Question

\(\int\limits_1^2 {\int\limits_{arctanx}^{\frac{\pi }{4}} {f(x,y)dy} dx} \)

Short answer

We should know how to do graphical representation.

Step-by-step solution

Given data:

On the region \(D\):

The variable \(y\) enters at \(y = {\tan ^{ - 1}}(x)\) and leaves at \(y = \frac{\pi }{4}\). And the variable \(x\) varies from \(x = 0\) to \(x = 1\). The region can be written as \(R = \left\{ {(x,y)|{{\tan }^{ - 1}}x \le y \le \frac{\pi }{4},0 \le x \le 1} \right\}\)

Graphical representation:

Now change the order of integration \(dxdy\).

Rewrite the equation \(y = {\tan ^{ - 1}}(x)\)as follows:

\(\begin{array}{l}y = {\tan ^{ - 1}}(x)\\x = \tan y\end{array}\)

When \(x = 0,\tan y = 0 \Rightarrow y = 0\)

When \(x = 1,\tan y = 1 \Rightarrow y = \frac{\pi }{4}\)

Region \(R\) is shown below:

Graph:

Double integral:

On the region \(D\):

The variable \(x\) enters at \(x = 0\)and leaves at \(x = \tan y\).

And the variable \(y\) varies from \(y = 0\) to \(y = \frac{\pi }{4}\).

The region can be written \(D = \left\{ {\left( {x,y} \right)/0 \le y \le \frac{\pi }{4},0 \le x \le \tan y} \right\}\).

Therefore, the double integral can be written as \(\int\limits_1^2 {\int\limits_{\arctan x}^{\frac{\pi }{4}} {f(x,y)dy} dx} = \int\limits_0^{\frac{\pi }{4}} {\int\limits_0^{\tan y} {f(x,y)dx} dy} \).