Question 1099377
{{{ V[L] = L*(di/dt) }}}

Which, after integration gives:
{{{  i = (1/L) * (i[0] + int( V(t), dt, 0,  T ))  }}}

We have 
 {{{ i[0] = 0 }}} amps
{{{ T = 20 }}} sec
{{{ V(t) = 10-(1/2)t }}} volts, assumed
{{{ L = 2 }}} Henries

So
{{{ i = (1/2) *( 0 + int(10-(1/2)t, dt, 0, 20)) }}} 
{{{ i =  (1/2) *(  10t - (1/4)t^2 ) }}} evaluated at t=20 and t=0
{{{  i = (1/2) * ((200 - (1/4)(20^2)) - (0 - 0)) }}}  for {{{ t>=0 }}}
 <b>{{{ highlight( i = 50) }}}   amps </b> 

—
Below is a graph of i (y-axis) vs t (x-axis)   (ignore the part to the left and below the origin, i[0] = 0).  
 It happens that i(t)  is maximal at t=20sec, after that the current decays back down toward zero, and would go below zero after t=40sec  IF the voltage source continued to obey 10-5t.   
{{{
graph( 500, 500, -4, 60, -10, 60, (1/2)*(10x - (1/4)x^2) ) 
}}}