$\int_{0}^{1} \sqrt{y} \, dy = \frac{2}{3}y^{3/2} \Big|_{0}^{1} = \frac{2}{3}$$Since the integral of $\sqrt{y}$ with respect to $x$ over the interval $[0,1]$ is simply $y^{1/2}$ times the length of the interval (which is 1), the result is the same as integrating $\sqrt{y}$ over $[0,1]$. Therefore, the total integral is:$$\frac{2}{3} + \frac{2}{3} = \frac{4}{3}$$

Integrate the Function F Over the Given Region f(x,y)=x+yf ( x , y ) = \sqrt { x } + \sqrt { y }f(x,y)=x​+y​

Integrate the Function F Over the Given Region $f ( x , y ) = \sqrt { x } + \sqrt { y }$