Find the interval on which the curve of y= the integral from 0 to x of 2/(1+3t+t^2)dt is concave up.

Question

Can someone please explain? I'm so confused about how to find the answer, all I know is if the 2nd derivative is positive then it's concave up.

Answer 1

Yes, when you find a point of inflection at which the second derivative is positive, then it is a point of minima, and the interval throughout which it is concave up will be between the next two inflection points on either side of the original point.

Which step are you confused about?

First, you have to integrate the given function (use completing the square to form a standard integral), and then you have to identify points of inflection on the curve by equating the first derivative with 0.

Answer 2

I was confused about which derivative the question has given me. Did they give me the integral of the first derivative? Or are they telling me to integrate the original function to find the first derivative? So all I need to do is take the derivative of 2/(1+3t+t^2)?

Answer 3

The given function is an integral.

So, the first derivative will be the function inside the integral

The second derivative will be the derivative of the function inside the integral.

Answer 4

In my original response, I wrote: "First, you have to integrate the given function"

I correct myself, this step is not required.

Answer 5

Ok this helps a lot, thank you!