• Calculons l'intégrale $\int_{1}^{2} 2x \, dx$:

$$\begin{align*} \int_{1}^{2} 2x \, dx &= \left[ x^2 \right]_{1}^{2} = 2^2 - 1^2 \\&= 4 - 1 = 3 \end{align*}$$

Donc,

$$\boxed{\int_{1}^{2} 2x \, dx = 3}$$

• Calculons l'intégrale $\int_{e}^{e^2} \dfrac{1}{t} \, dt$:

$$\begin{align*} \int_{e}^{e^2} \dfrac{1}{t} \, dt &= \left[ \ln |t| \right]_{e}^{e^2} = \ln |e^2| - \ln |e| \\ &= 2 - 1 = 1 \end{align*}$$

Donc,

$$\boxed{\int_{e}^{e^2} \dfrac{1}{t} \, dt = 1}$$

• Calculons l'intégrale $\int_{0}^{\dfrac{\pi}{6}} \cos(2\theta) \, d\theta$:

$$\begin{align*} \int_{0}^{\dfrac{\pi}{6}} \cos(2\theta) \, d\theta &= \left[ \dfrac{\sin(2\theta)}{2} \right]_{0}^{\dfrac{\pi}{6}} \\ &= \dfrac{\sin(\dfrac{\pi}{3})}{2} - \dfrac{\sin(0)}{2} \\&= \dfrac{\dfrac{\sqrt{3}}{2}}{2} - 0 = \dfrac{\sqrt{3}}{4} \end{align*}$$

Donc,

$$\boxed{ \int_{0}^{\dfrac{\pi}{6}} \cos(2\theta) \, d\theta = \dfrac{\sqrt{3}}{4}}$$