33. \( \int_{1}^{2}\left(\frac{x}{2}-\frac{2}{x}\right) d x \)

To tackle the integral \( \int_{1}^{2}\left(\frac{x}{2}-\frac{2}{x}\right) d x \), we can break it into two simpler integrals: \[ \int_{1}^{2} \frac{x}{2} \, dx - \int_{1}^{2} \frac{2}{x} \, dx. \] First, let's evaluate \( \int_{1}^{2} \frac{x}{2} \, dx \): \[ \int \frac{x}{2} \, dx = \frac{1}{2} \cdot \frac{x^2}{2} = \frac{x^2}{4}, \] therefore: \[ \int_{1}^{2} \frac{x}{2} \, dx = \left[\frac{x^2}{4}\right]_{1}^{2} = \frac{2^2}{4} - \frac{1^2}{4} = \frac{4}{4} - \frac{1}{4} = \frac{3}{4}. \] Next, we evaluate \( \int_{1}^{2} \frac{2}{x} \, dx \): \[ \int \frac{2}{x} \, dx = 2 \ln|x|, \] therefore: \[ \int_{1}^{2} \frac{2}{x} \, dx = \left[2 \ln|x|\right]_{1}^{2} = 2 \ln(2) - 2 \ln(1) = 2 \ln(2) - 0 = 2 \ln(2). \] Putting it all together, we have: \[ \int_{1}^{2}\left(\frac{x}{2}-\frac{2}{x}\right) d x = \frac{3}{4} - 2 \ln(2). \] Thus, the final result is: \[ \int_{1}^{2}\left(\frac{x}{2}-\frac{2}{x}\right) d x = \frac{3}{4} - 2 \ln(2). \]