Question

\(\int\limits_0^8 {\int\limits_{\sqrt[3]{y}}^2 {{e^{{x^4}}}dx} } dy\).

Short answer

The process of switching between \(dxdy\) order and \(dydx\) order in double integral is called changing the order of integration.

Step-by-step solution

Given Data.

Consider,

\(\int\limits_0^8 {\int\limits_{\sqrt[3]{y}}^2 {{e^{{x^4}}}dx} } dy\)

The inner limits of integration indicate that \(\sqrt[3]{y} \le x \le 2\). The outer limits of integration indicate that \(0 \le y \le 8\).

Since \(y = 8\) is the intersection of the limits on \(x\) it is an extraneous border to the region of integration.

Reverse the Order of Integration.

The limits of a vertical line passing through the region of integration are \(0 \le y \le {x^3}\). The extreme \(x\) values are \(0\) and \(2\).

Denote the desired integral I and reverse the order of integral.

\(\begin{array}{c}I = \int\limits_0^2 {\int\limits_0^{{x^3}} {{e^{{x^4}}}dy} dx} \\ = \int\limits_0^2 {{e^{{x^4}}}} \int\limits_0^{{x^3}} {dydx} \end{array}\)

Integrate on \(y\)

\(\begin{array}{c}I = \int\limits_0^2 {{e^{{x^4}}}} \left[ y \right]_0^{{x^3}}dx\\ = \int\limits_0^2 {{x^3}{e^{{x^4}}}} dx\end{array}\)

Put \(u = {x^4} \Rightarrow du = 4{x^3}dx\)

\( \Rightarrow {x^3}dx = \frac{1}{4}du\)

If \(x = 2\) then \(u = {2^4} = 16\)

If \(x = 0\) then \(u = 0\)

\(\begin{array}{c}I = \frac{1}{4}\left[ {{e^u}} \right]_0^{16}\\ = \frac{1}{4}\left( {{e^{16}} - 1} \right)\end{array}\)

Therefore, the value of integral is \(\frac{1}{4}\left( {{e^{16}} - 1} \right)\).