Problem 33 - Calculus Early Transcendentals

Question

Evaluate the integral. $$ \int_{0}^{1}\left(16 t^{3} \mathbf{i}-9 t^{2} \mathbf{j}+25 t^{4} \mathbf{k}\right) d t $$

Step-by-Step Solution

Answer

The integral evaluates to \( 4\mathbf{i} - 3\mathbf{j} + 5\mathbf{k} \).

Step 1: Identify the Function

The given integral is a vector-valued function: \( \int_{0}^{1}(16t^3 \mathbf{i} - 9t^2 \mathbf{j} + 25t^4 \mathbf{k}) \, dt \). The function consists of three components: \( 16t^3 \) for \( \mathbf{i} \), \( -9t^2 \) for \( \mathbf{j} \), and \( 25t^4 \) for \( \mathbf{k} \). We will integrate each component separately.

Step 2: Integrate the \( \mathbf{i} \) Component

Integrate the function \( 16t^3 \) with respect to \( t \) from 0 to 1: \[ \int_{0}^{1} 16t^3 \, dt = 16 \left[ \frac{t^4}{4} \right]_{0}^{1} = 16 \left( \frac{1}{4} \right) - 16 \left( 0 \right) = 4 \].

Step 3: Integrate the \( \mathbf{j} \) Component

Integrate the function \( -9t^2 \) with respect to \( t \) from 0 to 1: \[ \int_{0}^{1} -9t^2 \, dt = -9 \left[ \frac{t^3}{3} \right]_{0}^{1} = -9 \left( \frac{1}{3} \right) - (-9)(0) = -3 \].

Step 4: Integrate the \( \mathbf{k} \) Component

Integrate the function \( 25t^4 \) with respect to \( t \) from 0 to 1: \[ \int_{0}^{1} 25t^4 \, dt = 25 \left[ \frac{t^5}{5} \right]_{0}^{1} = 25 \left( \frac{1}{5} \right) - 25(0) = 5 \].

Step 5: Combine the Results

Combine the results of the three integrations to express the vector value: \( \int_{0}^{1}(16t^3 \mathbf{i} - 9t^2 \mathbf{j} + 25t^4 \mathbf{k}) \, dt = 4\mathbf{i} - 3\mathbf{j} + 5\mathbf{k} \).

Key Concepts

Vector Integration

Vector integration is a process where we integrate vector-valued functions over a specific interval. A vector-valued function, like the one in our exercise, consists of multiple components. Each component typically represents a different direction in a multidimensional space.

When dealing with vector integration, it is important to remember that each vector's component behaves like a separate scalar function. This allows us to apply the rules of integration to each component individually.

Key points to consider with vector integration:

• Break down the vector into its constituent components.
• Apply standard integration techniques to each component separately.
• Combine the results vectorially to form the integrated vector.

Vector integration is particularly useful in physics and engineering, where forces and fields are often represented as vectors.

Definite Integral

The definite integral is a fundamental concept used in calculus that computes the accumulation of quantities, which can be areas under curves, total outputs, etc. In our context, it helps to find the total effect of a function over a fixed interval.

When dealing with the definite integral, especially with vector functions, you need to:

• Choose the bounds of integration, which are the limits that define the interval.
• Calculate the integral of each function component between these limits.
• Consider the evaluation of the antiderivative at the upper and lower limits to find the net value.

The result of a definite integral gives a number for scalar functions and a vector for vector-valued functions, summarizing the incremental contributions over a range.

Component-wise Integration

Component-wise integration is the tactic of integrating each part of a vector-valued function independently. With vector calculus, you're often handling multi-dimensional data, with each axis of the vector needing separate attention.

The steps for component-wise integration are straightforward:

• Identify each component of the vector separately.
• Perform integration on each scalar component using the regular rules of integration.
• For definite integrals, evaluate each antiderivative at the upper and lower bounds and subtract.
• Reconstruct the vector using the integrated components.

This method simplifies the process of vector integration by letting you focus on each component one at a time. It makes complex functions manageable and ensures that you account for the contribution of each dimension comprehensively.