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Chapter 12: Problem 38

Evaluate the definite integral. $$ \int_{-3}^{3}\left\langle(3-t)^{3 / 2},(3+t)^{3 / 2}, 1\right\rangle d t $$

Short Answer

Expert verified
The integral evaluates to the vector \(\langle 0, 0, 6 \rangle\).

Step by step solution

01

Understand the Vector-Valued Integral

The integral given is a vector-valued integral: \[ \int_{-3}^{3} \langle (3-t)^{3/2}, (3+t)^{3/2}, 1 \rangle \, dt \]This means the integral is over each component of the vector separately from \(-3\) to \(3\). We'll evaluate each integral component-wise.
02

Evaluate Integral of First Component

The first component to evaluate is \[ \int_{-3}^{3} (3-t)^{3/2} \, dt. \]Due to symmetry and the fact it is an odd function over the interval, this integral evaluates to 0 because the functions cancel out over symmetric limits.
03

Evaluate Integral of Second Component

The second component is \[ \int_{-3}^{3} (3+t)^{3/2} \, dt. \]Similarly, like the first integral, this function is also symmetric (odd) over the interval \([-3, 3]\), so it evaluates to 0 because of the cancellation over symmetric limits.
04

Evaluate Integral of Third Component

The third component is straightforward:\[ \int_{-3}^{3} 1 \, dt. \]This integral equals the length of the interval, which is \[ 3 - (-3) = 6. \]
05

Combine Component Integrals

Combine the evaluated integrals for the vector: \[ \langle 0, 0, 6 \rangle. \]This is the result of the definite vector integral over the given interval.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector-Valued Integral
Vector-valued integrals extend the idea of integrating scalar functions to vector functions. A vector-valued integral can be broken down into individual integrals for each component within the vector. These components are evaluated separately and independently over the specified interval.
For instance, given a vector-valued function \[ \ \int_a^b \langle f(t), g(t), h(t) \rangle \, dt, \] \ each function \( f(t), g(t), \text{and} \ h(t) \) is integrated over the interval from \( a \) to \( b \).
  • The integral is computed for each separate part of the vector: \( \int_a^b f(t) \, dt \), \( \int_a^b g(t) \, dt \), \( \int_a^b h(t) \, dt \).
  • The results of these integrals are then combined back into a new vector to give the final result.
This approach allows us to handle multiple functions simultaneously, making vector-valued integrals a powerful tool in fields like physics and engineering.
Odd Function
An odd function has a unique property that reflects its symmetry about the origin. Essentially, a function \( f(t) \) is considered \("odd"\) if it satisfies the condition: \[ f(-t) = -f(t) \]. This inherent symmetry plays a key role when evaluating integrals over symmetric limits.
  • An odd function integrated over a symmetric interval \([-a, a]\) results in zero. This occurs because the areas above and below the x-axis cancel each other out.
  • Odd functions are visually symmetric about the origin, making them predictable when dealing with integrals across symmetric limits.
  • In the context of our specified exercise, recognizing the odd nature of functions like \((3-t)^{3/2}\) and \((3+t)^{3/2}\) helps simplify the evaluation process significantly.
Symmetric Limits
Symmetric limits refer to evaluating an integral over an interval that is centered around zero, i.e., \([-a, a]\). This symmetry can be particularly advantageous in reducing complexity during integral calculation, especially when dealing with even and odd functions.
When integrating functions over symmetric limits, certain properties apply:
  • If the function is odd, \( \int_{-a}^{a} f(t) \, dt = 0 \). The symmetry ensures that the negative and positive sides of the interval cancel each other.
  • If the function is even, such symmetry can be used to simplify calculations, as both halves of the interval contribute equally to the integral's value.
  • Understanding these properties can spare a lot of effort when calculating integrals over symmetrical bounds.
Using symmetric limits is a strategic choice in solving integrals, allowing for more straightforward calculations by leveraging the function's properties.

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