# Strange integral sum. Where did I go wrong?

Let $$f(x)=x^2$$. $$df(x)=2xdx$$

For $$x=0$$: $$f(0+dx)-f(0)=0$$. Thus $$f(dx)=0$$.

I want to find area under the curve $$f(x)=x^2$$ from $$0$$ to $$b$$: $$\int x^2dx = f(0)dx + f(0+dx)dx + f(0+2dx)dx + ... + f(0+(n-1)dx)dx$$ where $$dx=b/n$$, $$n=\infty$$.

But if $$f(0+dx) = 0$$ as I mentioned above, then I get wrong sum: $$f(0)dx + f(0+dx)dx + ... = 0dx + 0dx + ...$$

What is wrong with my reasoning?

Thanks.

• Could you explain how you get "$f(0 + dx) - f(0) = 0$"? Commented Dec 23, 2022 at 18:24
• In line 2 you seem to prove that if $f$ has zero derivative at a point $x$ then $f$ is identically $0$. The sum of infinitely many infinitely small terms is not something that is well-defined, either. Commented Dec 23, 2022 at 18:24
• @VTand $f(x+dx) = f(x) + f'(x)dx$ for $x=0$ : $f(0+dx) - f(0) = 0$ Commented Dec 23, 2022 at 18:39
• @MichalAdamaszek Ok. I read in one book that If we have point $B(dx,dx^2)$ on the graph of the function $f(x)=x^2$ then point B is upon the X axis - $B(dx,0)$. How is it possible? Thanks. Commented Dec 23, 2022 at 19:13

$$f(x+dx)-f(x)=(x+dx)²-x²=2xdx+dx²$$ For $$x=0$$ it's $$dx²$$, not $$0$$.
Now, when dealing with integration, the result of integration, let $$dx=\frac{b}{\omega}$$ for some arbitrary infinite natural numbers $$\omega$$. Then the integration, for every infenitesimal $$\varepsilon$$ (s.t $$-dx<\varepsilon) , will be equal to standard part (real number infinitely close to that) of:
$$\sum_{i=0}^\omega f(0+\varepsilon+i dx)dx$$
$$dx\sum_{i=0}^\omega \varepsilon²+2\varepsilon dx+i²dx²=dx( \varepsilon²\omega+2\varepsilon dx\sum_{i=0}^\omega i +dx²\sum_{i=0}^\omega i²)=dx(\varepsilon²\omega+2\varepsilon dx\frac{(\omega+1)\omega}{2}+dx²\frac{\omega(\omega+1)(2\omega+1)}{6})=(\text{substituting dx=\frac{b}{\omega}})=b\varepsilon²+2b²\varepsilon\frac{(1+\frac{1}{\omega})}{2}+b³\frac{(1+\frac{1}{\omega})(2+\frac{1}{\omega})}{6}\approx 0+0+b³\frac{(1+0)\cdot (2+0)}{6}=\frac{b³}{3}$$