\(-1\frac{1}{2}\)
\(-\frac{15}{16}\)
\(\frac{15}{16}\)
\(1\frac{1}{2}\)
Correct answer is D
\(\int \frac{4}{x^{3}} \mathrm {d} x = \int 4x^{-3} \mathrm {d} x\)
\(\frac{4x^{-3 + 1}}{-2} = -2x^{-2}|_{1}^{2} = \frac{-2}{x^{2}}|_{1}^{2}\)
= \(\frac{-2}{2^{2}} - \frac{-2}{1^{2}} = -\frac{1}{2} + 2 = 1\frac{1}{2}\)