∫ 0 ∞ e 2 y 3 e − e y 2 d y = ?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
The substitution u = e y 2 would be more natural, and simplifies the working.
I = ∫ 0 ∞ e 2 y 3 e − e y 2 d y Let x = e y 2 ⇒ d x = 2 e y d y = ∫ 0 ∞ 2 e y e 2 y 3 e − x d x = 2 1 ∫ 0 ∞ e y 2 e − x d x = 2 1 ∫ 0 ∞ x e − x d x = 2 Γ ( 2 ) = 2 1 ! = 0 . 5
Same solution!!
Problem Loading...
Note Loading...
Set Loading...
Let I = ∫ 0 ∞ e 2 y 3 e − e y 2 d y
= ∫ 0 ∞ e 2 − e y 2 y 3 d y
Now let 2 − e y 2 = − u
2 e y d y = d u
y 2 = e u + 2
So the Inegral is equal to : = ∫ 0 ∞ e 2 − e y 2 y 2 2 e 2 e y d y = ∫ 0 ∞ e − u 2 e 2 u + 2 d u [ Since y 2 = e u + 2 ]
= ∫ 0 ∞ e − ( u + 2 ) 2 u + 2 d u
= 2 1 ∫ 0 ∞ e − ( u + 2 ) ( u + 2 ) 2 − 1 d ( u + 2 )
= 2 1 Γ ( 2 )
= 2 1 [Since Γ ( 2 ) = 1 ]
= 0 . 5