Mentor

## Why do we use anti-derivatives to find the values of definite integrals?

 Quote by mishin050 ##\displaystyle\frac{d}{dx}=\lim_{\Delta x\to 0}\frac{\Delta}{\Delta x}##
 Quote by pwsnafu The right hand side is not mathematically well-defined, and furthermore that not the definition of d/dx. As Vorde said, you can't treat d/dx as a fraction.
And the right side isn't defined, either, as Δ by itself doesn't mean anything in this context.

 Quote by mishin050 $$\frac{d}{dx} \int_a^x f(t)~dt = f(x);$$ $$dx \cdot\frac{d}{dx} \int_a^x f(t)~dt = f(x)\cdot dx;$$ $$\int d \int_a^x f(t)~dt =\int f(x)~dx;$$ $$\int_a^x f(t)~dt =\int f(x)~dx;$$ $$\int f(x)~dx= F(x)-F(a)!$$
In your own (incorrect anyway) proof, you used the fact that $\int f(x) dx=F(x)$ between the third and fourth lines. You weren't correct getting to that line, but even if you were, you assume the exact fact that you are trying to prove wrong. Here's my proof...

Let the operator $\int dx$ denote the operation that sends a function $f$ to the equivalence class of antiderivatives $F+C$ where $F$ is any antiderivative and $C$ can be any element of the range of the function F.

Therefore,
$$\int f(x)dx=F(x)$$
because it is the definition! There's no math involved. The fundamental theorem makes this notation meaningful, but this is merely notation that was chosen because it makes sense.