The two functions x 3 {\displaystyle x^{3}} and sin ( 2 x ) {\displaystyle \sin(2x)} are odd because:
( − x ) 3 = − x 3 , sin ( − 2 x ) = − sin ( 2 x ) {\displaystyle (-x)^{3}=-x^{3},\ \sin(-2x)=-\sin(2x)} , so ∫ − 7 7 x 3 + sin ( 2 x ) d x = 0 {\displaystyle \int _{-7}^{7}x^{3}+\sin(2x)dx=0}
Therefore
∫ − 7 7 ( x 3 + sin ( 2 x ) + 1 ) d x = ∫ − 7 7 x 3 + sin ( 2 x ) d x + ∫ − 7 7 d x = 0 + x | − 7 7 = ( 7 − ( − 7 ) ) = 14 {\displaystyle {\begin{aligned}\int _{-7}^{7}(x^{3}+\sin(2x)+1)dx&=\int _{-7}^{7}x^{3}+\sin(2x)dx+\int _{-7}^{7}dx\\&=0+x|_{-7}^{7}\\&=(7-(-7))\\&=14\end{aligned}}}
∫ − 7 7 ( x 3 + sin ( 2 x ) + 1 ) d x = 14 {\displaystyle {\color {blue}\int _{-7}^{7}(x^{3}+\sin(2x)+1)dx=14}}
Note that without using the odd function property, if we use the direct computation of the integral we get the same result.
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