Q. 14 - Calculus

Question

Reversing the order of integration: Sketch the region determined by the limits of the given iterated integrals, and then evaluate the integrals by reversing the order of integration.

${\int }_0^{16}{\int }_{\sqrt[4]{x}}^2\frac{1}{1+y^5}dydx$

Step-by-Step Solution

Answer

${\int }_0^{y^4}{\int }_0^2\frac{1}{1+y^5}dydx=\frac{1}{5}\ln 33$

Step 1: Draw the region

The region determined by the limits of the given iterated integral is shown below,  

Step 2: Reversing the order of integration

${\int }_0^{16}{\int }_{\sqrt[4]{x}}^2\frac{1}{1+y^5}dydx \Rightarrow {\int }_0^{y^4}{\int }_0^2\frac{1}{1+y^5}dydx$

Step 3: Evaluate the integral

$$\begin{gathered} I={\int }_0^{y^4}dx{\int }_0^2\frac{1}{1+y^5}dy \\ I={\int }_0^2\frac{y^4}{1+y^5}dy \end{gathered}$$

Substitute,

 $$\begin{gathered} 1+y^5=t \\ 5y^{4}dy=dt \\ dy=\frac{1}{5y^4}dt \\ \mathrm{When} y=0, t=1+0=1 \\ \mathrm{When} y=2, t=1+2^5=33 \end{gathered}$$

Therefore,

$$\begin{gathered} I=\frac{1}{5}{\int }_1^{33}\frac{1}{t}dt \\ I=\frac{1}{5}{\left[\ln t\right]}_1^{33} \\ I=\frac{1}{5}\ln 33 \end{gathered}$$