Evaluate \(\int_{1}^{2} [\frac{x^{3} - 1}{x^{2}}] \mathrm {d} x\)

A.

0.5

B.

1.0

C.

1.5

D.

2.0

Correct answer is B

\(\frac{x^{3} - 1}{x^{2}} \equiv x - \frac{1}{x^{2}} = x - x^{-2}\)

\(\int_{1}^{2} (x - x^{-2}) \mathrm {d} x = (\frac{x^{2}}{2} + \frac{1}{x})|_{1}^{2}\)

= \((\frac{2^{2}}{2} + \frac{1}{2}) - (\frac{1^{2}}{2} + \frac{1}{1})\)

= \((2 + \frac{1}{2}) - (\frac{1}{2} + 1)\)

= \(2\frac{1}{2} - 1\frac{1}{2}\)

= \(1.0\)