Reverse the order of integration for the following integral. $\int _ { 1 } ^ { 3 } \int _ { x ^ { 2 } - 9 } ^ { 4 x - 12 } f ( x , y ) d y d x$
A) $\int _ { 0 } ^ { 8 } \int _ { ( y + 12 ) / 4 } ^ { \sqrt { y + 9 } } f ( x , y ) d x d y$
B) $\int _ { - 8 } ^ { 0 } \int _ { ( y + 12 ) / 4 } ^ { \sqrt { y + 9 } } f ( x , y ) d y d x$
C) $\int _ { - 8 } ^ { 0 } \int _ { ( y + 12 ) / 4 } ^ { \sqrt { y + 9 } } f ( x , y ) d x d y$
D) $\int _ { - 8 } ^ { 0 } \int _ { \sqrt { y + 9 } } ^ { ( y + 12 ) / 4 } f ( x , y ) d x d y$
E) $\int _ { - 8 } ^ { 0 } \int _ { ( y - 12 ) / 4 } ^ { \sqrt { y + 9 } } f ( x , y ) d x d y$

Question

Quizplus · Accepted Answer

To reverse the order of integration, we first identify the bounds for $y$ based on the given $x$ bounds. The original integral goes from $x^2 - 9$ to $4x - 12$ in $y$, for $x$ from 1 to 3. Solving $x^2 - 9 = 0$ gives $x = 3$, and $4x - 12 = 0$ gives $x = 3$, but these are the points where the functions intersect, not the $y$-bounds. The actual $y$-bounds are found by evaluating the $y$-functions at the $x$-bounds: $1^2 - 9 = -8$ and $4(3) - 12 = 0$, giving the $y$-range as $[-8, 0]$.Next, we solve for $x$ in terms of $y$ from the original bounds: $y = x^2 - 9$ gives $x = \sqrt{y + 9}$, and $y = 4x - 12$ gives $x = (y + 12)/4$. Therefore, the correct reversed order of integration is from $y = -8$ to $y = 0$, and for $x$, from $(y + 12)/4$ to $\sqrt{y + 9}$, which matches choice C.

Reverse the Order of Integration for the Following Integral ∫13∫x2−94x−12f(x,y)dydx\int _ { 1 } ^ { 3 } \int _ { x ^ { 2 } - 9 } ^ { 4 x - 12 } f ( x , y ) d y d x∫13​∫x2−94x−12​f(x,y)dydx

Reverse the Order of Integration for the Following Integral $\int _ { 1 } ^ { 3 } \int _ { x ^ { 2 } - 9 } ^ { 4 x - 12 } f ( x , y ) d y d x$