Find the integral on which the curve of y = the integral from 0 to x of (6/1+2t+t^2) dt
Can someone provide an explanation for this?
2 answers
Sorry, forgot to include the "is concave up" part of the question
y = ∫[0,x] (6/1+2t+t^2) dt = ∫[0,x] 6/(t+1)^2 dt
y' = 6/(x+1)^2
y" = -12/(x+1)^3
the graph is concave up where y" > 0
That is, where x+1 < 0, or x < -1
y' = 6/(x+1)^2
y" = -12/(x+1)^3
the graph is concave up where y" > 0
That is, where x+1 < 0, or x < -1