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.<br /><br />Given:<br />$\frac{dv}{dx} = \frac{2}{x} - \frac{1}{x^2}$<br />$v(-1) = -1$<br />$x = 1$<br /><br />Step 1: Integrate the differential equation.<br />$\int \left(\frac{2}{x} - \frac{1}{x^2}\right) dx = \int dv$<br />$2\ln|x| + \frac{1}{x} = v + C$<br />$2\ln|1| + \frac{1}{1} = -1 + C$<br />$C = -2$<br /><br />Step 2: Apply the initial condition to find the value of $C$.<br />$-1 = -2 + C$<br />$C = -1$<br /><br />Step 3: Substitute the value of $C$ back into the equation.<br />$2\ln|x| + \frac{1}{x} = v - 1$<br /><br />Therefore, the solution to the initial value problem is:<br />$v(x) = 2\ln|x| + \frac{1}{x} - 1$<br /><br />To find the average value $V$ of the function $y = 2x - x$, we need to integrate the function over the given interval and divide by the length of the interval.<br /><br />Given:<br />$y = 2x - x$<br />Interval: $[-2, 2]$<br /><br />Step 1: Integrate the function over the given interval.<br />$\int_{-2}^{2} (2x - x) dx = \int_{-2}^{2} x dx$<br />$\left[\frac{x^2}{2}\right]_{-2}^{2} = \frac{2^2}{2} - \frac{(-2)^2}{2}$<br />$= 2 - 2 = 0$<br /><br />Step 2: Divide the result by the length of the interval.<br />$V = \frac{0}{4} = 0$<br /><br />Therefore, the average value $V$ of the function $y = 2x - x$ is $0$.<br /><br />To compute the arc length of the curve $y = \ln(3e^{x})$, we need to use the arc length formula.<br /><br />Given 1: Find the derivative of$ to $x$.<br />$\frac{dy}{dx} = \frac{d}{dx} \ln(3e^{x}) = \frac{1}{e^{x}} \cdot 3e^{x} = 3$<br /><br />Step 2: Apply the arc length formula.<br />$S = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$<br />$S = \int_{a}^{b} \sqrt{1 + 3^2} dx$<br />$S = \int_{a}^{b} \sqrt{10} dx$<br />$S = \sqrt{10} \cdot (b - a)$<br /><br />Therefore, the arc length of the curve $y = \ln(3e^{x})$ is $\sqrt{10} \cdot (b - a)$.<br /><br />To verify the calculus for $f(x) = x^2 + 2x - 8$, we need to find the derivative and integral of the function.<br /><br />Given:<br />$f(x) = x^2 + 2x - 8$<br /><br />Step 1: Find the derivative of $f(x)$.<br />$f'(x) = \frac{d}{dx} (x^2 + 2x -) = 2x + 2$<br /><br />Step 2: Find the integral of $f(x)$.<br />$\int f(x) dx = \int (x^2 + 2x - 8) dx$<br />$= \frac{x^3}{3} + x^2 - 8x + C$<br /><br />Therefore, the derivative of $f(x) = x^2 + 2x - 8$ is $f'(x) = 2x + 2$, and the integral of $f(x)$ is $\int f(x) dx = \frac{x^3}{3} + x^2 - 8x + C$.