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Chapter 10: Problem 51

Evaluate the definite integral. $$ \int_{0}^{\pi / 2}[(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j}+\mathbf{k}] d t $$

Short Answer

Expert verified
The integral of the vector-valued function \( [a \cos(t), a \sin(t), 1] \) over the interval [0, \( \pi / 2 \)] is \( [a, a, \pi / 2] \)

Step by step solution

01

Define the vector-valued function

The vector-valued function given in the problem is \( [a \cos t, a \sin t, 1] \). The goal is to integrate this function over the interval [0, \( \pi / 2 \)].
02

Separate the integral into components

Start by breaking down the integral into its parts. We can treat each component separately, so we want to find: \n 1. \( \int_{0}^{\pi / 2} a \cos(t) dt \) 2. \( \int_{0}^{\pi / 2} a \sin(t) dt \) 3. \( \int_{0}^{\pi / 2} dt \)
03

Compute each integral

Upon integrating, we get: 1. \( \int_{0}^{\pi / 2} a \cos(t) dt = [a \sin(t)]_{0}^{\pi / 2} = a(1-0) = a \) 2. \( \int_{0}^{\pi / 2} a \sin(t) dt = [-a \cos(t)]_{0}^{\pi / 2} = a(0-(-1)) = a \) 3. \( \int_{0}^{\pi / 2} dt = [t]_{0}^{\pi / 2}= (\pi / 2)- 0 = \pi / 2 \)
04

Combine the parts

Now combine these results to get the integral of the entire vector-valued function as a vector itself: \( \int_{0}^{\pi / 2} [a \cos(t), a \sin(t), 1] dt = [a, a, \pi / 2] \)

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