Question

Evaluate the following definite integrals. $$\int_{-1}^{1}\left(\mathbf{i}+t \mathbf{j}+3 t^{2} \mathbf{k}\right) d t$$

Short answer

Question: Evaluate the definite integral of the vector function F(t) = i + t*j + 3t^2*k from -1 to 1. Answer: The definite integral of the given vector function from -1 to 1 is 2*i + 2*k.

Step-by-step solution

Identify the vector components

Begin by identifying the components of the given vector function which are: $$\mathbf{F}(t)=\mathbf{i}+t \mathbf{j}+3 t^{2} \mathbf{k}$$

Integrate each component

Now, integrate the i, j, and k components separately with respect to t: For the i component: $$\int (\mathbf{i})\, d t = t\mathbf{i} + C_{1}$$ For the j component: $$\int (t \mathbf{j})\, d t = \frac{1}{2}t^{2}\mathbf{j} + C_{2}$$ For the k component: $$\int (3 t^{2} \mathbf{k})\, d t = t^{3}\mathbf{k} + C_{3}$$

Combine the integrated components

Combine the integrated components to form a new vector function: $$\mathbf{G}(t)=\left(t\mathbf{i} + \frac{1}{2}t^{2}\mathbf{j} + t^{3}\mathbf{k}\right) + \mathbf{C}$$ Where \(\mathbf{C}=C_{1}\mathbf{i}+C_{2}\mathbf{j}+C_{3}\mathbf{k}\) is the constant vector from each individual integration.

Evaluate the definite integral

We now need to evaluate the definite integral within the given limits, -1 to 1. $$\int_{-1}^{1}\left(\mathbf{i}+t \mathbf{j}+3 t^{2} \mathbf{k}\right) d t= \left[\mathbf{G}(t)\right]_{-1}^{1}$$ $$\left[\left(t\mathbf{i}+\frac{1}{2}t^{2}\mathbf{j}+t^{3}\mathbf{k}\right) + \mathbf{C}\right]_{-1}^{1}$$ Evaluate the integrand for the upper limit, t = 1: $$\mathbf{G}(1)=\left(1\mathbf{i}+\frac{1}{2}\mathbf{j}+1\mathbf{k}\right) + \mathbf{C}$$ Evaluate the integrand for the lower limit, t = -1: $$\mathbf{G}(-1)=\left(-1\mathbf{i}+\frac{1}{2}\mathbf{j}-1\mathbf{k}\right) + \mathbf{C}$$

Calculate the definite integral result

Calculate the definite integral by subtracting the result of the lower limit from the result of the upper limit, which cancels the constant vector: $$\mathbf{G}(1) - \mathbf{G}(-1) = \left(1\mathbf{i}+\frac{1}{2}\mathbf{j}+1\mathbf{k}\right) - \left(-1\mathbf{i}+\frac{1}{2}\mathbf{j}-1\mathbf{k}\right)$$ $$=2\mathbf{i} +0\mathbf{j} +2\mathbf{k}$$ Therefore, the definite integral of the given vector function from -1 to 1 is: $$\int_{-1}^{1}\left(\mathbf{i}+t \mathbf{j}+3 t^{2} \mathbf{k}\right) d t = 2\mathbf{i} + 2\mathbf{k}$$

Key concepts

Vector Calculus

Understanding vector calculus is essential for tackling problems involving vector-valued functions and their integrals, especially within the realm of physics and engineering. A vector function, such as \( \mathbf{F}(t)=\mathbf{i}+t \mathbf{j}+3 t^{2} \mathbf{k} \), maps a scalar to a vector and is comprised of component functions aligning with the base vectors \( \mathbf{i}, \mathbf{j} \), and \( \mathbf{k} \).

In the context of our exercise, when performing the integration of a vector function over a certain interval, we deal with each component independently. This approach simplifies the problem into manageable parts. Additionally, understanding the properties of vector functions is crucial, such as the fact that you can add vectors or scalar multiply vectors but cannot multiply vectors directly in standard algebraic ways.

Integration Techniques

Numerous integration techniques exist for solving various kinds of functions. The key to mastering integration is to recognize which technique is best suited for the given function. For polynomials, a straightforward antiderivative application suffices, as seen in our exercise for the components of the vector function. For instance, integrating the \( t \mathbf{j} \) term involves finding a function whose derivative gives back \( t \mathbf{j} \)—in this case, \( \frac{1}{2}t^{2}\mathbf{j} \).

It's essential to note that constants of integration \( C \) are added when finding indefinite integrals. However, when evaluating definite integrals, the constants cancel out, making them unnecessary for the final answer. This concept is a pivotal part of finding the solution in definite integral problems.

Multivariable Calculus

When dealing with functions of several variables, we enter the realm of multivariable calculus. Integrals in this field may be over curves, surfaces, or volumes, depending on the number of variables. For a single-variable vector function, like our exercise presents, the integration is similar to standard calculus but performed on each component of the vector.

In more complex scenarios, such as when functions depend on two or more variables, the integration techniques employed may involve double or triple integrals. In such cases, students might encounter tools like Jacobian determinants, which help when changing variables during integration.

Building a solid foundation in single-variable integrals—as practiced in our exercise—is a prerequisite before diving into more complex multivariable scenarios.