Find the derivative of the integral

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SUMMARY

The discussion focuses on finding the interval where the curve defined by the integral \( y = \int_0^{x} \frac{1}{1 + t + t^2} dt \) is concave upward. Participants emphasize the necessity of applying the Fundamental Theorem of Calculus, specifically Part 2, to derive the first and second derivatives of the integral. The correct approach involves differentiating the integral directly to obtain \( y' = \frac{1}{1 + x + x^2} \) and subsequently analyzing the second derivative to determine concavity.

PREREQUISITES
  • Understanding of the Fundamental Theorem of Calculus
  • Knowledge of derivatives and concavity analysis
  • Familiarity with integral calculus
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the Fundamental Theorem of Calculus, particularly Part 2
  • Learn how to compute second derivatives for concavity tests
  • Explore techniques for evaluating definite integrals
  • Investigate applications of concavity in real-world scenarios
USEFUL FOR

Students and educators in calculus, mathematicians focusing on integral calculus, and anyone interested in understanding the properties of functions derived from integrals.

merced
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Find the interval on which the curve y = [tex]\int_0^{x}[/tex](1/(1 + t + t2))dt
is concave upward.

I know that I need to find the derivative of the integral.
But I can't figure out how.

I tried to use substitution, letting u = (1 + t + t 2).
Then du = (1 + 2t)dt. But that doesn't work.
 
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merced said:
Find the interval on which the curve y = [tex]\int_0^{x}[/tex](1/(1 + t + t2))dt
is concave upward.

I know that I need to find the derivative of the integral.
But I can't figure out how.

I tried to use substitution, letting u = (1 + t + t 2).
Then du = (1 + 2t)dt. But that doesn't work.

You need to find the derivative, and the second derivative of the integral, you don't actually need to find the integral, maybe the second fundamental theorem of calculus would be useful here..
 
Ok, the Fundamental Theorem of Calculus Part 2 says to take the antiderivative of the function evaluated at b and a...
i.e.: F(b) -F(a)

I don't know how to find the antiderivative any other way than to evaluate the integral...(because the integral is the antiderivative of the function, right?).

Or are you saying to take the derivative of the integral to get the function back?

So, would y' = 1/(1 + t + t2)?
 

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