A.P Cal AB: Integral problem?
Let y = the integral (dt) / (t^2 + t + 1) from 1 (which is at the bottom of the integral sign) to 3x (at top). Find (d^2y) / (dx^2)....how do you solve this problem? Please explain step by step, thanks :)
More Questions From This User See All
Answers & Comments
Verified answer
... y = ∫ [1, 3x] { 1 / (t²+t+1) } dt
dy/dx = { 1 / [ (3x)² + (3x) + 1 ] } · d/dx( 3x )
dy/dx = 3 [ 1 / (9x² + 3x + 1) ]
....................................................................
d²y/dx² = 3 [ -1 / (9x² + 3x + 1)² ]. d/dx( 9x² + 3x + 1)
. . . . . .= [ -3 / (9x² + 3x + 1)² ]· [ 9(2x) + 3(1) + 0 ]
. . . . . .= -9 (6x+1) / ( 9x² + 3x + 1 )² ............................... Ans.
__________________________________