# What does the derivative of an integral give you?

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?

• – Jan Feb 6 '17 at 7:06
• Concavity is measured with the second derivative test. When you find the zero of a second derivative you're finding inflection points, and that's where it tells where the function is concave up or down. Read this math.hmc.edu/calculus/tutorials/secondderiv – blade Feb 6 '17 at 7:32
• Well the function of f''(x) is only negative on $x \in$ ($-\infty$, -4], so doesn't that mean it will be concave down at x = -4? – user349557 Feb 6 '17 at 7:37

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 $I$

• No, not "fuction". And not "intervall". It is "function" og "interval", respectively. – Peter Mortensen Mar 16 '17 at 17:08
• Oh yeah ! A very important comment. – Fred Mar 16 '17 at 20:53

That's exactly the fundamental theorem of calculus, part 1. I'd recommend you to read it on Wikipedia.

• 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. – user349557 Feb 6 '17 at 7:13