The area of a region R is given by the iterated integrals $\int _ { 0 } ^ { 2 x } 16 d y d x + \int _ { 2 } ^ { 4 } \int _ { 0 } ^ { 4 - x } 16 d y d x$ Switch the order of integration and show that both orders yield the same area. What is this area?
A) 32
B) 1025
C) 64
D) 771
E) 63

Question

The area of a region R is given by the iterated integrals $\int _ { 0 } ^ { 2 x } 16 d y d x + \int _ { 2 } ^ { 4 } \int _ { 0 } ^ { 4 - x } 16 d y d x$ Switch the order of integration and show that both orders yield the same area. What is this area? 
A) 32
B) 1025
C) 64
D) 771
E) 63

Quizplus · Accepted Answer

The area of the region R can be calculated using the given iterated integrals. First, we evaluate the given integrals as they are:1. $\int _ { 0 } ^ { 2 } 16 d y d x = 16y \Big|_{0}^{2x} = 32x$, then $\int _ { 0 } ^ { 2 } 32x dx = 32 \times \frac{x^2}{2} \Big|_{0}^{2} = 32 \times 2 = 64$.2. $\int _ { 2 } ^ { 4 } \int _ { 0 } ^ { 4 - x } 16 d y d x = 16y \Big|_{0}^{4-x} = 16(4-x)$, then $\int _ { 2 } ^ { 4 } 16(4-x) dx = 16(4x - \frac{x^2}{2}) \Big|_{2}^{4} = 16(8 - 8 - (8 - 2)) = 0$.Adding both parts, the total area is $64 + 0 = 64$.To switch the order of integration, we need to understand the region R. The first integral covers a triangular region from x=0 to x=2 with y going from 0 to 2x, and the second integral covers a trapezoidal region from x=2 to x=4 with y going from 0 to 4-x. Switching the order of integration, we consider y as the outer variable. For y=0 to y=4, x varies differently in two segments: from y=0 to y=4, x goes from y/2 to 4-y for the combined region.The new integral becomes $\int _ { 0 } ^ { 4 } \int _ { y/2 } ^ { 4-y } 16 d x d y$. Evaluating this:1. For the inner integral, $16x \Big|_{y/2}^{4-y} = 16(4-y) - 16(y/2) = 64 - 24y$.2. For the outer integral, $\int _ { 0 } ^ { 4 } (64 - 24y) dy = 64y - 12y^2 \Big|_{0}^{4} = 256 - 192 = 64$.Both orders of integration yield the same area of 64, confirming that the area of the region R is 64.

The Area of a Region R Is Given by the Iterated