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

Question: Compute the indefinite integral of the vector function \(\mathbf{r}(t) = e^{3t}\mathbf{i} + \frac{1}{1+t^2}\mathbf{j} - \frac{1}{\sqrt{2t}}\mathbf{k}\). Answer: The indefinite integral of the vector function \(\mathbf{r}(t)\) is given by: \(\mathbf{R}(t) = \frac{e^{3t}}{3}\mathbf{i}+\arctan(t)\mathbf{j}-\sqrt{t}\mathbf{k} + \mathbf{C}\), where \(\mathbf{C} = C_1\mathbf{i} + C_2\mathbf{j} + C_3\mathbf{k}\) is the vector constant of integration.

Step-by-step solution

Integrate the \(\mathbf{i}\) component

To integrate the \(\mathbf{i}\) component function, \(e^{3t}\), we will use the power rule for integration which states that \(\int e^{at} dt = \frac{e^{at}}{a} + C\). In this case, \(a=3\), so we have: $$\int e^{3t} dt = \frac{e^{3t}}{3} + C_1$$

Integrate the \(\mathbf{j}\) component

Now, let's integrate the \(\mathbf{j}\) component function, \(\frac{1}{1+t^2}\). This is a well-known integral and results in the arctangent function: $$\int \frac{1}{1+t^2} dt = \arctan(t) + C_2$$

Integrate the \(\mathbf{k}\) component

Finally, we need to integrate the \(\mathbf{k}\) component function, \(-\frac{1}{\sqrt{2t}}\). To do this, we rewrite the function as \(-\frac{1}{\sqrt{2}}t^{-\frac{1}{2}}\). Now we can use the power rule for integration, which states that \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\): $$\int -\frac{1}{\sqrt{2}}t^{-\frac{1}{2}} dt = \frac{-\sqrt{2}t^{\frac{1}{2}}}{\sqrt{2}} + C_3$$

Combine the integrated components

We have found the indefinite integral of each component function. Now, we will combine them to find the indefinite integral of the vector function \(\mathbf{r}(t)\): $$\int \mathbf{r}(t) dt = \left(\frac{e^{3t}}{3} + C_1\right)\mathbf{i} + \left(\arctan(t) + C_2\right)\mathbf{j} + \left(\frac{-\sqrt{2}t^{\frac{1}{2}}}{\sqrt{2}} + C_3\right)\mathbf{k}$$ Thus, the indefinite integral of the given vector function is: $$\mathbf{R}(t) = \frac{e^{3t}}{3}\mathbf{i}+\arctan(t)\mathbf{j}-\sqrt{t}\mathbf{k} + \mathbf{C},$$ where \(\mathbf{C} = C_1\mathbf{i} + C_2\mathbf{j} + C_3\mathbf{k}\) is the vector constant of integration.