Solve the problem.
-Write
$$\int _ { 0 } ^ { 5 } \int _ { 0 } ^ { y } f ( x , y ) d x d y + \int _ { 5 } ^ { 10 } \int _ { 0 } ^ { 10 - y } f ( x , y ) d x d y + \int _ { 0 } ^ { 5 } \int _ { y } ^ { 5 } f ( x , y ) d x d y$$
as a single iterated integral.
A) $\int _ { 0 } ^ { 5 } \int _ { 5 } ^ { 10 - x } f ( x , y ) d y d x$
B) $\int _ { - 5 } ^ { 5 } \int _ { 5 } ^ { 10 - x } f ( x , y ) d y d x$
C) $\int _ { - 5 } ^ { 5 } \int _ { x } ^ { 10 - x } f ( x , y ) d y d x$
D) $\int _ { 0 } ^ { 5 } \int _ { 0 } ^ { 10 - x } f ( x , y ) d y d x$

Question

Quizplus · Accepted Answer

First, let's look at the limits of integration for each of the three integrals given in the problem:

1. $\int _ { 0 } ^ { 5 } \int _ { 0 } ^ { y } f ( x , y ) d x d y$: Integrating over the region below the line $y=x$ and above the x-axis, from x=0 to x=5 and from y=0 to y=x.

2. $\int _ { 5 } ^ { 10 } \int _ { 0 } ^ { 10 - y } f ( x , y ) d x d y$: Integrating over the region above the line y=10-x and below the line x=10, from x=5 to x=10 and from y=0 to y=10-x.

3. $\int _ { 0 } ^ { 5 } \int _ { y } ^ { 5 } f ( x , y ) d x d y$: Integrating over the region above the line x=y and below the line x=5, from x=y to x=5 and from y=0 to y=5.

To write a single iterated integral that covers the entire region, we need to find one set of limits of integration that encompasses all three regions. Looking at the given limits, we can see that the x-limits are always 0 to 5 or 5 to 10. We can split the integral into two parts based on x limits, and then use the given y-limits for each part.

So, we can write the iterated integral as:

$$\int _ { 0 } ^ { 10 } \int _ { \ldots } ^ { \ldots } f ( x , y ) d y d x$$

To find the missing limits, we can draw the lines x=0, x=5, and x=10, and see how they split the given regions.

Region 1 is above x=0 and below x=5, so its limits are y=0 to y=x. 
Region 2 is above x=5 and below x=10, so its limits are y=0 to y=10-x. 
Region 3 is between x=0 and x=5, so its limits are y=x to y=5.

Putting it all together, we get:

$$\int _ { 0 } ^ { 10 } \int _ { 0 } ^ { x } f ( x , y ) d y d x + \int _ { 0 } ^ { 10 } \int _ { 10-x } ^ { 5 } f ( x , y ) d y d x$$

This is equivalent to the integral in choice D, $\int _ { 0 } ^ { 5 } \int _ { 0 } ^ { 10 - x } f ( x , y ) d y d x$.

Solve the Problem As a Single Iterated Integral ∫05∫510−xf(x,y)dydx\int _ { 0 } ^ { 5 } \int _ { 5 } ^ { 10 - x } f ( x , y ) d y d x∫05​∫510−x​f(x,y)dydx

Solve the Problem As a Single Iterated Integral $\int _ { 0 } ^ { 5 } \int _ { 5 } ^ { 10 - x } f ( x , y ) d y d x$