x is + over the interval so the function we want to integrate is always + over the interval
when you see something squared on the bottom think:
d/du (v/u) = [ u dv/du - v ] / u^2
so
here let
u = (3x+1)
if v were 1
d/du(v/u) = [0-3]/(3x+1)^2
so let v = -1/3
and
integral = -(1/3)/(3x+1) + c
now of course check that by taking the derivative
now evaluate that at 1
-(1/3)/4 = -1/12 + c
evaluate that at 0
-(1/3)/1 = -1/3 + c
subtract
-1/12 - (-4/12) = 3/12 = 1/4
find the area under the curve y= 1/(3x+1)^2 over the interval [0,1]. please show all steps
1 answer