Cho hàm số $f\left(x\right)$ liên tục trên $\left[ 0;10\right]$ thỏa mãn $\displaystyle\displaystyle\int\limits_{0}^{10}f\left(x\right)\mathrm{d}x=7$, $\displaystyle\displaystyle\int\limits_{2}^{6}f\left(x\right)\mathrm{d}x=3$. Tính $P=\displaystyle\displaystyle\int\limits_{0}^{2}f\left(x\right)\mathrm{d}x+\displaystyle\displaystyle\int\limits_{6}^{10}f\left(x\right)\mathrm{d}x$.

	$P=4$
	$P=-4$
	$P=5$
	$P=7$