Usando o contexto: C´ALCULO 2-A/ TURMA 9-11 – VE1 (GABARITO)(1) Calcule:(a)(1.5 pts) Zπ0sen2(x) cos2(x)dx ;(b)(2 pts) Zdxx3+x2+ 2x;(c)(1 pt) limx→0...

Usando o contexto: C´ALCULO 2-A/ TURMA 9-11 – VE1 (GABARITO)(1) Calcule:(a)(1.5 pts) Zπ0sen2(x) cos2(x)dx ;(b)(2 pts) Zdxx3+x2+ 2x;(c)(1 pt) limx→0Rx2xet2dxsen(2x);(a) TemosZπ0sen2(x) cos2(x)dx =Zπ0(1 −cos(2x))(1 + cos(2x))4dx =Zπ01−cos2(2x)4dx =hx4iπ0−14Zπ01 + cos(4x)2dx =π4−14x2+sen(4x)8π0=π4−π8=π8.(b) Escrevemos1x3+x2+ 2x=Ax+B x +Cx2+x+ 2 .Se encontra A= 1/2 e B=C=−1/2. LogoZdxx3+x2+x=12Zdxx−12Z(x+ 1)dxx2+x+ 2 =ln|x|2−14Z(2x+ 1)dxx2+x+ 2 −14Zdxx2+x+ 2 .Temos Z(2x+ 1)dxx2+x+ 2 = ln|x2+x+ 2|+cZdxx2+x+ 2 =Zdx(x+ 1/2)2+ 7/4=47Zdxx+1/2√7/22+ 1=2√77arctg x+ 1/2√7/2+ccon c∈R. LogoZdxx3+x2+ 2x=ln|x|2−14ln|x2+x+ 2| − √714 arctg x+ 1/2√7/2+c.(c) Como o limite ´e indeterminado, tentamos usar de l’ Hospital. Vamos usar oteorema fundamental do calculo: se F(x) ´e uma antiderivada de f(x) = ex2,limx→0Rx2xet2dxsen(2x)= limx→0ddx Rx2xet2dxddx (sen(2x)) = limx→0ddx (F(x2)−F(x))2cos(x)=limx→0(2xF 0(x2)−F0(x))2cos(x)= limx→02xex4−ex22cos(x)=−12 Responda: