Answer:
After reversing the order of integration and evaluating , we get:
[tex]\frac{e}{2}-1[/tex]
Correction in Problem Statement:
Since the problem statement doesn't contain the proper integral, so I am going to take the following integral and evaluate it by reversing the order of integration.
[tex]\int\limits^1_0\int\limits^z_x {e^{\frac{x}{y}}} \,dydx[/tex]
here, [tex]z=\sqrt{x}[/tex]
Step-by-step explanation:
As initially we are given:
[tex]\int\limits^1_0\int\limits^z_x {e^{\frac{x}{y}}} \,dydx[/tex]
here, [tex]z=\sqrt{x}[/tex]
Now reversing the order:
[tex]\int\limits^1_0\int\limits^y_m {e^{\frac{x}{y}}} \,dydx[/tex]
here, [tex]m=y^{2}[/tex]
=[tex]\int\limits^1_0({e^{\frac{x}{y}}}y)\left\|{{y}\atop{y^{2} }} \right.\,dy[/tex]
=[tex]\int\limits^1_0{(ey)-(e^{y}y) }\,dy[/tex]
For next part, we use integration by parts,
u=y,
[tex]dv=e^{y}[/tex]
So
du=1
[tex]v=e^{y}[/tex]
=[tex]((\frac{ey^{2} }{2})-(ye^{y}))\left||{{{1}\atop{0}}\right.+\int\limits^1_0 {e^{y}} \, dy[/tex]
=[tex]\frac{e}{2} -e+e-1=\frac{e}{2}-1[/tex]
Which is the integral evaluated by reversing the order of integration.
