Evaluate the following definite integrals: (a) \(\int_{0}^{1} x \cos 2 x \mathrm{~d} x\) (b) \(\int_{0}^{\pi / 2} x \sin 2 x \mathrm{~d} x\) (c) \(\int_{-1}^{1} t \mathrm{e}^{2 t} \mathrm{~d} t\)

We will use integration by parts, let u = x and dv = cos(2x) dx. Then, du = dx and to find v, we can integrate dv: $$v = \int \cos(2x) dx = \frac{1}{2} \sin(2x)$$ Now we can apply the integration by parts formula: $$\int u \: dv = u \cdot v - \int v \: du$$ So, in this case, we have: $$\int_{0}^{1} x \cos(2x) dx = [x\frac{1}{2}\sin(2x)]_{0}^{1} - \int_{0}^{1} \frac{1}{2}\sin(2x) dx$$ Now, we need to compute the integral of the sine function and apply the limits: $$\int_{0}^{1} \frac{1}{2}\sin(2x) dx = [-\frac{1}{4}\cos(2x)]_{0}^{1}$$ Now, we can plug the limits back into our previously found expression and simplify: $$[x\frac{1}{2}\sin(2x) - (-\frac{1}{4}\cos(2x))]_{0}^{1} = (\frac{1}{2}\sin(2) + \frac{1}{4}\cos(2)) - \frac{1}{4}$$ So, the result for (a) is: $$\frac{1}{2}\sin(2) + \frac{1}{4}\cos(2)-\frac{1}{4}$$

We will use integration by parts, let u = x and dv = sin(2x) dx. Then, du = dx and to find v, we can integrate dv: $$v = \int \sin(2x) dx = -\frac{1}{2} \cos(2x)$$ Now we can apply the integration by parts formula: $$\int u \: dv = u \cdot v - \int v \: du$$ So, in this case, we have: $$\int_{0}^{\pi/2} x \sin(2x) dx = [-x\frac{1}{2}\cos(2x)]_{0}^{\pi/2} + \int_{0}^{\pi/2} \frac{1}{2}\cos(2x) dx$$ Now, we need to compute the integral of the cosine function and apply the limits: $$\int_{0}^{\pi/2} \frac{1}{2}\cos(2x) dx = [\frac{1}{4}\sin(2x)]_{0}^{\pi/2}$$ Now, we can plug the limits back into our previously found expression and simplify: $$[-x\frac{1}{2}\cos(2x) + \frac{1}{4}\sin(2x)]_{0}^{\pi/2} = (-\frac{\pi}{4} + \frac{1}{4})$$ So, the result for (b) is: $$-\frac{\pi}{4}+\frac{1}{4}$$

We will use integration by parts, let u = t and dv = e^(2t) dt. Then, du = dt and to find v, we can integrate dv: $$v = \int \mathrm{e}^{2t} dt = \frac{1}{2} \mathrm{e}^{2t}$$ Now we can apply the integration by parts formula: $$\int u \: dv = u \cdot v - \int v \: du$$ So, in this case, we have: $$\int_{-1}^{1} t \mathrm{e}^{2t} dt = [t\frac{1}{2}\mathrm{e}^{2t}]_{-1}^{1} - \int_{-1}^{1} \frac{1}{2}\mathrm{e}^{2t} dt$$ Now, we need to compute the integral of the exponential function and apply the limits: $$\int_{-1}^{1} \frac{1}{2}\mathrm{e}^{2t} dt = [\frac{1}{4}\mathrm{e}^{2t}]_{-1}^{1}$$ Now, we can plug the limits back into our previously found expression and simplify: $$[t\frac{1}{2}\mathrm{e}^{2t} - \frac{1}{4}\mathrm{e}^{2t}]_{-1}^{1} = (\frac{1}{2}\mathrm{e}^{2} - \frac{1}{4}\mathrm{e}^{2}) - ((-\frac{1}{2}\mathrm{e}^{-2} + \frac{1}{4}\mathrm{e}^{-2}))$$ So, the result for (c) is: $$\frac{1}{4}\mathrm{e}^{2}+\frac{1}{4}\mathrm{e}^{-2}$$