Evaluate \(\int_{1}^{2} [\frac{x^{3} - 1}{x^{2}}] \mathrm {d} x\).
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option B
Explanation:
\(\frac{x^{3} - 1}{x^{2}} \equiv x - \frac{1}{x^{2}} = x - x^{-2}\)
\(\int_{1}^{2} (x - x^{-2}) \mathrm {d} x = (\frac{x^{2}}{2} + \frac{1}{x})|_{1}^{2}\)
= \((\frac{2^{2}}{2} + \frac{1}{2}) - (\frac{1^{2}}{2} + \frac{1}{1})\)
= \((2 + \frac{1}{2}) - (\frac{1}{2} + 1)\)
= \(2\frac{1}{2} - 1\frac{1}{2}\)
= \(1.0\)
\(\frac{x^{3} - 1}{x^{2}} \equiv x - \frac{1}{x^{2}} = x - x^{-2}\)
\(\int_{1}^{2} (x - x^{-2}) \mathrm {d} x = (\frac{x^{2}}{2} + \frac{1}{x})|_{1}^{2}\)
= \((\frac{2^{2}}{2} + \frac{1}{2}) - (\frac{1^{2}}{2} + \frac{1}{1})\)
= \((2 + \frac{1}{2}) - (\frac{1}{2} + 1)\)
= \(2\frac{1}{2} - 1\frac{1}{2}\)
= \(1.0\)