evaluate the improper integral whenever it is convergent
∫0,-00 1/(4-x)^(3/2) dx
What are the limits? The following is not clear.
-00
∫ dx/(4-x)(3/2)
0
Ooh, I get it. its from 0 to -∞ (or -inf if you want)
-∞
∫ dx/(4-x)(3/2)
0
Proceed with normal integration
=[2/sqrt(4-x)] from 0 to -∞
=[2/sqrt(4-(-inf)) - 2/sqrt(4-0)]
=[0 - 2/2)
=-1
To evaluate the improper integral ∫₀⁻∞ (1/(4-x)^(3/2)) dx, we can follow these steps:
Step 1: Determine the Domain of the Integral
Since this integral extends to negative infinity, the domain of the integral is (-∞, 0].
Step 2: Check for Convergence
To determine whether the integral is convergent, we need to check if the integrand approaches a finite limit as x approaches the endpoint of the domain (in this case, as x approaches 0).
Let's evaluate the limit of the integrand as x approaches 0:
lim (x → 0) (1/(4-x)^(3/2))
Substituting x = 0 into the expression gives us:
lim (x → 0) (1/(4-0)^(3/2))
= lim (x → 0) (1/4^(3/2))
= lim (x → 0) (1/8)
= 1/8
Since the limit exists and is finite, the integrand approaches a finite limit as x approaches 0, indicating that the integral is convergent.
Step 3: Evaluate the Integral
∫₀⁻∞ (1/(4-x)^(3/2)) dx = ∫₀⁻∞ (1/(√(4-x))^3) dx
We can simplify the expression by using a substitution: let u = √(4-x), then du = (-1/2)(1/√(4-x)) dx.
When x is the lower limit (0), u = √(4-0) = 2, and when x is the upper limit (-∞), u = √(4-(-∞)) = √(4+∞) = √∞ = ∞.
Now, rewrite the integral in terms of u:
∫₀² -(1/u^3) du
Integrating -(1/u^3) with respect to u, we get:
-[(1/u^2) / 2] from 0 to 2
= -[(1/2u^2)] from 0 to 2
= -[(1/2(2)^2)] - (-[(1/2(0)^2)])
= -[(1/8)] - 0
= -1/8
Therefore, the value of the improper integral ∫₀⁻∞ (1/(4-x)^(3/2)) dx is -1/8.