Question

The value of the integral ${\int }_0^1(t+2i{)}^{3}dt$ will be

• A. $-\frac{23}{4}-6i$
• B. $-\frac{23}{6}+6i$
• C. $-6+\frac{23}{6}i$
• D. $-6-\frac{23}{6}i$

Answer 1

The correct option is A $-\frac{23}{4}-6i$

${\int }_0^1{(t+2\iota )}^{3}dt$

$={\int }_0^1{(t}^3+{\left(2\iota \right)}^3+3t^2\left(2\iota \right)+3t{\left(2\iota \right)}^2)dt$

$={\int }_0^1{(t}^3-8\iota +6t^2\iota -12t)dt$

$=\frac{t^4}{4}-8t\iota +2t^3\iota -6t^2$

$=\frac{1}{4}-8\iota +2\iota -6=\frac{-23}{4}-6\iota$