Find the area of the region specified by the integral(s).
-$\int _ { 0 } ^ { 3 } \int _ { 3 - x } ^ { 4 - x } d y d x + \int _ { 3 } ^ { 4 } \int _ { 0 } ^ { 4 - x } d y d x$
A) $\frac { 25 } { 2 }$
B) 8
C) $\frac { 9 } { 2 }$
D) $\frac { 7 } { 2 }$

Question

Quizplus · Accepted Answer

The region can be visualized as a right triangle with vertices at (0,0), (3,0) and (4,1). Hence, we can split the region into two triangles and find their areas separately.
The first integral represents the area of the triangle with vertices (0,0), (3,0) and (4,-1). The limits of integration for this integral are y = 3-x to y = 4-x and x = 0 to x = 3. Hence, we can write:
$$\int _ { 0 } ^ { 3 } \int _ { 3 - x } ^ { 4 - x } d y d x = \int _ { 0 } ^ { 3 } (4-2x) dx = 6$$
The second integral represents the area of the triangle with vertices (3,0), (4,-1) and (4,0). The limits of integration for this integral are y = 0 to y = 4-x and x = 3 to x = 4. Hence, we can write:
$$\int _ { 3 } ^ { 4 } \int _ { 0 } ^ { 4 - x } d y d x = \int _ { 3 } ^ { 4 } (4-x) dx = 5$$
Therefore, the total area of the region is 6+5 = 11, which is equivalent to option D.

Find the Area of the Region Specified by the Integral(s) ∫03∫3−x4−xdydx+∫34∫04−xdydx\int _ { 0 } ^ { 3 } \int _ { 3 - x } ^ { 4 - x } d y d x + \int _ { 3 } ^ { 4 } \int _ { 0 } ^ { 4 - x } d y d x∫03​∫3−x4−x​dydx+∫34​∫04−x​dydx

Find the Area of the Region Specified by the Integral(s) $\int _ { 0 } ^ { 3 } \int _ { 3 - x } ^ { 4 - x } d y d x + \int _ { 3 } ^ { 4 } \int _ { 0 } ^ { 4 - x } d y d x$