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Chapter 12: Problem 64

Evaluate the following definite integrals. $$\int_{0}^{\ln 2}\left(e^{-t} \mathbf{i}+2 e^{2 t} \mathbf{j}-4 e^{t} \mathbf{k}\right) d t$$

Short Answer

Expert verified
Answer: $$\frac{1}{2}\mathbf{i}+3\mathbf{j}-4\mathbf{k}$$

Step by step solution

01

Calculate the integral of e^{-t} with respect to t

To calculate the definite integral of the first component with respect to t, we will apply the following formula: $$\int e^{-t} dt = -e^{-t} + C$$ Now, let's evaluate the definite integral: $$\int_{0}^{\ln 2} e^{-t} dt = \left[-e^{-t}\right]_{0}^{\ln 2} = -e^{-\ln 2} + e^0 = -\frac{1}{2} + 1 = \frac{1}{2}$$
02

Calculate the integral of 2e^{2t} with respect to t

To calculate the definite integral of the second component with respect to t, we will apply the following formula: $$\int 2e^{2t} dt = e^{2t} + C$$ Now, let's evaluate the definite integral: $$\int_{0}^{\ln 2} 2e^{2t} dt = \left[e^{2t}\right]_{0}^{\ln 2} = e^{2\ln 2} - e^0 = 2^2 - 1 = 3$$
03

Calculate the integral of -4e^{t} with respect to t

To calculate the definite integral of the third component with respect to t, we will apply the following formula: $$\int -4e^{t} dt = -4e^{t} + C$$ Now, let's evaluate the definite integral: $$\int_{0}^{\ln 2} -4e^{t} dt = \left[-4e^{t}\right]_{0}^{\ln 2} = -4e^{\ln 2} + 4e^0 = -4\cdot 2 + 4 = -4$$
04

Combine the integrated components

Now that we have integrated each component of the vector function, we combine them to form the integrated vector function: $$\int_{0}^{\ln 2}\left(e^{-t} \mathbf{i}+2 e^{2 t} \mathbf{j}-4 e^{t}\mathbf{k}\right) d t = \left(\frac{1}{2}\mathbf{i}+3\mathbf{j}-4\mathbf{k}\right)$$ So, the evaluated definite integral of the given vector function is: $$\frac{1}{2}\mathbf{i}+3\mathbf{j}-4\mathbf{k}$$

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Most popular questions from this chapter

Evaluate the following limits. $$\lim _{t \rightarrow 0}\left(\frac{\sin t}{t} \mathbf{i}-\frac{e^{t}-t-1}{t} \mathbf{j}+\frac{\cos t+t^{2} / 2-1}{t^{2}} \mathbf{k}\right)$$

Show that the two-dimensional trajectory $$x(t)=u_{0} t+x_{0}\( and \)y(t)=-\frac{g t^{2}}{2}+v_{0} t+y_{0},\( for \)0 \leq t \leq T$$ of an object moving in a gravitational field is a segment of a parabola for some value of \(T>0 .\) Find \(T\) such that \(y(T)=0\)

Find the domains of the following vector-valued functions. $$\mathbf{r}(t)=\sqrt{t+2} \mathbf{i}+\sqrt{2-t} \mathbf{j}$$

\(\mathbb{R}^{3}\) Consider the vectors \(\mathbf{I}=\langle 1 / 2,1 / 2,1 / \sqrt{2}), \mathbf{J}=\langle-1 / \sqrt{2}, 1 / \sqrt{2}, 0\rangle,\) and \(\mathbf{K}=\langle 1 / 2,1 / 2,-1 / \sqrt{2}\rangle\) a. Sketch I, J, and K and show that they are unit vectors. b. Show that \(\mathbf{I}, \mathbf{J},\) and \(\mathbf{K}\) are pairwise orthogonal. c. Express the vector \langle 1,0,0\rangle in terms of \(\mathbf{I}, \mathbf{J},\) and \(\mathbf{K}\).

The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|(\text {because}|\cos \theta| \leq 1) .\) This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. Show that for real numbers \(u_{1}, u_{2},\) and \(u_{3},\) it is true that \(\left(u_{1}+u_{2}+u_{3}\right)^{2} \leq 3\left(u_{1}^{2}+u_{2}^{2}+u_{3}^{2}\right)\). (Hint: Use the Cauchy-Schwarz Inequality in three dimensions with \(\mathbf{u}=\left\langle u_{1}, u_{2}, u_{3}\right\rangle\) and choose \(\mathbf{v}\) in the right way.)
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