# From the second fundamental theorem of calculus to the first

Let's say I have a function $$f(x) = \int_{0}^{1} \frac{\partial f}{\partial x}(\sigma x)\cdot x d\sigma$$

with $$\frac{\partial f}{\partial x}(\sigma x)\cdot x d\sigma$$ that satisfy all the hypothesis for the Fundamental theorem of calculus.

Let's also define $$g(\sigma) = \sigma x$$

Now my question Is it mathematically correct the following implication?

$$f(x) = \int_{0}^{1} \frac{\partial f}{\partial x}(\sigma x)\cdot x d\sigma = \int_{0}^{1} \frac{\partial f}{\partial x}(g(\sigma))\cdot x d\sigma \Rightarrow f(x,g(\sigma)) = \int_{0}^{\sigma} \frac{\partial f}{\partial x}(g(\sigma))\cdot x d\sigma$$ for $$\sigma = [0,1]$$

and thus

$$\frac{\text{d}f}{\text{d}\sigma}(x,g(\sigma)) = \frac{\partial f}{\partial x}(g(\sigma))\cdot x d\sigma$$

again for $$\sigma = [0,1]$$

ALL FROM HERE IS EDITED

Thanks the answering I realized that my implication was wrong.

$$f(x) = \int_{0}^{1} \frac{\partial f}{\partial x}(\sigma x)\cdot x d\sigma = \int_{0}^{1} \frac{\partial f}{\partial x}(g(\sigma))\cdot x d\sigma \Rightarrow f(g(\sigma)) = \int_{0}^{\sigma} \frac{\partial f}{\partial x}(g(\sigma))\cdot x d\sigma$$ for $$\sigma = [0,1]$$

Now makes more sense because if for instance we set sigma = 1 $$f(g(\sigma)) = f(g(1)) = f(x)$$.

But I'm not completely convinced. Is it wright my (edited) implication or still wrong?

What is $f$ ?
-(In the beginning)A function from $\mathbb{R} \to \mathbb{R}$.
-(In the end) A function from $\mathbb{R}^2 \to \mathbb{R}$.
• Do you realise that $g(\sigma) = \sigma x$, but you still write $f(g(\sigma))$ in the LHS of $$f(g(\sigma)) = \int_{0}^{\sigma} \frac{\partial f}{\partial x}(g(\sigma))\cdot x d\sigma$$ – onurcanbektas Jul 21 '18 at 13:20