Question

Value of the definite integral ${\int }_{-3}^1\left(2\right(t+1{)}^5-5(t+1{)}^3+t+3)dt$ is equal to -

• A. $8$
• B. $5$
• C. $20$
• D. $\frac{15}{4}$

Answer 1

The correct option is A $8$

$t+1=z$
$dt=dz$
${\int }_{-2}^2(2z^5-5z^3+2+z)dz$
${\int }_{-2}^2(2z^5-5z^3+z)dz+{\int }_{-2}^2\left(2\right)dz=0+8=8$