Answer
$\displaystyle\int\limits_0^1(1-8v^{3}+16v^{7})dv=1$
Work Step by Step
$\displaystyle\int\limits_0^1(1-8v^{3}+16v^{7})dv$
Integrate each term separately:
$\displaystyle\int\limits_0^1dv-\int\limits_0^18v^{3}dv+\int\limits_0^116v^{7}dv=...$
Take the constants out of the integrals and continue with the process:
$...=\displaystyle\int\limits_0^1dv-8\int\limits_0^1v^{3}dv+16\int\limits_0^1v^{7}dv=...$
$...=v-8(\dfrac{1}{4})v^{4}+16(\dfrac{1}{8})v^{8}\Big|_0^1=v-2v^{4}+2v^{8}\Big|_0^1=...$
Apply the fundamental theorem of calculus to get the answer:
$...=\Big[1-2(1)^{4}+2(1)^{8}\Big]-\Big[0-2(0)^{4}+2(0)^{8}\Big]=...$
$...=1-2+2=1$
