Question

The function f is continuous and ₀∫⁸ f(u)du = 6. What is the value ₁∫³ xf(x²ˣ⁻¹)dx?

A. 3/2
B. 3
C. 6
D. 12
E.24

Answer 1

To find the integral from 1 to 3 of xf(x^2x^(-1))dx, we can make a substitution using the equation u = x^2x^(-1). After making the substitution, we can solve the integral using u-substitution.

Step-by-step explanation:

We are given that the function f is continuous and that the integral from 0 to 8 of f(u)du is equal to 6. We are asked to find the integral from 1 to 3 of xf(x^2x^(-1))dx.

Let's start by making a substitution. Let u = x^2x^(-1), which means x = u^(1/(2x-1)).

Now we can find the integral from 1 to 3 of xf(x^2x^(-1))dx using the substitution u = x^2x^(-1):

1. Find the derivative of u with respect to x: du/dx = (1/(2x-1))x^(2x-1)
2. Solve for dx: dx = (2x-1)/x^(2x-1) du
3. Substitute u and dx into the integral:

∫(1 to 3) xf(x^2x^(-1))dx = ∫(u) (u^(1/(2x-1)))f(u) (2x-1)/u^(2x-1) du.