Question

Compute the indefinite integral of the following functions. $$\mathbf{r}(t)=e^{3 t} \mathbf{i}+\frac{1}{1+t^{2}} \mathbf{j}-\frac{1}{\sqrt{2 t}} \mathbf{k}$$

Short answer

Answer: The indefinite integral of the given vector function is \(\int \mathbf{r}(t) \, dt = \left(\frac{1}{3}e^{3t} + C_1\right)\mathbf{i} + \left(\arctan(t) + C_2\right)\mathbf{j} + \left(-\sqrt{2t} + C_3\right)\mathbf{k}\).

Step-by-step solution

Identify the Components

First, identify the components of the vector function: $$\mathbf{r}(t) = e^{3t}\mathbf{i} + \frac{1}{1+t^2}\mathbf{j} - \frac{1}{\sqrt{2t}}\mathbf{k}$$ The components are: - \(f(t) = e^{3t}\) - \(g(t) = \frac{1}{1+t^2}\) - \(h(t) = -\frac{1}{\sqrt{2t}}\)

Integrate the Components

Integrate each component with respect to \(t\): 1. \(\int f(t) \, dt= \int e^{3t} \, dt\) 2. \(\int g(t) \, dt= \int \frac{1}{1+t^2} \,dt\) 3. \(\int h(t) \, dt= \int -\frac{1}{\sqrt{2t}} \, dt\)

Find the Indefinite Integrals of the Components

Find the indefinite integral for each component: 1. \(\int e^{3t} \, dt = \frac{1}{3}e^{3t} + C_1\) 2. \(\int \frac{1}{1+t^2} \, dt = \arctan(t) + C_2\) 3. \(\int -\frac{1}{\sqrt{2t}} \, dt = -\sqrt{2t} + C_3\)

Write the Vector Function

Combine the integrals of the components to create the integral vector function: $$\int \mathbf{r}(t) \, dt = \left(\frac{1}{3}e^{3t} + C_1\right)\mathbf{i} + \left(\arctan(t) + C_2\right)\mathbf{j} + \left(-\sqrt{2t} + C_3\right)\mathbf{k}$$ The indefinite integral of the given vector function is: $$\int \mathbf{r}(t) \, dt = \left(\frac{1}{3}e^{3t} + C_1\right)\mathbf{i} + \left(\arctan(t) + C_2\right)\mathbf{j} + \left(-\sqrt{2t} + C_3\right)\mathbf{k}$$