integration of limit 0 to 2 x(8-x^3)^1/3dx =1/9.gama of 1/3.gama of 2/3
Answers
Answer:
Step-by-step explanation:
I tried substituting x3=8u but I just got stuck.
Answer: To solve this integral, we can use the substitution u = 8 - x^3. Then, du/dx = -3x^2, so dx = -du/(3x^2).
When x = 0, u = 8, and when x = 2, u = 0.
Substituting these limits and the expressions for x and dx in terms of u and du, we get:
∫[0,2] x(8-x^3)^(1/3) dx = -1/3 ∫[8,0] (8-u)^(1/3) du/u^(2/3)
Next, we can use the identity for the beta function, which is defined as:
B(p, q) = ∫[0,1] t^(p-1) (1-t)^(q-1) dt
to write the integral in terms of the beta function:
∫[0,2] x(8-x^3)^(1/3) dx = -1/3 B(1/3, 2/3)
Now, we can use the property of the beta function that relates it to the gamma function:
B(p, q) = Γ(p)Γ(q) / Γ(p+q)
to obtain:
∫[0,2] x(8-x^3)^(1/3) dx = -1/3 Γ(1/3)Γ(2/3) / Γ(1)
Since Γ(1) = 1, we can simplify to get:
∫[0,2] x(8-x^3)^(1/3) dx = -1/3 Γ(1/3)Γ(2/3)
Finally, we can use the relationship between the gamma function and the beta function again to get:
Γ(1/3)Γ(2/3) = B(1/3, 2/3) = π / sin(π/3) = 2π / √3
Substituting this back into the expression for the integral, we get:
∫[0,2] x(8-x^3)^(1/3) dx = -1/3 (2π / √3) = -2π / (3√3)
Therefore, the value of the integral is -2π / (3√3).
Learn more about the beta function here
https://brainly.in/question/50596720
Learn more about the gamma function here
https://brainly.in/question/38249041
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