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.
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.
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.
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)?
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.
In my original response, I wrote: "First, you have to integrate the given function"
I correct myself, this step is not required.
Ok this helps a lot, thank you!
To determine the interval on which the curve is concave up, we need to find the second derivative of the given function first.
Let's start with the given function:
y = ∫[0 to x] 2/(1+3t+t^2)dt
To find the second derivative, we start by finding the first derivative.
First, we apply the Fundamental Theorem of Calculus to find the derivative of y with respect to x:
dy/dx = d/dx [∫[0 to x] 2/(1+3t+t^2)dt]
Now, to evaluate this derivative, we'll use the Leibniz rule for differentiating an integral with a variable limit. This rule states that if the limit of integration is a function of x, you can differentiate the integral by differentiating the integrand with respect to x and then multiplying by the derivative of the upper limit of integration:
dy/dx = 2/(1+3x+x^2)
Now, let's find the second derivative, which is d^2y/dx^2:
d^2y/dx^2 = d/dx [dy/dx]
Applying the chain rule for differentiation, we get:
d^2y/dx^2 = d/dx [2/(1+3x+x^2)]
To find this derivative, we can use the quotient rule:
d^2y/dx^2 = [ (0 - 2 * (1+3x+x^2)') / (1+3x+x^2)^2 ]
Simplifying the numerator and denominator, we have:
d^2y/dx^2 = [ -2(1+3x+x^2)' / (1+3x+x^2)^2 ]
Now, we can simplify the numerator, which is the derivative of 1+3x+x^2:
(1+3x+x^2)' = 3 + 2x
Substituting this back into the expression for the second derivative:
d^2y/dx^2 = [ -2(3 + 2x) / (1+3x+x^2)^2 ]
Now, we want to find the interval on which this second derivative is positive, which indicates concavity upward.
For the second derivative to be positive, we need the numerator to be negative and the denominator to be positive.
Since the numerator is -2(3 + 2x), it will always be negative for any value of x.
Now we need to determine when the denominator, (1+3x+x^2)^2, is positive. Since it's a squared term, it will always be positive or zero.
So, the second derivative, d^2y/dx^2, is negative for all values of x.
Therefore, the curve of the function y = ∫[0 to x] 2/(1+3t+t^2)dt is concave down for all values of x; it is not concave up on any interval.
In summary, the curve is not concave up on any interval.