PROBLEM 1: Find the interval on which the curve y= integral of (1)/(1+t+t^2) dt

Q: PROBLEM 1: Find the interval on which the curve y= integral of (1)/(1+t+t^2) dt

1)I didnt find anything wrong with your logic. the manipulation could be wrong, Since y''(x)= -(1+2x)/(1+x+x^2)^2 hence for the expression to remain negative x should take values between (-1/2,+infinity) 2)no problem

ID: 2834253 • Letter: P

Question

PROBLEM 1:

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

This is the work I've shown so far:

By FTOC, y'(x)= (1)/(1+x+x^2).. where y''(x)= -(1+2x)/(1+x+x^2)^2. Y'' is 0 when x= -1/2.

Therefore, y''(x) is concave down when x < -1/2 ... so then the interval is (-infinity, -1/2)?

PROBLEM 2:

Suppose the curve y=f(x) passes through the origin and the point (1,1). Find the value of the integral of f'(x) dx from 0 to 1.

It's just 1 according to the FTOC. Here's my work: integral of f"(x)dx from 0 to 1 is just f(1)-f(0)=1-0=1.

Explanation / Answer

1)I didnt find anything wrong with your logic.
the manipulation could be wrong,
Since
y''(x)= -(1+2x)/(1+x+x^2)^2

hence for the expression to remain negative
x should take values between (-1/2,+infinity)

2)no problem

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PROBLEM 1: Find the interval on which the curve y= integral of (1)/(1+t+t^2) dt

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