(a) Which property of integrals proves

(a) Which property of integrals proves
${\int }_0^{1}v(x)dx\le {\int }_0^1|v(x)|dx$?
(b) Which property proves $-{\int }_0^{1}v(x)dx\le {\int }_0^1|v(x)|dx$?
Together these are Property 8: $|{\int }_0^{1}v(x)dx|\le {\int }_0^1|v(x)|dx$.

Gilbert Strang, Calculus, 3rd edition (Wellesley-Cambridge Press; MIT OpenCourseWare RES.18-001) · §5.6 Properties of the Integral and Average Value, #20