Question

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

Short answer

The indefinite integral of the given vector function is $$\mathbf{R}(t) =\left(\frac{2^{t}}{\ln(2)} + C_{1}\right)\mathbf{i} + \left(\frac{1}{2}\ln(1+2t) + C_{2}\right)\mathbf{j} + \left(t \ln(t) - t + C_{3}\right)\mathbf{k}$$, where $$C_1$$, $$C_2$$, and $$C_3$$ are constants of integration.

Step-by-step solution

Integrate the i-component

First, we will find the antiderivative of the scalar function $$2^{t}$$ with respect to t. The integral is given by: $$\int 2^{t} dt$$ We can use a substitution method to solve this integral; let $$u = 2^t$$ and therefore, $$\frac{du}{dt} = \ln(2) 2^{t} $$ ⟹ $$dt = \frac{du}{\ln(2) \cdot u}$$. Now, substitute this into the integral: $$\int \frac{u}{\ln(2) \cdot u} du = \frac{1}{\ln(2)}\int \,du$$ Integrating, we get: $$\frac{u}{\ln(2)}+ C_{1} = \frac{2^{t}}{\ln(2)} + C_{1}$$ So, the antiderivative of $$2^t$$ is $$\frac{2^{t}}{\ln(2)} + C_{1}$$.

Integrate the j-component

Next, find the antiderivative of the scalar function $$\frac{1}{1+2 t}$$ with respect to t: $$\int \frac{1}{1+2t} dt$$ This is a basic integration as it is a simple rational function. Use the substitution method and let $$ u = 1+2t $$, thus, $$\frac{du}{dt} = 2$$, and $$dt = \frac{du}{2}$$. Substitute this into the integral: $$\int \frac{1}{u} \cdot \frac{du}{2} = \frac{1}{2}\int \frac{1}{u} du$$ The integral of $$\frac{1}{u}$$ is $$\ln(u)$$, so we get: $$\frac{1}{2}\ln(u) + C_{2} = \frac{1}{2}\ln(1+2t) + C_{2}$$ Thus, the antiderivative of $$\frac{1}{1+2t}$$ is $$\frac{1}{2}\ln(1+2t) + C_{2}$$.

Integrate the k-component

Lastly, find the antiderivative of the scalar function $$\ln(t)$$ with respect to t: $$\int\ln(t)dt$$ We can integrate by parts, using the formula: $$\int u dv = uv - \int v du$$ Let $$u = \ln(t)$$ and $$dv = dt$$. Then, $$du = \frac{1}{t}dt$$ and $$v=t$$. Substitute these into the integration by parts formula: $$\int \ln(t) dt = t\ln(t) - \int t\frac{1}{t} dt = t \ln(t) - \int dt$$ Now, integrate with respect to t: $$t \ln(t) - t + C_{3}$$ Thus, the antiderivative of $$\ln(t)$$ is $$t \ln(t) - t + C_{3}$$.

Assemble the antiderivative as a vector function

Combine the antiderivatives found in Steps 1, 2, and 3 to create the antiderivative as a vector function: $$\mathbf{R}(t) =\left(\frac{2^{t}}{\ln(2)} + C_{1}\right)\mathbf{i} + \left(\frac{1}{2}\ln(1+2t) + C_{2}\right)\mathbf{j} + \left(t \ln(t) - t + C_{3}\right)\mathbf{k}$$ This is the indefinite integral (antiderivative) of the given vector function $$\mathbf{r}(t)$$.

Key concepts

Vector Calculus

Vector calculus is a field of mathematics that deals with vector fields and operations on them. In this domain, we work with functions that output vectors instead of scalars. A vector function, like \(\mathbf{r}(t)\), maps a scalar input \(t\) to a vector output, which includes multiple components. For example, \(\mathbf{r}(t)=2^{t} \, \mathbf{i}+\frac{1}{1+2 t} \, \mathbf{j}+\ln t \, \mathbf{k}\) is represented in terms of standard basis vectors \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\). These correspond to the x, y, and z directions, respectively. The main goal is to analyze changes in these vector fields by finding derivatives and integrals of vector functions. When finding the integral or antiderivative of a vector function, we perform the operation on each component of the vector separately. This step-by-step integration helps us understand how each part of the function behaves over time or with respect to another variable.

Integration by Parts

Integration by parts is a valuable technique used in calculus, particularly when handling products of functions. This technique is derived from the product rule for differentiation and is formulated as: \[\int u \, dv = uv - \int v \, du\].By choosing suitable \(u\) and \(dv\), we can simplify complex integrals. In the context of vector calculus, we can use this method to evaluate integrals component by component. For example, when integrating \(\ln(t)\), we choose \(u = \ln(t)\) and \(dv = dt\). This selection gives us \(du = \frac{1}{t}dt\) and \(v = t\). Applying the integration by parts formula results in:- \(t \ln(t)\) for the \(uv\) part.- The integral of 1 with respect to \(t\) for \(\int v \, du\), which is \(t\).This simplification results in \(t \ln(t) - t + C\), where \(C\) is the integration constant. Each use case can vary, so it's crucial to select \(u\) and \(dv\) appropriately.

Substitution Method

The substitution method is another key tool for finding integrals, especially when dealing with composite functions. This method simplifies an integral by changing variables, transforming a difficult integral into a simpler one.When using substitution in vector calculus, we solve each component of the vector function separately. For instance, to integrate \(2^{t}\), we set \(u = 2^t\). This implies \( \frac{du}{dt} = \ln(2) 2^t\), hence \(dt = \frac{du}{\ln(2) \cdot u}\). Substituting these into the integral creates a simpler integral: \[\int \frac{u}{\ln(2) \cdot u} \, du = \frac{1}{\ln(2)} \int \, du\]. Integration then becomes straightforward, resulting in \(\frac{2^{t}}{\ln(2)} + C\). Overall, substitution is beneficial for breaking down integrals that are initially challenging to integrate directly.

Antiderivative

An antiderivative, also known as an indefinite integral, represents a function whose derivative is the original function. For example, if \(F(t)\) is the antiderivative of \(f(t)\), then \(\frac{d}{dt}[F(t)] = f(t)\). This relationship helps us understand the accumulation of quantities or the area under curves.In the case of vector functions, we compute the antiderivative for each component of the vector. With our function \(\mathbf{r}(t)\), we calculate antiderivatives separately for the \(\mathbf{i}, \mathbf{j},\) and \(\mathbf{k}\) components. Each step involves common integration techniques, from simple power rules to more complex methods like substitution and integration by parts.The sake of completeness and precision, it is important to always include an integration constant \(C\) (or \(C_1, C_2, C_3\) for each component) in the final expression. These constants represent the family of functions that make up the antiderivative. The final result is presented as a vector function that includes these individually integrated components.