Q. 13 - 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.

${\int }_0^9{\int }_{\sqrt{y}}^3\frac{1}{1+x^3}dxdy$

Step-by-Step Solution

Answer

${\int }_0^{x^2}{\int }_0^3\frac{1}{1+x^3}dxdy=\frac{1}{3}\ln 28$

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

From the above diagram, reversing the order of integration. 

${\int }_0^9{\int }_{\sqrt{y}}^3\frac{1}{1+x^3}dxdy \Rightarrow {\int }_0^{x^2}{\int }_0^3\frac{1}{1+x^3}dxdy$

Step 3: Evaluate the integral

$$\begin{gathered} I={\int }_0^{x^2}{\int }_0^3\frac{1}{1+x^3}dxdy \\ I={\int }_0^{x^2}dy{\int }_0^3\frac{1}{1+x^3}dx \\ I={\int }_0^3\frac{x^2}{1+x^3}dx \end{gathered}$$

Substitute,

$$\begin{gathered} 1+x^3=t \\ 3x^{2}dx=dt \\ dx=\frac{1}{3x^2}dt \\ \mathrm{When} x=0, t=1+0=1 \\ \mathrm{When} x=3, t=1+3^3=28 \end{gathered}$$

Therefore,

$$\begin{gathered} I=\frac{1}{3}{\int }_1^{28}\frac{1}{t}dt \\ I=\frac{1}{3}{\left[\ln t\right]}_1^{28} \\ I=\frac{1}{3}\ln 28 \end{gathered}$$