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91Ó°ÊÓ

Evaluate the double integral over the specified region.

∬Ωxex3dAwhere Ωis the triangular region in the first quadrant bounded below by the x-axis, bounded above y=mx,wherem>0and bounded on the

right by the line with equationx=1.

Short Answer

Expert verified

The required integral ism3(e-1).

Step by step solution

01

Given Information

We are given following double integral as ∬Ωxex3dA

Ωis triangular region in first quadrant bounded by y=mx, slope m>0

bounded on right by x=1

02

Simplification

Region of integration is region between line y=mx,y=0,x=1

It can be seen that for type I integral, 0≤x≤1,0≤y≤mx

Therefore

∬Ωxex3dA=∫01∫0mxxex3dydx

∫01∫0mxxex3dydx=∫01[y]0mxxex3dx

=m∫01x2ex3dx

Using x3=t

⇒3x2dx=dt

Limits are

t=03=0,t=13=1

∫01∫0mxxex3dydx=m3∫01etdt

=m3et01

=m3(e-1)

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