# Need help understanding Fundamental Theorem of Calculus

#### JudoHonor

##### New member
I was trying to understand, how does the proof of the FTC1 ties to FTC2. Can someone please correct me, if my reasoning is wrong?

The first part tells us, that if function $f$ is continous over the interval $[a,b]$, then:

$$\displaystyle f(x) = F'(x)$$, where $$\displaystyle F(x) = \int_a^xf(t)dt$$, for every $$\displaystyle x \in [a,b]$$

Now would it be correct (?) to write this as:

$$\displaystyle F'(x) = f(x)$$ // $$\displaystyle \int$$

antiderivative of both sides, to get rid of the derivative on the left, and we get:

$$\displaystyle F(x) = \int f(x)$$

And then just plug it in:

$$\displaystyle \int_a^bf(t)dt = \int_a^bf(t)dt - \int_a^af(t)dt = \int f(b) - \int f(a)$$

So we are left with the notion that a definite integral of function $$\displaystyle f$$ from $$\displaystyle a$$ to $$\displaystyle b$$ is the antiderivative of $$\displaystyle f$$evaluated at $$\displaystyle b - f$$ evaluated at $$\displaystyle a$$: which is how we calculate integrals

Is my reasoning good? ( for a beginner of course, this is not meant to be a professional proof )

#### Jomo

##### Elite Member
I have no idea what the 5th line from the top means.

Your last equation is meaningless as there is no differential (dx, dt, dq, ...)

It appears that you are saying that $$\displaystyle \int_a^b f(t)dt = \int f(b) dt(?) = f(b)*t + c.\ Why\ do\ you\ have \int_a^af(t)dt??\ You\ do\ know\ that\ integral\ equals\ 0?$$

#### JudoHonor

##### New member
Well okay then, I guess my question is if, as stated in FTC2:

$$\displaystyle \int_a^bf(x)dx = F(b) - F(a)$$ where $$\displaystyle F' = f$$

Then how we figure out what $$\displaystyle F$$ is, if we only know what $$\displaystyle F'$$ is ? - we need to know this, to be able to calculate $$\displaystyle F(b)$$ an $$\displaystyle F(a)$$

And so, I thought that to get $$\displaystyle F$$ out of the equation $$\displaystyle F' = f$$, we could take the antiderivative of $$\displaystyle F'$$ (and consequently we need to do this on the other side of the equation)

this sort of makes sense because if we substitue $$\displaystyle F$$ with antiderivative of $$\displaystyle f$$, we get:

antiderivative of derivative of $$\displaystyle f= f$$, left side cancels out and we have:

$$\displaystyle f= f$$

going back we are left with:

antiderivative of $$\displaystyle F'$$ (teh derivative and antiderivative cancel out) = any antiderivative of $$\displaystyle f$$, since a constant would not make any difference

And this in my head makes sense because going back to:

$$\displaystyle \int_a^bf(x)dx = F(b) - F(a)$$

to calculate $$\displaystyle \int_a^bf(x)dx$$ we subtract $$\displaystyle F(b)$$ which is any antiderivative of $$\displaystyle f$$ evaluated at $$\displaystyle b$$ and subtrack $$\displaystyle f$$ at $$\displaystyle a$$
and this is the algorithm used to calculate definite integrals

I am clearly missing something here 