Step 1 The givens and requirements: It is required to obtain the charge q(t) flowing through a device, given that the current is: (a) i(t)=3A, q(0)=1C (b) i(t)=(2t+5) mA, q(0)=0 (c) i(t)=20 cos(10t+pi/6) mu A, q(0)=2 muC (d) i(t)= 10 e^(−30t) sin 40tA, q(0)=0 Step 2 Mathematically, the relationship between current ii, charge q, and time t is i(t) = (dq(t))/(dt) By integrating both sides of the previous equation, we obtain q(t) = q(0) + int_(0)^(t) i(tau) d tau (a) q(t)=q(0)+int_(0)^(t)i(tau)dtau =1+int_(0)^(t)3dtau =1+(3tau)(|)^(t)_(0) =(1+3t)C To check: q(t)=1+3t =1+3*0 =1C

Step 4 (b) q(t)=q(0)+int_(0)^(t)i(tau)dtau =0+int_(0)^(t) (2tau+5)dtau =(2(tau^(2))/(2)+5t)mid^(t)_(0) =(t^(2)+5t)mC To check: q(0)=t^(2)+5t =0^(2)+5*0 =0

Step 5 (c) q(t)=q(0)+int_(0)^(t)i(tau)dtau =2+int_(0)^(t) 20cos(10tau+pi/6)dtau =2+20*(sin(10tau+pi/6))/(10)mid^(t)_(0) =2+2(sin(10t+pi/6)-(1)/(2)) =(1+2sin(10t+pi/6))mu C To check: q(0)=1+2sin(10t+pi/6) =1+2sin(10*0+pi/6) =1+2*(1)/(2) =2mu C Step 6 (d) q(t)=q(0)+int_(0)^(t)i(tau)dtau =0+int_(0)^(t) 10e^(-30tau)sin40tau dtau In Appendix (C.5) in Page (A-19), we can see that int(e^(ax)sin bx)dx=(e^(ax))/(a^(2)+b^(2))*(asin bx-b cos bx) Therefore we can obtain the integration int_(0)^(t) 10e^(-30tau)sin40tau dtau directly without using by-parts integration. Thus use a=-30, b=40, and xarrow tau q(t)=int_(0)^(t) 10e^(-30tau)sin40tau dtau =10*(e^(-30tau))/((-30)^(2)+40^(2))*(-30sin 40tau- 40cos 40tau)mid^(t)_(0) =4*10^(-2)(-3e^(-30tau)sin 40tau - 4e^(-30tau)cos 40tau)mid^(t)_(0) =4*10^(-2)[(-3e^(-30t)sin 40t - 4e^(-30t)cos 40t)-(0-4)] =(160-e^(-30t)(120sin 40t+160cos 40t))mC To check: q(0)=160-e^(-30t)(120sin40t+160cos40t) =160-e^(-30*0)(120sin(40*0)+160cos(40*0)) =160-1*(120*0+160*1) =160-160 =0