The notation of the second fundamental theorem of Calculus

Question

I am self studying calculus, and just finished the lesson on the second fundamental theorem of calculus.

the way the theorem is described is:

$\Large\frac{d}{dx}(\int_a^x f(t)\,dt) = f(x)$

and it was told that the meaning is that the derivative of an integral of a function is the function itself.

I don't get how you can get that from this. the expression that I would think suggests this is:

$\Large\frac{d}{dx}(\int f(x)\,dt) = f(x)$

so the derivative of an indefinite integral (as oppose to integrating over a range) of a function is the function itself.

another interpretation of the FToC2 I read here, is that it means that the derivative of the functions that gives the area under the curve of a different function is the different function. this is also something I don't understand how the FToC2 suggests of?

to me, it seems like what this means:

$\Large\frac{d}{dx}(\int_a^x f(t)\,dt) = f(x)$

is how a very small change in $x$ affects that area under $f(t)$ between $a$ (a constant) and $x$. how do I get from that to the right interpretation?

Answer 1

It should be $$\frac{\text{d}}{\text{d}x} \int_a^x f(t)\:\text{d}t=f(x).$$ The variable $t$ has no meaning outside the context of the integral.

Concerning your question about indefinite integrals, they may be a convenient abuse of notation if you're trying to solve an ODE, but they are confusing at best and misleading at worst and aren't/shoudn't be a thing. The integral is designed to produce a number given a function (and maybe a set to integrate over, but that one can be done away with), and conditioning students to mechanically add a $+C$ to integrals does way more harm than good in my experience.

Answer 2

The two are the same statements just expressed in different ways:

Use your first theorem as $\newcommand{\d}{\text{d}}$ $$\int_a^b f(x)\,\text{d}x=F(b)-F(a)$$ Set $b=x$ and take derivatives: $$\frac{\text{d}}{\text{d}x}\int_a^xf(x)\,\d x=f(x)-0=f(x)$$ which is the mathematical expression of the quoted sentence of the theorem.

The point is, the antiderivative $F(x)=\displaystyle\int f(x)\,\d x$ attains a constant value for the second limit and is the same function at the first limit; the derivative of constant is zero, which produces the result.

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You may wonder why it is expressed so indirectly.

The first and second theorem have many different expressions. For example, an other version of the second theorem states

If $f$ is a function and $c$ is any constant, then there exists an antiderivative $A$ such that $A(c)=0$ and is given by $$A(x)=\int_c^xf(x)\,\d x$$

So, don't worry about the wording. Just understand the meaning!

Hope this helps. Ask anything if not clear :)

EDIT

You may wish to visit this as well.

Answer 3

Suppose $ f $ is continuous on $ I \subset\mathbb{R}$, let $ a,x_{0}\in I $.

Let $ \varepsilon >0 $, there exists some $ \eta >0 $, such that $ \left(\forall y\in\left(x_{0}-\eta,x_{0}+\eta\right)\cap I\right),\ \left\vert f\left(y\right)-f\left(x_{0}\right)\right\vert<\varepsilon$

We have for any $ x\in \left(x_{0}-\eta,x_{0}+\eta\right)\cap I $ : \begin{aligned}\left|\int_{a}^{x}{f\left(y\right)\mathrm{d}y}-\int_{a}^{x_{0}}{f\left(y\right)\mathrm{d}y}-\left(x-x_{0}\right)f\left(x_{0}\right)\right|&=\left|\int_{x_{0}}^{x}{\left(f\left(y\right)-f\left(x_{0}\right)\right)\mathrm{d}y}\right|\\&\leq\int_{\min\left(x,x_{0}\right)}^{\max\left(x,x_{0}\right)}{\left|f\left(y\right)-f\left(x_{0}\right)\right|\mathrm{d}y}\\ &\leq\left|x-x_{0}\right|\varepsilon\end{aligned}

Since $ x_{0} $ is arbitrary, for any $ x_{0}\in I $, we have : $$ \int_{a}^{x}{f\left(y\right)\mathrm{d}y}=\int_{a}^{x_{0}}{f\left(y\right)\mathrm{d}y}+\left(x-x_{0}\right)f\left(x_{0}\right)+\underset{\overset{x\to x_{0}}{}}{\scriptsize{\mathcal{O}}}\left(x-x_{0}\right) $$

Which means $ x\mapsto f\left(x\right) $ is the derivative of $ x\mapsto\int_{a}^{x}{f\left(y\right)\mathrm{d}y} $.