Use a double integral to find the area of R. R is the region bounded by y = 3x,y = ln x,y = 0,and y = 1.
A) $\int _ { y - 0 } ^ { y - 1 } \int _ { x - \frac { y } { 3 } } ^ { x - e ^ { y } } 1 d x d y = e - \frac { 7 } { 6 }$
B) $\int _ { y - 0 } ^ { y - 1 } \int _ { x - 0 } ^ { x - e ^ { y } } 1 d x d y = e - 1$
C) $\int _ { x = 0 } ^ { x - e } \int _ { y - \ln x } ^ { y - 1 } 1 d y d x = e$
D) $\int _ { x = 0 } ^ { x - 2 } \int _ { y - \operatorname { m } x } ^ { y - 3 x } 1 d y d x = \frac { 3 e ^ { 2 } } { 2 }$

Question

Quizplus · Accepted Answer

The correct answer is A because the limits of integration for the inner integral are the x-coordinates of the left and right boundaries of R, and the limits of integration for the outer integral are the y-coordinates of the bottom and top boundaries of R.

Use a Double Integral to Find the Area of R ∫y−0y−1∫x−y3x−ey1dxdy=e−76\int _ { y - 0 } ^ { y - 1 } \int _ { x - \frac { y } { 3 } } ^ { x - e ^ { y } } 1 d x d y = e - \frac { 7 } { 6 }∫y−0y−1​∫x−3y​x−ey​1dxdy=e−67​

Use a Double Integral to Find the Area of R $\int _ { y - 0 } ^ { y - 1 } \int _ { x - \frac { y } { 3 } } ^ { x - e ^ { y } } 1 d x d y = e - \frac { 7 } { 6 }$