Aparticle travels along a straight line with a velocity of v(t)=3e^(-1/2)sin(2t) meters per second. what is the total distance in meters, traveled by the particle during the time interval 0

Solve for v(t)=0 over the interval [0,2] 

sin(pi)<0, try t=pi/2, 

3e^(-pi/4)sin(pi)= 0, so the interval 0 to pi/2 the particle travels forward [0,pi/2), then it goes negative for the rest of the interval (pi/2,2) and travels backward. 

total distance = |integral 0 to pi/2 v(t) dt| + |integral pi/2 to 2 v(t) dt|
∫[0 to π/2] v(t) dt = ∫[0 to π/2] (3 e^(−t/2) sin(2t)) dt 
. . . . . . . . . . . . . = ∫[0 to π/2] (3 e^(−t/2) sin(2t)) dt 
. . . . . . . . . . . . . = −6/17 e^(−t/2) (sin(2t) + 4cos(2t)) |[0 to π/2] 
. . . . . . . . . . . . . = −6/17 (e^(−π/4) (sin(π) + 4cos(π)) − e^(0) (sin(0) + 4cos(0))) 
. . . . . . . . . . . . . = −6/17 (e^(−π/4) (0 − 4) − (0 + 4)) 
. . . . . . . . . . . . . = −6/17 (−4e^(−π/4) − 4) 
. . . . . . . . . . . . . = 24/17 (e^(−π/4) + 1) 
. . . . . . . . . . . . . = 2.05544 

∫[π/2 to 2] v(t) dt = ∫[π/2 to 2] (3 e^(−t/2) sin(2t)) dt 
. . . . . . . . . . . . . = ∫[π/2 to 2] (3 e^(−t/2) sin(2t)) dt 
. . . . . . . . . . . . . = −6/17 e^(−t/2) (sin(2t) + 4cos(2t)) |[π/2 to 2] 
. . . . . . . . . . . . . = −6/17 (e^(−1) (sin(4) + 4cos(4)) − e^(−π/4) (sin(π) + 4cos(π))) 
. . . . . . . . . . . . . = −6/17 ( (sin(4)+4cos(4))/e − e^(−π/4) (0 − 4)) 
. . . . . . . . . . . . . = −6/17 ( (sin(4)+4cos(4))/e + 4e^(−π/4)) 
. . . . . . . . . . . . . = −0.205938  

when you add absolute values of area you get actual distance traveled: 
∫[0 to π/2] v(t) dt − ∫[π/2 to 2] v(t) dt 
= 2.05544 − (−0.205938) 
= 2.05544 + 0.205938 
= 2.261378 

0=6x+2y+10 is equivalent to the simplified form: y= -3x - 5• parallel lines always have the same slope, different y intercept knowing this, we can say the slope, or m of the line parallel to y= -3x - 5 is -3.m= -3 hope this could !