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I just find this so confusing. So if you evaluate the integral you are given its anti derivative.

So like if you have $f(x)$, $f'(x)$, the integral of $f'(x)$ is $f(x)$.

What does the derivative of an integral give you? For example, if I am at $$\frac{d}{dx}\big(\int_{0}^{x}\frac{t^2}{t^2+t+2} dt\big)$$, what does this give you if $\frac{t^2}{t^2+t+2}$ is $f(x)$?

Okay, let's use my exact function above $y$, for example $$y = \int_{0}^{x}\frac{t^2}{t^2+t+2} dt$$ Find the interval where it is concave down.

How would I do this?

I did: $$y' = \frac{x^2}{x^2+x+2}$$

$$y'' = \frac{x(x+4)}{(x^2+x+2)^2}$$

The two points at zero are $x = 0$ and $x = -4$. I completely forgot how to get concavity, but I'm pretty sure $x = -4$ would give us the concave down. So $$\int_{0}^{-4} \frac{t^2}{t^2+t+2} dt$$, would give me the correct answer, right?

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If $I$ is an intervall in $ \mathbb R$ and $f:I \to \mathbb R$ is continuous and if $a \in I$, then the fuction

$F(x)=\int_{a}^{x}f(t) dt$ $\quad $ ($x \in I$) $\quad $

has the following properties:

$F \in C^1(I)$ and $F'=f$ on $I$

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  • $\begingroup$ No, not "fuction". And not "intervall". It is "function" og "interval", respectively. $\endgroup$ – Peter Mortensen Mar 16 '17 at 17:08
  • $\begingroup$ Oh yeah ! A very important comment. $\endgroup$ – Fred Mar 16 '17 at 20:53
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That's exactly the fundamental theorem of calculus, part 1. I'd recommend you to read it on Wikipedia.

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  • $\begingroup$ Okay. Look at my edit, I'm confused on how I would do that. My thought process was finding f(x), then manually going to the second derivative and plotting the values , but I'm just confused atm. $\endgroup$ – user349557 Feb 6 '17 at 7:13

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