Write an equivalent double integral with the order of integration reversed.
-$\int _ { 0 } ^ { 1 } \int _ { 0 } ^ { \tan ^ { - 1 } x } d y d x$
A) $\int _ { 0 } ^ { \pi / 4 } \int _ { \tan y } ^ { \pi / 2 } d x d y$
B) $\int _ { 0 } ^ { \pi / 4 } \int _ { \tan ^ { - 1 } y } ^ { \pi / 2 } d x d y$
C) $\int _ { 0 } ^ { \pi / 4 } \int _ { \tan y } ^ { 1 } d x d y$
D) $\int _ { 0 } ^ { \pi / 4 } \int _ { \tan ^ { - 1 } y } ^ { 1 } d x d y$

Question

Quizplus · Accepted Answer

The original integral has $x$ ranging from 0 to 1 and $y$ ranging from 0 to $\tan^{-1}(x)$. When reversing the order of integration, $y$ becomes the outer variable, ranging from 0 to $\pi/4$ (since $\tan^{-1}(1) = \pi/4$). For each $y$, $x$ ranges from $\tan(y)$ to 1, making option C correct.

Write an Equivalent Double Integral with the Order of Integration ∫01∫0tan⁡−1xdydx\int _ { 0 } ^ { 1 } \int _ { 0 } ^ { \tan ^ { - 1 } x } d y d x∫01​∫0tan−1x​dydx

Write an Equivalent Double Integral with the Order of Integration $\int _ { 0 } ^ { 1 } \int _ { 0 } ^ { \tan ^ { - 1 } x } d y d x$