Question

Evaluate : ${\int }_1^4(\frac{8}{\sqrt{t}}-12\sqrt{t^3})dt$

• A. $-\frac{640}{5}$
• B. $-\frac{664}{5}$
• C. $-\frac{32}{5}$
• D. $-\frac{608}{5}$

Answer 1

The correct option is B $-\frac{664}{5}$

${\int }_1^4(\frac{8}{\sqrt{t}}-12\sqrt{t^3})dt={\int }_1^4(8t^{-\frac{1}{2}}-12t^{\frac{3}{2}})dt$
$={(16t^{\frac{1}{2}}-\frac{24}{5}{t}^{\frac{5}{2}})}_1^4=-\frac{664}{5}$