ol Solve the initial value problem (dv)/(dx)=(2)/(x)-(1)/(x^2) v=-1 x=1 y=sqrt (4-x^2) about the xaxis from x=-2 x=2 (3 marks) Find the average value Vofithe is y=2x-x (3 marks) Compute the are length of the curve y=ln(3ecx) (3 marks) Venty the fur calculus for f(x)=x^2+2x-8 (3 marks)

To solve the initial value problem, we need to integrate the given differential equation with the initial condition provided.Given: Step 1: Integrate the differential equation. Step 2: Apply the initial condition to find the value of . Step 3: Substitute the value of back into the equation. Therefore, the solution to the initial value problem is: To find the average value of the function , we need to integrate the function over the given interval and divide by the length of the interval.Given: Interval: Step 1: Integrate the function over the given interval. Step 2: Divide the result by the length of the interval. Therefore, the average value of the function is .To compute the arc length of the curve , we need to use the arc length formula.Given 1: Find the derivative of x \frac{dy}{dx} = \frac{d}{dx} \ln(3e^{x}) = \frac{1}{e^{x}} \cdot 3e^{x} = 3 S = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx S = \int_{a}^{b} \sqrt{1 + 3^2} dx S = \int_{a}^{b} \sqrt{10} dx S = \sqrt{10} \cdot (b - a) y = \ln(3e^{x}) \sqrt{10} \cdot (b - a) f(x) = x^2 + 2x - 8 f(x) = x^2 + 2x - 8 f(x) f'(x) = \frac{d}{dx} (x^2 + 2x -) = 2x + 2 f(x) \int f(x) dx = \int (x^2 + 2x - 8) dx = \frac{x^3}{3} + x^2 - 8x + C f(x) = x^2 + 2x - 8 f'(x) = 2x + 2 f(x) \int f(x) dx = \frac{x^3}{3} + x^2 - 8x + C$.