Question

$(\sum _{n=1}^{10}{\int }_{-2n-1}^{-2n}{\sin }^{27}xdx)+(\sum _{n=1}^{10}{\int }_{2n}^{2n+1}si{n}^{27}$ $xdx)=$

• A. ${27}^2$
• B. -54
• C. 54
• D. 0

Answer 1

The correct option is D 0

${\sum }_{n=1}^{10}{\int }_{-2n-1}^{-2n}si{n}^{27}xdx+{\sum }_{n=1}^{10}{\int }_{2n+1}^{2n}si{n}^{27}xdx$
$I={\int }_{-21}^{21}si{n}^{27}xdx$
$I={\int }_{-21}^{21}si{n}^{27}xdx-{\int }_{-21}^{21}si{n}^{27}xdx$
$\to I=-I\Rightarrow I=0$