Q. 16 - 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.

${\int }_0^1{\int }_{\sqrt{x}}^1{e}^{\frac{x}{y^2}} dydx$

Step-by-Step Solution

Answer

${\int }_0^1{\int }_{\sqrt{x}}^1{e}^{\frac{x}{y^2}} dydx=\frac{1}{3}(e-1)$

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 diagram, the order of differentiation is changed,

${\int }_0^1{\int }_{\sqrt{x}}^1{e}^{\frac{x}{y^2}} dydx \Rightarrow {\int }_0^1{\int }_0^{y^2}{e}^{\frac{x}{y^2}} dydx$

Step 3: Evaluate the integral

$$\begin{gathered} I= {\int }_0^1\left[{\int }_0^{y^2}{e}^{\frac{x}{y^2}} dx\right]dy \\ I={\int }_0^1{\left[\frac{e^{{\displaystyle \frac{x}{y^2}}}}{1/y^2}\right]}_0^{y^2}dy \\ I={\int }_0^1{y}^2\left[e-1\right]dy \\ I=(e-1){\left[\frac{y^3}{3}\right]}_0^1 \\ I=\frac{1}{3}(e-1) \end{gathered}$$