Find the interval on which the curve of y equals the integral from 0 to x of 2 divided by the quantity 1 plus 3 times t plus t squared, dt is concave up.

Question



Answer 1

y = ∫_{(0 to x)} [2/(1+3t+t²)]dt

Using the Fundamental Theorem of Calculus, y' = 2/(1+3x+x²) = 2(1 +3x + x²)^-1

So, y" = -2(1 + 3x + x²)^-2 (3+2x)

y" = 0 when x = -3/2

If x < -3/2, then y" > 0 So, the graph of y is concave up

If x > -3/2, then y" < 0 So , the graph of y is concave down