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Let \( \vec{r}(t)=\left\langle\sqrt[3]{t-4}, 4 e^{3 t}, \frac{2}{t-11}\right\rangle \). Evaluate the definite integral. \( \int_{12}^{31} \vec{r}(t) d t=\langle\square \),

Ask by Delgado Ruiz. in the United States
Mar 12,2025

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The definite integral of \( \vec{r}(t) \) from 12 to 31 is: \[ \int_{12}^{31} \vec{r}(t) dt = \left\langle \frac{195}{4}, \frac{4e^{93}-4e^{36}}{3}, 2\ln{(20)} \right\rangle \]

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To evaluate the definite integral \(\int_{12}^{31} \vec{r}(t) d t\), we need to integrate each component of the vector function \(\vec{r}(t)=\left\langle\sqrt[3]{t-4}, 4 e^{3 t}, \frac{2}{t-11}\right\rangle\) separately. 1. **First component:** \[ \int_{12}^{31} \sqrt[3]{t-4} \, dt \] Using the substitution \(u = t - 4\), we get \[ \int u^{1/3} \, du = \frac{3}{4} u^{4/3} + C = \frac{3}{4} (t-4)^{4/3} + C. \] Evaluating from 12 to 31: \[ \left[ \frac{3}{4} (31-4)^{4/3} \right] - \left[ \frac{3}{4} (12-4)^{4/3} \right] = \frac{3}{4} (27)^{4/3} - \frac{3}{4} (8)^{4/3} = \frac{3}{4} (81 - 48) = \frac{3 \cdot 33}{4} = \frac{99}{4}. \] 2. **Second component:** \[ \int_{12}^{31} 4 e^{3t} \, dt = 4 \left[ \frac{1}{3} e^{3t} \right]_{12}^{31} = \frac{4}{3} \left( e^{93} - e^{36} \right). \] 3. **Third component:** \[ \int_{12}^{31} \frac{2}{t-11} \, dt = 2 \left[ \ln|t-11| \right]_{12}^{31} = 2 \left( \ln(20) - \ln(1) \right) = 2 \ln(20). \] As \(\ln(1) = 0\). Finally, combining the results, the definite integral evaluates to: \[ \int_{12}^{31} \vec{r}(t) \, dt = \left\langle \frac{99}{4}, \frac{4}{3} (e^{93} - e^{36}), 2 \ln(20) \right\rangle. \] Thus, we can write: \[ \int_{12}^{31} \vec{r}(t) d t=\left\langle \frac{99}{4}, \frac{4}{3} (e^{93} - e^{36}), 2 \ln(20) \right\rangle. \]

Related Questions

\[ y=x^{3}, \quad 0 \leq x \leq 2 \] Step 1 We are asked to find the surface area of the curve defined by \( y=x^{3} \) rotated about the \( x \)-axis over the interv \( 0 \leq x \leq 2 \). Recall the following formula for the surface area of a function of \( x \) rotated about the \( x \)-axis. Note t as the curve rotates in a circular manner about the \( x \)-axis, the expression \( 2 \pi y \) is the circumference of radius and the radical measures the arc length that is the width of a band, \[ S=\int_{a}^{b} 2 \pi y \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x \] We begin by substituting \( y=x^{3} \) and its derivative in the surface area formula and simplifying, \[ \begin{aligned} S & \left.=\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+\left(\sqrt{3 x^{2}}\right.} \sqrt{3 x^{2}}\right)^{2} d x \\ & =\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+9} \sqrt[9]{ } x^{4} d x \end{aligned} \] Step 2 We have found the following integral for the surface area. \[ S=\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+9 x^{4}} d x \] To evaluate the integral we will first make the substitution \( u=1+9 x^{4} \). We also will need the following to complete the substitution. \[ \begin{array}{l} d u=36 x^{3} \\ x=0 \rightarrow u=1 \\ x=2 \rightarrow u=\square 14 \end{array} \] Step 3 We can now make the substitution \( u=1+(9 x)^{4} \) and evaluate the definite integral with respect to \( u \). \[ \begin{aligned} \int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+(9 x)^{4}} d x & =\int_{1}^{145} 2 \pi \sqrt{u}\left(\frac{d u}{36}\right) \\ & =\frac{2 \pi}{36} \int_{1}^{145} \sqrt{u} d u \end{aligned} \] \[ =\frac{2 \pi}{36}\left[\frac{2}{3} u^{\frac{2}{3}}\right]_{1}^{145} \]
Step 1 We are asked to find the surface area of the curve defined by \( y=x^{3} \) rotated about the \( x \)-axis over the interval \( 0 \leq x \leq 2 \). Recall the following formula for the surface area of a function of \( x \) rotated about the \( x \)-axis. Note that as the curve rotates in a circular manner about the \( x \)-axis, the expression \( 2 \pi y \) is the circumference of radius \( y \) and the radical measures the arc length that is the width of a band. \[ S=\int_{a}^{b} 2 \pi y \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x \] We begin by substituting \( y=x^{3} \) and its derivative in the surface area formula and simplifying. \[ \begin{aligned} S & \left.=\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+\left(\sqrt{3 x^{2}}\right.} \sqrt{3 x^{2}}\right)^{2} d x \\ & =\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+9}, 9^{x^{4}} d x \end{aligned} \] Step 2 We have found the following integral for the surface area. \[ S=\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+9 x^{4}} d x \] To evaluate the integral we will first make the substitution \( u=1+9 x^{4} \). We also will need the following to complete the substitution. \[ \begin{array}{l} d u=36 x^{3} \\ x=0 \rightarrow u=1 \\ x=2 \rightarrow u=\square 146 x^{d x} \end{array} \] Step 3 We can now make the substitution \( u=1+(9 x)^{4} \) and evaluate the definite integral with respect to \( u \). \[ \begin{aligned} \int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+(9 x)^{4}} d x & =\int_{1}^{145} 2 \pi \sqrt{u}\left(\frac{d u}{36}\right) \\ & =\frac{2 \pi}{36} \int_{1}^{145} \sqrt{u} d u \end{aligned} \] \[ =\frac{2 \pi}{36}\left[\frac{2}{3} u^{\frac{2}{3}}\right]_{1}^{145} \]
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