Use a double integral to find the area of R. R is the region bounded by y = 9x, y = ln x, y = 0, and y = 1.
A) $\int _ { x = 0 } ^ { x - 2 } \int _ { y - lnx } ^ { y - 9 x } 1 d y d x = \frac { 9 e ^ { 2 } } { 2 }$

B) $\int _ { y = 0 } ^ { y - 1 } \int _ { x - \frac { y } { 9 } } ^ { x - e ^ { y } } 1 d x d y = e - \frac { 19 } { 18 }$

C) $\int _ { y - 0 } ^ { y - 1 } \int _ { x - 0 } ^ { x - e ^ { y } } 1 d x d y = e - 1$

D) $\int _ { x - 0 } ^ { x - 2 } \int _ { y - \ln x } ^ { y - 1 } 1 d y d x = e$

Question

Use a double integral to find the area of R. R is the region bounded by y = 9x, y = ln x, y = 0, and y = 1.
A) $\int _ { x = 0 } ^ { x - 2 } \int _ { y - lnx } ^ { y - 9 x } 1 d y d x = \frac { 9 e ^ { 2 } } { 2 }$

B) $\int _ { y = 0 } ^ { y - 1 } \int _ { x - \frac { y } { 9 } } ^ { x - e ^ { y } } 1 d x d y = e - \frac { 19 } { 18 }$

C) $\int _ { y - 0 } ^ { y - 1 } \int _ { x - 0 } ^ { x - e ^ { y } } 1 d x d y = e - 1$

D) $\int _ { x - 0 } ^ { x - 2 } \int _ { y - \ln x } ^ { y - 1 } 1 d y d x = e$

Quizplus · Accepted Answer

The correct integral setup to find the area of the region R, bounded by the given curves, involves integrating with respect to x first (inner integral) from the curve on the left ($x = \frac{y}{9}$) to the curve on the right ($x = e^y$), and then integrating with respect to y (outer integral) from the bottom boundary (y = 0) to the top boundary (y = 1). This setup correctly accounts for the area of the region described, using the given boundaries.

Use a Double Integral to Find the Area of R ∫x=0x−2∫y−lnxy−9x1dydx=9e22\int _ { x = 0 } ^ { x - 2 } \int _ { y - lnx } ^ { y - 9 x } 1 d y d x = \frac { 9 e ^ { 2 } } { 2 }∫x=0x−2​∫y−lnxy−9x​1dydx=29e2​

Use a Double Integral to Find the Area of R $\int _ { x = 0 } ^ { x - 2 } \int _ { y - lnx } ^ { y - 9 x } 1 d y d x = \frac { 9 e ^ { 2 } } { 2 }$