Let g(x)=∫1xt2 dtg(x) = \int_1^x t^2\,dtg(x)=∫1xt2dt. Then g(2)=g(2) =g(2)=A333B13\tfrac{1}{3}31C888D73\tfrac{7}{3}37check_circleExplanationg(2)=[t3/3]12=8/3−1/3=7/3g(2) = [t^3/3]_1^2 = 8/3 - 1/3 = 7/3g(2)=[t3/3]12=8/3−1/3=7/3.