91Ó°ÊÓ

A vector-valued function is a crucial concept in vector calculus. Unlike scalar functions that only provide magnitude, vector-valued functions describe both magnitude and direction. These functions map real numbers to vector space, providing components along multiple dimensions, which are often denoted as \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) for the x, y, and z axes, respectively.
In this exercise, the function is given by \( te^{t^2} \mathbf{i} + \sqrt{t} \mathbf{j} + (t^2+1)^{-1} \mathbf{k} \). Each component function corresponds to a different basis vector in 3D space.

• The \( \mathbf{i} \) component \( te^{t^2} \) affects the x-direction.
• The \( \mathbf{j} \) component \( \sqrt{t} \) affects the y-direction.
• The \( \mathbf{k} \) component \( (t^2+1)^{-1} \) impacts the z-direction.

Such decompositions help us understand how each part changes independently, making it easier to solve and combine results to form a complete vector.

Definite integrals are integrals evaluated over a specific interval, providing a numerical value representing the accumulated quantity, such as area under a curve. When dealing with vector-valued functions, we perform definite integrals on each component individually over the given interval.
In this particular problem, we dealt with three definite integrals, one for each component from 0 to 1:

• \( \int_0^1 te^{t^2} \, dt \) for the \( \mathbf{i} \) component.
• \( \int_0^1 \sqrt{t} \, dt \) for the \( \mathbf{j} \) component.
• \( \int_0^1 \frac{1}{t^2+1} \, dt \) for the \( \mathbf{k} \) component.

These integrals sum the contributions of each function part over the interval, providing a vector result that combines the areas of all components.

Three primary integration techniques were applied in this solution. Each technique matched the corresponding function's form, showing the diversity of methods available in calculus.

Substitution Method

This technique is useful when a function includes a composite function. For the \( \mathbf{i} \) component, \( te^{t^2} \), substitution simplified the expression. By letting \( u = t^2 \), the integral transformed, making it easier to evaluate.

Power Rule

This rule simplifies integrals of the form \( t^{n} \). Used for the \( \mathbf{j} \) component, \( \sqrt{t} \) translated to \( t^{1/2} \). The power rule allows raising \( t \'s\) exponent and dividing by the new exponent for a quick result.

Integration of Special Functions

The \( \mathbf{k} \) component involved \( 1/(t^2+1) \), recognized as the derivative of \( \tan^{-1}(t) \). Such special functions are crucial and require familiarity with their integral forms to solve effectively.Each technique tackles specific challenges posed by diverse integrands, showcasing the adaptive nature of integration in calculus.