# Solución

Apartado a

Usamos la primera parte de la definición de la integral de una curva espacial:

$$\int [(3t^2+2t)\bold{i}+(3t−6)\bold{j}+(6t^3+5t^2−4)\bold{k}]dt$$ $$\begin{aligned} &= \bigg[\int (3t^2+2t)dt\bigg]\bold{i} + \bigg[\int (3t−6)dt\bigg]\bold{j}+\bigg[\int (6t^3+5t^2−4)dt\bigg]\bold{k}\\ &= (t^3+t^2)\bold{i}+\left(\frac{3}{2}t^2−6t\right)\bold{j}+\left(\frac{3}{2}t^4+\frac{5}{3}t^3−4t\right)\bold{k}+\bold{C} \end{aligned}$$

Apartado

Primero calcula $\lang t, t^2, t^3\rang\times\lang t^3, t^2, t\rang$:

$$\begin{aligned} \lang t, t^2, t^3\rang\times\lang t^3, t^2, t\rang &= \begin{vmatrix} \bold{i} & \bold{j} & \bold{k}\\ t & t^2 & t^3\\ t^3 & t^2 & t \end{vmatrix}\\ &= (t^2(t)−t^3(t^2))\bold{i}−(t^2−t^3(t^3))\bold{j}+(t(t^2)−t^2(t^3))\bold{k}\\ &= (t^3−t^5)\bold{i}+(t^6−t^2)\bold{j}+(t^3−t^5)\bold{k} \end{aligned}$$

Luego, sustituye esto nuevamente en la integral e integra:

$$\begin{aligned} \lang t, t^2, t^3\rang\times\lang t^3, t^2, t\rang &= \int (t^3−t^5)\bold{i} + (t^6−t^2)\bold{j} + (t^3−t^5)\bold{k}dt\\ &= \bigg(\frac{t^4}{4}−\frac{t^6}{6}\bigg)\bold{i}+\bigg(\frac{t^7}{7}−\frac{t^3}{3}\bigg)\bold{j} + \bigg(\frac{t^4}{4}−\frac{t^6}{6}\bigg)\bold{k}+\bold{C} \end{aligned}$$

Apartado c

Usa la segunda parte de la definición de la integral de una curva espacial:

$$\int_0^{\pi/3}[sen^2t\bold{i}+tant\bold{j}+e^{−2t}\bold{k}]dt$$ $$\begin{aligned} &= \bigg[\int_0^{\pi/3}sen^2tdt\bigg]\bold{i} + \bigg[\int_0^{\pi/3}tantdt\bigg]\bold{j} + \bigg[\int_0^{\pi/3}e^{−2t}dt\bigg]\bold{k}\\ &= \bigg(\frac{t}{2}−\frac{1}{4}sen2t\bigg)\bigg|_0^{\pi/3}\bold{i} − \bigg(ln(cost)\bigg)\bigg|_0^{\pi/3}\bold{j} − \bigg(\frac{1}{2}e^{−2t}\bigg)\bigg|_0^{\pi/3}\bold{k}\\ &= \bigg(\frac{\pi}{6}-\frac{\sqrt{3}}{8}\bigg)\bold{i} − \bigg(ln(cos\frac{\pi}{3})−ln(cos0)\bigg)\bold{j} − \bigg(\frac{1}{2}e^{−2\pi/3}−\frac{1}{2}e^{−2(0)}\bigg)\bold{k}\\ &= \bigg(\frac{\pi}{6}-\frac{\sqrt{3}}{8}\bigg)\bold{i} − \bigg(−ln2\bigg)\bold{j} − \bigg(\frac{1}{2}e^{−2\pi/3}−\frac{1}{2}\bigg)\bold{k}\\ &= \bigg(\frac{\pi}{6}-\frac{\sqrt{3}}{8}\bigg)\bold{i} +(ln2)\bold{j} +\bigg(\frac{1}{2}−\frac{1}{2}e^{−2\pi/3}\bigg)\bold{k} \end{aligned}$$