Write an equivalent double integral with the order of integration reversed.
-$$\int _ { 4 } ^ { 6 } \int _ { 4 } ^ { 10 - y } d x d y$$
A) $\int _ { 4 } ^ { 6 } \int _ { 6 } ^ { 10 - x } d y d x$
B) $\int _ { 4 } ^ { 2 } \int _ { 4 } ^ { 10 - x } d y d x$
C) $\int _ { 4 } ^ { 2 } \int _ { 6 } ^ { 10 - x } d y d x$
D) $\int _ { 4 } ^ { 6 } \int _ { 4 } ^ { 10 - x } d y d x$

Question

Quizplus · Accepted Answer

To reverse the order of integration, we need to change the limits of integration accordingly.
The limits on x go from 4 to 6, and the limits on y go from 4 to 10-x. This means that the region of integration is a triangular region above the line y=4 and below the line x+ y=10.
If we reverse the order of integration, we need to integrate over y first and then x. This means that the limits on y need to come first. The limits on y are from 4 to 10-x, and the limits on x are from 4 to 6. Thus, the equivalent double integral with the order of integration reversed is:
$$\int _ { 4 } ^ 6 \int _ { 4 } ^ {10-x} dydx$$

Write an Equivalent Double Integral with the Order of Integration ∫46∫410−ydxdy\int _ { 4 } ^ { 6 } \int _ { 4 } ^ { 10 - y } d x d y∫46​∫410−y​dxdy

Write an Equivalent Double Integral with the Order of Integration $\int _ { 4 } ^ { 6 } \int _ { 4 } ^ { 10 - y } d x d y$