Solve the initial value v=-1 x=1 Find the volume of the solid of re olution formed by rotating y=sqrt (4-x^3) about the xaxis from x=-2 to x=2 (3 marks) Find the is value Vof the iunction y=2x-x (3 marky) Compute the are length y=ln(xecx) 13 marks Venty the theoren f(x)x^2+2x-8.+1 int _(1)^2(1)/(lnx) Trapezoidal rule (iii) (3 marki)

To solve the initial value problem, we need to find the solution to the differential equation $v' = -1$ with the initial condition $v(1) = -1$.<br /><br />The general solution to the differential equation $v' = -1$ is $v(x) = -x + C$, where $C$ is the constant of integration.<br /><br />Using the initial condition $v(1) = -1$, we can substitute $x = 1$ into the general solution to find the value of $C$:<br /><br />$-1 = -(1) + C$<br /><br />$C = 0$<br /><br />Therefore, the solution to the initial value problem is $v(x) = -x$.<br /><br />To find the volume of the solid of revolution formed by rotating the region $y = \sqrt{4 - x^3}$ about the x-axis from $x = -2$ to $x = 2$, we can use the disk method.<br /><br />The volume $V$ is given by the integral:<br /><br />$V = \pi \int_{-2}^{2} (\sqrt{4 - x^3})^2 dx$<br /><br />Simplifying the integrand:<br /><br />$V = \pi \int_{-2}^{2} (4 - x^3) dx$<br /><br />Integrating term by term:<br /><br />$V = \pi \left[ 4x - \frac{x^4}{4} \right]_{-2}^{2}$<br /><br />Evaluating the integral:<br /><br />$V = \pi \left[ 8 - \frac{16}{4} - (-8 + \frac{16}{4}) \right]$<br /><br />$V = \pi \left[ 8 - 4 + 8 - 4 \right]$<br /><br />$V = \pi \cdot 8$<br /><br />$V = 8\pi$<br /><br />Therefore, the volume of the solid of revolution is $8\pi$.<br /><br />To find the solution to the equation $y = 2x - x$, we can simplify the equation:<br /><br />$y = x$<br /><br />Therefore, the solution to the equation is $y = x$.<br /><br />To compute the area under the curve $y = \ln(x)$, we can use the definite integral:<br /><br />$\int_{1}^{2} \ln(x) dx$<br /><br />Using integration by parts, let $u = \ln(x)$ and $dv = dx$. Then $du = \frac{1}{x} dx$ and $v = x$.<br /><br />$\int_{1}^{2} \ln(x) dx = \left[ x \ln(x) - \int_{1}^{2} x \cdot \frac{1}{x} dx \right]_{1}^{2}$<br /><br />Simplifying the integral:<br /><br />$\int_{1}^{2} \ln(x) dx = \left[ x \ln(x) - x \right]_{1}^{2}$<br /><br />Evaluating the integral:<br /><br />$\int_{1}^{2} \ln(x) dx = \left[ 2 \ln(2) - 2 - (1 \ln(1) - 1) \right]$<br /><br />$\int_{1}^{2} \ln(x) dx = 2 \ln(2) - 2 + 1$<br /><br />$\int_{1}^{2} \ln(x) dx = 2 \ln(2) - 1$<br /><br />Therefore, the area under the curve $y = \ln(x)$ from $x = 1$ to $x = 2$ is $2 \ln(2) - 1$.<br /><br />To evaluate the integral $\int_{1}^{2} \frac{1}{\ln(x)}$ using the trapezoidal rule, we can divide the interval $[1, 2]$ into $n$ subintervals of equal width $\Delta x = \frac{2 - 1}{n} = \frac{1}{n}$.<br /><br />The trapezoidal rule formula is:<br /><br />$\int_{a}^{b} f(x) dx \approx \frac{\Delta x}{2} \left[ f(a) + 2 \sum_{i=1}^{n-1} f(x_i) + f(b) \right]$<br /><br />In this case, $f(x) = \frac{1}{\ln(x)}$, $a = 1$, and $b = 2$. We need to evaluate $f(x)$ at $n$ points $x_0, x_1, \ldots, x_n$, where $x_i = 1 + i \Delta x$ for $i = 0, 1, \ldots, n-1$.<br /><br />Using the trapezoidal rule, we can approximate the integral:<br /><br />$\int_{1}^{2} \frac{1}{\ln(x)} dx \approx \frac{1}{2n} \left[