Determine whether the integral is convergent or divergent.If it is convergent, evaluate it.
from 0 to infinity
e^(-y^1/2)dy
To evaluate
∫e^(-√x) dx
let
z^2 = x
2z dz = dx
and now you have
∫e^-z 2z dz
Now let
u = z
dv = e^-z dz
du = dz
v = -e^-z dz
∫u dv = uv - ∫v du, so
∫2z e^-z dz = -2ze^-z + 2∫e^-z dz
= -2ze^-z - 2e^-z
Now evaluate that at the limits and you get
∫[0,∞] e^(-√x) dx = 2
To determine whether the integral ∫e^(-y^(1/2))dy from 0 to infinity is convergent or divergent, we can use the limit comparison test.
1. Start by selecting a function g(y) that is positive, continuous, and has a known convergent or divergent integral.
In this case, we can choose g(y) = 1/y^(1/4). The integral ∫(1/y^(1/4))dy is a known convergent integral.
2. Take the limit as y approaches infinity of f(y)/g(y), where f(y) is the original function.
Taking the limit of e^(-y^(1/2))/1/y^(1/4) as y approaches infinity, we get:
lim (y → ∞) e^(-y^(1/2)) / (1/y^(1/4))
Using L'Hôpital's rule, we can differentiate the numerator and denominator with respect to y:
lim (y → ∞) (-1/2)e^(-y^(1/2)) / (-1/4)y^(-3/4)
Simplifying, we get:
lim (y → ∞) (2/4) y^(3/4) e^(-y^(1/2))
By rearranging, we can write this limit as:
lim (y → ∞) (2y^(3/4) / 4e^(y^(1/2)))
3. Evaluate the limit.
We can see that as y approaches infinity, the numerator (2y^(3/4)) will grow faster than the denominator (4e^(y^(1/2))). This means that the limit will approach ∞.
Since the limit is not a finite number, the integral ∫e^(-y^(1/2))dy is divergent.
Therefore, the integral is divergent and does not have a calculable value.