My people always say multiply and conquer

$\int_{-7}^7f(x)dx$

$=\int_{-6-1}^{-6+1}f(x)dx+\int_{-4-1}^{-4+1}f(x)dx$

$\quad +\int_{-2-1}^{-2+1}f(x)dx+\int_{0-1}^{0+1}f(x)dx+\int_{2-1}^{2+1}f(x)dx$

$\quad+\int_{4-1}^{4+1}f(x)dx+\int_{6-1}^{6+1}f(x)dx$

$=(6^2+4^2+2^2)\times2$

$=3136$

(Ans.)

As value is n^4, so it is symmetric about 0. So we search for sections. Finding suitables 6,4,2 and a middle one for 0, we get and=2*{(6^4)+(4^4)+(2^4)}+0^4=3136