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Chapter 12: Problem 63

Evaluate the following definite integrals. $$\int_{-\pi}^{\pi}(\sin t \mathbf{i}+\cos t \mathbf{j}+2 t \mathbf{k}) d t$$

Short Answer

Expert verified
Question: Evaluate the definite integral of the vector function $$\vec{r}(t) = \sin(t) \mathbf{i} + \cos(t) \mathbf{j} + 2t \mathbf{k}$$ from \(t = -\pi\) to \(t = \pi\). Answer: After evaluating the definite integral, the resulting vector is $$2\mathbf{i}.$$

Step by step solution

01

Write down the vector function

The given vector function is $$\vec{r}(t) = \sin(t) \mathbf{i} + \cos(t) \mathbf{j} + 2t \mathbf{k}.$$ We need to find the definite integral of this vector function from \(t = -\pi\) to \(t = \pi\).
02

Integrate each component separately

We will integrate each component of the vector function separately: $$\int_{-\pi}^{\pi} \sin(t) dt, \int_{-\pi}^{\pi} \cos(t) dt, \text{ and } \int_{-\pi}^{\pi} 2t dt.$$
03

Evaluate the integrals

Now, evaluate each integral: 1. \(\int_{-\pi}^{\pi} \sin(t) dt\) The integral of \(\sin(t)\) is \(-\cos(t)\). So, we get: $$-\cos(t) \Big|_{-\pi}^{\pi} = -\cos(\pi) - (-\cos(-\pi)) = -(-1) - (-1) = 2.$$ 2. \(\int_{-\pi}^{\pi} \cos(t) dt\) The integral of \(\cos(t)\) is \(\sin(t)\). So, we get: $$\sin(t) \Big|_{-\pi}^{\pi} = \sin(\pi) - \sin(-\pi) = 0 - 0 = 0.$$ 3. \(\int_{-\pi}^{\pi} 2t dt\) The integral of \(2t\) is \(t^2\). So, we get: $$t^2 \Big|_{-\pi}^{\pi} = \pi^2 - (-\pi)^2 = \pi^2 - \pi^2 = 0.$$
04

Write down the result as a vector

Combine the results from step 3 into a vector: $$\int_{-\pi}^{\pi}(\sin t \mathbf{i}+\cos t \mathbf{j}+2 t \mathbf{k}) d t = 2\mathbf{i} + 0 \mathbf{j} + 0 \mathbf{k}$$ Thus, the resulting vector after evaluating the definite integral is $$2\mathbf{i}.$$

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