Use a double integral to find the area of R.R is the triangle with vertices (-2, 1), (2, 1), and (0, -1).
A) $\int _ { x - - 2 } ^ { x - 2 } \int _ { y - - x - 1 } ^ { y - x - 1 } 1 d y d x = 4$

B) $\int _ { y -- 1 } ^ { y - 1 } \int _ { x - - y - 1 } ^ { x - y + 1 } 1 d x d y = 4$

C) $\int _ { x - - 2 } ^ { x - 2 } \int _ { y - - x - 1 } ^ { y - x - 1 } 1 d y d x = 8$

D) $\int _ { y - - 1 } ^ { y - 1 } \int _ { x- - y - 1 } ^ { x - y + 1 } 1 d x d y = 8$

Question

Use a double integral to find the area of R.R is the triangle with vertices (-2, 1), (2, 1), and (0, -1).
A) $\int _ { x - - 2 } ^ { x - 2 } \int _ { y - - x - 1 } ^ { y - x - 1 } 1 d y d x = 4$

B) $\int _ { y -- 1 } ^ { y - 1 } \int _ { x - - y - 1 } ^ { x - y + 1 } 1 d x d y = 4$

C) $\int _ { x - - 2 } ^ { x - 2 } \int _ { y - - x - 1 } ^ { y - x - 1 } 1 d y d x = 8$

D) $\int _ { y - - 1 } ^ { y - 1 } \int _ { x- - y - 1 } ^ { x - y + 1 } 1 d x d y = 8$

Quizplus · Accepted Answer

The correct setup for the double integral to find the area of the given triangle involves integrating over y first, from -1 to 1, and then over x, from the left line equation to the right line equation. The equations of the lines bounding the triangle can be derived from the given vertices, and the correct limits of integration reflect these equations. The area calculation does not depend on the order of integration, but the limits must correctly represent the region's boundaries. Option B correctly sets up these limits for integrating first with respect to x and then with respect to y, covering the entire triangular area.

Use a Double Integral to Find the Area of R ∫x−−2x−2∫y−−x−1y−x−11dydx=4\int _ { x - - 2 } ^ { x - 2 } \int _ { y - - x - 1 } ^ { y - x - 1 } 1 d y d x = 4∫x−−2x−2​∫y−−x−1y−x−1​1dydx=4

Use a Double Integral to Find the Area of R $\int _ { x - - 2 } ^ { x - 2 } \int _ { y - - x - 1 } ^ { y - x - 1 } 1 d y d x = 4$